Chapter 4 Random Variables and Probability Distributions 4.1

4.2

a.

The number of newspapers sold by New York Times each month can take on a countable number of values. Thus, this is a discrete random variable.

b.

The amount of ink used in printing the Sunday edition of the New York Times can take on an infinite number of different values. Thus, this is a continuous random variable.

c.

The actual number of ounces in a one gallon bottle of laundry detergent can take on an infinite number of different values. Thus, this is a continuous random variable.

d.

The number of defective parts in a shipment of nuts and bolts can take on a countable number of values. Thus, this is a discrete random variable.

e.

The number of people collecting unemployment insurance each month can take on a countable number of values. Thus, this is a discrete random variable.

a.

The closing price of a particular stock on the New York Stock Exchange is discrete. It can take on only a countable number of values.

b.

The number of shares of a particular stock that are traded on a particular day is discrete. It can take on only a countable number of values.

c.

The quarterly earnings of a particular firm is discrete. It can take on only a countable number of values.

d.

The percentage change in yearly earnings between 2008 and 2009 for a particular firm is continuous. It can take on any value in an interval.

e.

The number of new products introduced per year by a firm is discrete. It can take on only a countable number of values.

f.

The time until a pharmaceutical company gains approval from the U.S. Food and Drug Administration to market a new drug is continuous. It can take on any value in an interval of time.

4.3

Since there are only a fixed number of outcomes to the experiment, the random variable, x, the number of stars in the rating, is discrete.

4.4

The number of customers, x, waiting in line can take on values 0, 1, 2, 3, … . Even though the list is never ending, we call this list countable. Thus, the random variable is discrete.

4.5

The variable x, total compensation in 2008 (in $ millions), is reported in whole number dollars. Since there are a countable number of possible outcomes, this variable is discrete.

4.6

A banker might be interested in the number of new accounts opened in a month, or the number of mortgages it currently has, both of which are discrete random variables.

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Random Variables and Probability Distributions

153

4.7

An economist might be interested in the percentage of the work force that is unemployed, or the current inflation rate, both of which are continuous random variables.

4.8

The manager of a hotel might be concerned with the number of employees on duty at a specific time, or the number of vacancies there are on a certain night.

4.9

The manager of a clothing store might be concerned with the number of employees on duty at a specific time of day, or the number of articles of a particular type of clothing that are on hand.

4.10

A stockbroker might be interested in the length of time until the stockmarket is closed for the day.

4.11

a.

When a die is tossed, the number of spots observed on the upturned face can be 1, 2, 3, 4, 5, or 6. Since the six sample points are equally likely, each one has a probability of 1/6. The probability distribution of x may be summarized in tabular form:

4.12

x

1

2

3

4

5

6

p(x)

1 6

1 6

1 6

1 6

1 6

1 6

b.

The probability distribution of x may also be presented in graphical form:

a.

The variable x can take on values 1, 3, 5, 7, and 9.

b.

The value of x that has the highest probability associated with it is 5. It has a probability of .4.

c.

Using MINITAB, the probability distribution of x as a graph is:

d.

P(x = 7) = .2

e.

P(x ≥ 5) = p(5) + p(7) + p(9) = .4 + .2 + .1 = .7

f.

P(x > 2) = p(3) + p(5) + p(7) + p(9) = .2 + .4 + .2 + .1 = .9

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154

4.13

Chapter 4

a.

We know

 p( x) = 1. Thus, p(2) + p(3) + p(5) + p(8) + p(10) = 1

 p(5) = 1  p(2)  p(3)  p(8)  p(10) = 1  .15  .10  .25  .25 = .25

4.14

b.

P(x = 2 or x = 10) = P(x = 2) + P(x = 10) = .15 + .25 = .40

c.

P(x ≤ 8) = P(x = 2) + P(x = 3) + P(x = 5) + P(x = 8) = .15 + .10 + .25 + .25 = .75

a.

This is not a valid distribution because

b.

This is a valid distribution because 0 ≤ p(x) ≤ 1 for all values of x and

c.

This is not a valid distribution because p(4) = .3 < 0.

d.

The sum of the probabilities over all possible values of the random variable is

 p( x) = .9 ≠ 1.  p( x) = 1.  p( x) = 1.1 > 1, so

this is not a valid probability distribution. 4.15

a.

The sample points are (where H = head, T = tail): HHH HHT HTH THH HTT THT TTH TTT 3 2 2 2 1 1 1 0 1 1 If each event is equally likely, then P(sample point) =  n 8 x = # heads

b.

p(3) =

4.16

1 1 1 1 3 1 1 1 3 1 , p(2) =    , p(1) =    , and p(0) = 8 8 8 8 8 8 8 8 8 8

c.

Using Minitab, the graph of p(x) is:

d.

P(x = 2 or x = 3) = p(2) + p(3) =

a.

μ = E ( x) =

3 1 4 1    8 8 8 2

 xp( x)

= 10(.05) + 20(.20) + 30(.30) + 40(.25) + 50(.10) + 60(.10) = .5 + 4 + 9 + 10 + 5 + 6 = 34.5

σ2 = E(x − μ)2 =

σ =

 (x  μ )

2

p ( x)

= (10  34.5)2(.05) + (20  34.5)2(.20) + (30  34.5)2(.30) + (40  34.5)2(.25) + (50  34.5)2(.10) + (60  34.5)2(.10) = 30.0125 + 42.05 + 6.075 + 7.5625 + 24.025 + 65.025 = 174.75 174.75 = 13.219

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Random Variables and Probability Distributions

b.

c.

μ ± 2σ  34.5 ± 2(13.219)  34.5 ± 26.438  (8.062, 60.938) P(8.062 < x < 60.938) = p(10) + p(20) + p(30) + p(40) + p(50) + p(60) = .05 + .20 + .30 + .25 + .10 + .10 = 1.00

4.17

a.

μ = E(x) =

 xp( x) = 4(.02) + (3)(.07) + (−2)(.10) + (−1)(.15) + 0(.3) + 1(.18) + 2(.10) + 3(.06) + 4(.02) = −.08 − .21 − .2 − .15 + 0 + .18 + .2 + .18 + .08 = 0

σ2 = E[(x − μ)2] =

σ=

 ( x  μ ) p( x) 2

= (−4 − 0)2(.02) + (−3 − 0)2(.07) + (−2 − 0)2(.10) + (−1 − 0)2(.15) + (0 − 0)2(.30) + (1 − 0)2(.18) + (2 − 0)2(.10) + (3 − 0)2(.06) + (4 − 0)2(.02) = .32 + .63 + .4 + .15 + 0 + .18 + .4 + .54 + .32 = 2.94 2.94 = 1.715

b.

μ ± 2σ  0 ± 2(1.715)  0 ± 3.430  (−3.430, 3.430)

c.

P(−3.430 < x < 3.430) = p(−3) + p(−2) + p(−1) + p(0) + p(1) + p(2) + p(3) = .07 + .10 + .15 + .30 + .18 + .10 + .06 = .96

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155

156

4.18

Chapter 4

a.

It would seem that the mean of both would be 1 since they both are symmetric distributions centered at 1.

b.

P(x) seems more variable since there appears to be greater probability for the two extreme values of 0 and 2 than there is in the distribution of y.

c.

For x: μ = E(x) =

σ2

 xp( x) = 0(.3) + 1(.4) + 2(.3) = 0 + .4 + .6 = 1 = E[(x − μ) ] =  ( x  μ ) p ( x) 2

2

= (0 − 1)2(.3) + (1 − 1)2(.4) + (2 − 1)2(.3) = .3 + 0 + .3 = .6

 yp( y) = 0(.1) + 1(.8) + 2(.1) = 0 + .8 + .2 = 1 = E[(y − μ) ] =  ( y  μ ) p( y )

For y: μ = E(y) =

σ2

2

2

= (0 − 1)2(.1) + (1 − 1)2(.8) + (2 − 1)2(.1) = .1 + 0 + .1 = .2 The variance for x is larger than that for y. 4.19

a.

The probability distribution for x is found by converting the Percent column to a probability column by dividing the percents by 100. The probability distribution of x is: x 2 3 4 5

b.

P(x = 5) = p(5) = .1837.

c.

P(x ≤ 2) = p(2) = .0408.

d.

μ  E ( x) 

p(x) .0408 .1735 .6020 .1837

4

 x p( x ) 2(.0408) 3(.1735)  4(.6020)  5(.1837) i

i

i 1

 .0816  .5205  2.4080  .9185  3.9286  3.93 The average star rating for a car’s drivers-side star rating is 3.9286. 4.20

a.

Yes. Relative frequencies are observed values from a sample. Relative frequencies are commonly used to estimate unknown probabilities. In addition, relative frequencies have the same properties as the probabilities in a probability distribution, namely 1. all relative frequencies are greater than or equal to zero 2. the sum of all the relative frequencies is 1

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Random Variables and Probability Distributions

b.

157

Using MINITAB, the graph of the probability distribution is: 0.15

p(age)

0.10

0.05

0.00 20

25

30

age

c.

Let x = age of employee. Then P(x > 30) = .13 + .15 + .12 = .40. P(x > 40) = 0 P(x < 30) = .02 + .04 + .05 + .07 + .04 + .02 + .07 + .02 + .11 + .07 = .51

4.21

d.

P(x = 25 or x = 26) = .02 + .07 = .09

a.

Yes. For all values of x, 0 ≤ p(x) ≤ 1 and

 p( x) = .01 + .02 + .03 + .05 + .08 + .09 + .11 + .13 +

.12 + .10 + .08 + .06 + .05 + .03 + .02 + .01 + .01 = 1.00.

4.22

4.23

b.

P(x = 16) = .06

c.

P(x ≤ 10) = p(5) + p(6) + p(7) + p(8) + p(9) + p(10) = .01 + .02 + .03 + .05 + .08 + .09 = .28

d.

P(5 ≤ x ≤ 15) = p(5) + p(6) + p(7) + p(8) + p(9) + p(10) + p(11) + p(12) + p(13) + p(14) + p(15) = .01 + .02 + .03 + .05 + .08 + .09 + .11 + .13 + .12 + .10 + .08 = .82

a.

The probability distribution for x is: Grill Display Combination 1-2-3

x 6

p(x) 35 / 124 = .282

1-2-4

7

8 / 124 = .065

1-2-5

8

42 / 124 = .339

2-3-4

9

4 / 124 = .032

2-3-5

10

1 / 124 = .008

2-4-5

11

34 / 124 = .274

b.

P(x > 10) = p(10) + p(11) = .008 + .274 = .282

a.

X is a discrete random variable because it can take on only values 0, 1, 2, 3, 4, or 5 in this example.

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158

Chapter 4

b.

c.

p(0) 

5!(.35)0 (.65)5 0 5  4  3  2 1(1)(.65)5   .655  .1160 0!(5  0)! 1  5  4  3  2 1

p(1) 

5!(.35)1 (.65)51 5  4  3  2 1(.35)1 (.65) 4   5(.35)(.65) 4  .3124 1!(5  1)! 1  4  3  2 1

p(2) 

5!(.35)2 (.65)5 2 5  4  3  2 1(.35)2 (.65)3   10(.35)2 (.65)3  .3364 2!(5  2)! 2 1  3  2  1

p(3) 

5!(.35)3 (.65)53 5  4  3  2 1(.35)3 (.65)2   10(.35)3 (.65)2  .1811 3!(5  3)! 3  2  1  2 1

p(4) 

5!(.35)4 (.65)5 4 5  4  3  2 1(.35)4 (.65)1   5(.35)4 (.65)1  .0488 4!(5  4)! 4  3  2 11

p(5) 

5!(.35)5 (.65)5 5 5  4  3  2 1(.35)5 (.65)0   (.35)5  .0053 5!(5  5)! 5  4  3  2 11

The two properties of discrete random variables are that p(x) ≥ 0 for all x and Σp(x) = 1. From above, all probabilities are greater than 0 and Σp(x) = .1160 + .3124 + .3364 + .1811 + .0488 + .0053 = 1

4.24

d.

P(x ≥ 4) = p(4) + p(5) = .0488 + .0053 = .0541

a.

First, we must find the probability distribution of x. Define the following events: C: {Chicken is contaminated} N: {Chicken is not contaminated} If 3 slaughtered chickens are randomly selected, then the possible outcomes are: CCC, CCN, CNC, NCC, CNN, NCN, NNC, and NNN Each of these outcomes are NOT equally likely since P(C) = 1/100 = .01. P(N) = 1 – P(C) = 1-−.01 = .99. P(CCC) = P(C ∩ C ∩ C ) = P(C) P(C) P(C) = .01(.01)(.01) = .000001 P(CCN) = P(CNC) = P(NCC) = P(C ∩ C ∩ N ) = P(C) P(C) P(N) = .01(.01)(.99) = .000099 P(CNN) = P(NCN) = P(NNC) = P(C ∩ N ∩ N ) = P(C) P(N) P(N) = .01(.99)(.99) = .009801 P(NNN) = P(N ∩ N ∩ N ) = P(N) P(N) P(N) = .99(.99)(.99) = .970299.

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Random Variables and Probability Distributions

159

The variable x is defined as the number of contaminated chickens in the sample. The value of x for each of the outcomes is: Event CCC CCN CNC NCC CNN NCN NNC NNN

x 3 2 2 2 1 1 1 0

p(x) .000001 .000099 .000099 .000099 .009801 .009801 .009801 .970299

The probability distribution of x is: x 3 2 1 0

b.

p(x) .000001 .000297 .029403 .970299

Using MINITAB, the probability graph for x is:

1.0 0.9 0.8 0.7

p(x)

0.6 0.5 0.4 0.3 0.2 0.1 0.0 0

1

2

3

x

4.25

c.

P(x ≤ 1) = P(x = 0) + P(x = 1) = .970299 + .029403 = .999702

a.

p(1) = (.23)(.77)1−1 = (.23)(.77)0 = .23. The probability that one would encounter a contaminated cartridge on the first trial is .23.

b.

c.

p(5) = (.23)(.77)5−1 = (.23)(.77)4 = .0809. The probability that one would encounter a the first contaminated cartridge on the fifth trial is .0809. P(x ≥ 2) = 1 – P(x ≤ 1) = 1 – P(x = 1) = 1 − .23 = .77. The probability that the first contaminated cartridge is found on the second trial or later is .77.

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160

4.26

Chapter 4

To find the probability distribution of x, we first list the possible values of x. For this exercise, the possible values of x are −3, −1, and 5. Next, we list the number of cases, f(x), that result in the particular values of x. To find the probability distribution of x, we divide the number of cases for each value of x, f(x), by the total number of cases, 678. For x = −3, the probability is p(−3) = 68 / 678 = .100. For x = −1, the probability is p(−1) = 71 / 678 = .105. For x = 5, the probability is p(5) = 539 / 678 = .795. The probability distribution of x is:

x −3 −1 5 Total

f(x) 68 71 539 678

p(x) .100 .105 .795 1.000

Using MINITAB, the graph of the probability distribution is:

0.8 0.7 0.6

p(x)

0.5 0.4 0.3 0.2 0.1 0.0 -3

-2

-1

0

1

2

3

4

5

x

4.27

a.

 20  100  20  20! 80! 20! 80! 80  79  78 1  0   3 - 0  0!(20  0)! 3!(80  3)! 0!20! 3!77! 3 2    p(0)  100! 100! 100  99  98 100   3  3!(100  3)! 3!97! 3 2 

b.

82,160  .508 161, 700

 20  100  20  20! 80! 20! 80! 80  79 20   1   3 - 1  1!(20  1)! 2!(80  2)! 1!19! 2!78! 2    p(1)  100! 100! 100  99  98 100   3  3!(100  3)! 3!97! 3 2 

63, 200  .391 161, 700

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Random Variables and Probability Distributions

c.

 20  100  20  20! 80! 20! 80! 20 19  80  2   3 - 2  2!(20  2)! 1!(80  1)! 2!18! 1!79! 2    p(2)  100! 100! 100  99  98 100   3  3!(100  3)! 3!97! 3 2 

d.

a.

15, 200  .094 161, 700

 20  100  20  20! 80! 20! 20 19 18 1 1  3   3 - 0  3!(20  3)! 0!(80  0)! 3!17! 3 2    p(3)  100! 100! 100  99  98 100   3  3!(100  3)! 3!97! 3 2 

4.28

161

E(x) =

1,140  .007 161, 700

 xp( x)

All x

Firm A: E(x) = 0(.01) + 500(.01) + 1000(.01) + 1500(.02) + 2000(.35) + 2500(.30) + 3000(.25) + 3500(.02) + 4000(.01) + 4500(.01) + 5000(.01) = 0 + 5 + 10 + 30 + 700 + 750 + 750 + 70 + 40 + 45 + 50 = 2450 Firm B: E(x) = 0(.00) + 200(.01) + 700(.02) + 1200(.02) + 1700(.15) + 2200(.30) + 2700(.30) + 3200(.15) + 3700(.02) + 4200(.02) + 4700(.01) = 0 + 2 + 14 + 24 + 255 + 660 + 810 + 480 + 74 + 84 + 47 = 2450 b.

σ = σ2

σ2 =

 (x  μ)

2

p( x)

All x

Firm A: σ2 = (0 − 2450)2(.01) + (500 − 2450)2(.01) + ⋅⋅⋅ + (5000 − 2450)2(.01) = 60,025 + 38,025 + 21,025 + 18,050 + 70,875 + 750 + 75,625 + 22,050 + 24,025 + 42,025 + 65,025 = 437,500 σ = 661.44 Firm B: σ2 = (0 − 2450)2(.00) + (200 − 2450)2(.01) + ⋅⋅⋅ + (4700 − 2450)2(.01) = 0 + 50,625 + 61,250 + 31,250 + 84,375 + 18,750 + 84,375 + 31,250 + 61,250 + 50,625 = 492,500 σ = 701.78 Firm B faces greater risk of physical damage because it has a higher variance and standard deviation.

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162

4.29

Chapter 4

a.

The properties of valid probability distributions are:

 p( x) = 1 and 0 ≤ p(x) ≤ 1 for all x. For ARC a1:0 ≤ p(x) ≤ 1 for all x and

 p( x) = .6 + .25 + .1 + .05 = 1.00

Thus, this is a valid probability distribution. For ARC a2:0 ≤ p(x) ≤ 1 for all x and

 p( x) = .6 + .3 + .1 = 1.00

Thus, this is a valid probability distribution. For ARC a3:0 ≤ p(x) ≤ 1 for all x and

 p( x) = .9 + .1 = 1.00

Thus, this is a valid probability distribution. For ARC a4:0 ≤ p(x) ≤ 1 for all x and

 p( x) = .9 + .1 = 1.00

Thus, this is a valid probability distribution. For ARC a5:0 ≤ p(x) ≤ 1 for all x and

 p( x) = .9 + .1 = 1.00

Thus, this is a valid probability distribution. For ARC a6:0 ≤ p(x) ≤ 1 for all x and

 p( x) = .7 + .25 + .05 = 1.00

Thus, this is a valid probability distribution. b.

For Arc a1, P(x > 1) = P(x = 2) + P(x = 3) = .25 + .6 = .85

c.

For Arc a2, P(x > 1) = P(x = 2) = .6 For Arc a3, P(x > 1) = 0 For Arc a4, P(x > 1) = 0 For Arc a5, P(x > 1) = 0 For Arc a6, P(x > 1) = P(x = 2) = .7

d.

For Arc a1, E ( x) 

 xp( x)  3(.60)  2(.25)  1(.10)  0(.05)  1.80  .50  .1  0  2.40

The average capacity of Arc a1 is 2.40. For Arc a2, E ( x) 

 xp( x)  2(.60)  1(.30)  0(.10)  1.20  .30  0  1.50

The average capacity of Arc a2 is 1.50. For Arcs a3, a4, and a5, E ( x) 

 xp( x)  1(.90)  0(.10)  .90  0  .90

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Random Variables and Probability Distributions

The average capacity of Arc a3 is 0.90. The average capacity of Arc a4 is 0.90. The average capacity of Arc a5 is 0.90. For Arc a6, E ( x) 

 xp( x)  2(.70)  1(.25)  0(.10)  1.40  .25  0  1.65

The average capacity of Arc a6 is 1.65. e.

For Arc a1,

σ 2  E ( x  μ )   2

 (x  μ)

2

p( x)

 (3  2.4) 2 (.60)  (2  2.4) 2 (.25)  (1  2.4) 2 (.10)  (0  2.4) 2 (.05)  (.6) 2 (.60)  ( .4) 2 (.25)  (1.4)2 (.10)  ( 2.4)2 (.05)  .216  .04  .196  .288  .74

σ  .74  .86 We would expect most observations to fall within 2 standard deviations of the mean or 2.40 ± 2(.86)  2.40 ± 1.72  (.68, 4.12) For Arc a2,

σ 2  E ( x  μ )   2

 (x  μ)

2

p( x)

 (2  1.5) (.60)  (1  1.5) (.30)  (0  1.5)2 (.10) 2

2

 (.5)2 (.60)  (.5)2 (.30)  (1.5) 2 (.10)  .15  .075  .225  .45

σ  .45  .67 We would expect most observations to fall within 2 standard deviations of the mean or 1.50 ± 2(.67)  1.50 ± 1.34  (.16, 2.84) For Arcs a3, a4, and a5,

σ 2  E ( x  μ )   2

 (x  μ)

2

p( x)

 (1  .9)2 (.90)  (0  .9)2 (.10)  (.1)2 (.90)  (.9)2 (.10)  .009  .081  .090 σ  .09  .30 We would expect most observations to fall within 2 standard deviations of the mean or .90 ± 2(.30)  .90 ± .60  (.30, 1.50)

For Arc a6,

σ 2  E ( x  μ )   2

 (x  μ)

2

p( x)

 (2  1.65)2 (.70)  (1  1.65)2 (.25)  (0  1.65)2 (.05)  (.35)2 (.70)  (.65) 2 (.25)  (1.65) 2 (.05)  .08575  .105625  .136125  .3275 σ  .3275  .57 We would expect most observations to fall within 2 standard deviations of the mean or 1.65 ± 2(.57)  1.65 ± 1.14  (.51, 2.79)

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163

164

4.30

Chapter 4

a.

If a large number of measurements are observed, then the relative frequencies should be very good estimators of the probabilities.

b.

E(x) =

 xp( x) = 1(.01) + 2(.04) + 3(.04) + 4(.08) + 5(.10) + 6(.15) + 7(.25) + 8(.20) + 9(.08) + 10(.05) = .01 + .08 + .12 + .32 + .50 + .90 + 1.75 + 1.60 + .72 + .50 = 6.50

The average number of checkout lanes per store is 6.5. c.

σ2 =

 (x  μ)

2

p( x) = (1 − 6.5)2(.01) + (2 − 6.5)2(.04) + (3 − 6.5)2(.04)

All x

+ (4 − 6.5)2(.08) + (5 − 6.5)2(.10) + (6 − 6.5)2(.15) + (7 − 6.5)2(.25) + (8 − 6.5)2(.20) + (9 − 6.5)2(.08) + (10 − 6.5)2(.05) = .3025 + .8100 + .4900 + .5000 + .2250 + .0375 + .0625 + .4500 + .5000 + .6125 = 3.99

σ= d.

3.99 = 1.9975

Chebyshev's Rule says that at least 0 of the observations should fall in the interval μ ± σ. Chebyshev's Rule says that at least 75% of the observations should fall in the interval μ ± 2σ.

e.

μ ± σ  6.5 ± 1.9975  (4.5025, 8.4975)

P(4.5025 ≤ x ≤ 8.4975) = .10 + .15 + .25 + .20 = .70 This is at least 0.

μ ± 2σ  6.5 ± 2(1.9975)  6.5 ± 3.995  (2.505, 10,495) P(2.505 ≤ x ≤ 10.495) = .04 + .08 + .10 + .15 + .25 + .20 + .08 + .05 = .95 This is at least .75 or 75%. 4.31

a.

Let x = the potential flood damages. Since we are assuming if it rains the business will incur damages and if it does not rain the business will not incur any damages, the probability distribution of x is: x p(x)

b.

0 .7

300,000 .3

The expected loss due to flood damage is E(x) =

 xp(x) = 0(.7) + 300,000(.3) = 0 + 90,000 = $90,000

All x

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Random Variables and Probability Distributions

4.32

165

Let x = winnings in the Florida lottery. The probability distribution for x is: x p(x) −$1 22,999,999/23,000,000 $6,999,999 1/23,000,000 The expected net winnings would be:

μ = E(x) = (−1)(22,999,999/23,000,000) + 6,999,999(1/23,000,000) = −$.70 The average winnings of all those who play the lottery is −$.70. 4.33

a.

Since there are 20 possible outcomes that are all equally likely, the probability of any of the 20 numbers is 1/20. The probability distribution of x is: P(x = 5) = 1/20 = .05; P(x = 10) = 1/20 = .05; etc. x

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

p(x) .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 b.

E(x) =

 xp( x) = 5(.05) + 10(.05) + 15(.05) + 20(.05) + 25(.05) + 30(.05) + 35(.05)

+ 40(.05) + 45(.05) + 50(.05) + 55(.05) + 60(.05) + 65(.05) + 70(.05) + 75(.05) + 80(.05) + 85(.05) + 90(.05) + 95(.05) + 100(.05) = 52.5 c.

σ2 = E(x − μ)2 =

 ( x  μ ) p( x) = (5 − 52.5) (.05) + (10 - 52.5) (.05) 2

2

2

+ (15 − 52.5)2(.05) + (20 − 52.5)2(.05) + (25 − 52.5)2(.05) + (30 − 52.5)2(.05) + (35 − 52.5)2(.05) + (40 − 52.5)2(.05) + (45 − 52.5)2(.05) + (50 − 52.5)2(.05) + (55 − 52.5)2(.05) + (60 − 52.5)2(.05) + (65 − 52.5)2(.05) + (70 − 52.5)2(.05) + (75 − 52.5)2(.05) + (80 − 52.5)2(.05) + (85 − 52.5)2(.05) + (90 − 52.5)2(.05) + (95 − 52.5)2(.05) + (100 − 52.5)2(.05) = 831.25

σ = σ 2  831.25 = 28.83 Since the uniform distribution is not mound-shaped, we will use Chebyshev's theorem to describe the data. We know that at least 8/9 of the observations will fall with 3 standard deviations of the mean and at least 3/4 of the observations will fall within 2 standard deviations of the mean. For this problem,

μ ± 2σ  52.5 ± 2(28.83)  52.5 ± 57.66  (−5.16, 110.16). Thus, at least 3/4 of the data will fall

between −5.16 and 110.16. For our problem, all of the observations will fall within 2 standard deviations of the mean. Thus, x is just as likely to fall within any interval of equal length.

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166

Chapter 4

d.

If a player spins the wheel twice, the total number of outcomes will be 20(20) = 400. The sample space is: 5, 5 10, 5 5,10 10,10 5,15 10,15 . . . . . . 5,100 10,100

15, 5 15,10 15,15 . . . 15,100

20, 5 20,10 20,15 . . . 20,100

25, 5... 100, 5 25,10... 100,10 25,15... 100,15 . . . . . . 25,100... 100,100

Each of these outcomes are equally likely, so each has a probability of 1/400 = .0025. Now, let x equal the sum of the two numbers in each sample. There is one sample with a sum of 10, two samples with a sum of 15, three samples with a sum of 20, etc. If the sum of the two numbers exceeds 100, then x is zero. The probability distribution of x is:

e.

x p(x) 0 .5250 10 .0025 15 .0050 20 .0075 25 .0100 30 .0125 35 .0150 40 .0175 45 .0200 50 .0225 55 .0250 60 .0275 65 .0300 70 .0325 75 .0350 80 .0375 85 .0400 90 .0425 95 .0450 100 .0475 We assumed that the wheel is fair, or that all outcomes are equally likely.

f.

μ = E(x) =

 xp( x) = 0(.5250) + 10(.0025) + 15(.0050) + 20(.0075) + ...+ 100(.0475)

= 33.25

σ2 = E(x − μ)2 =

 (x - μ)

2

p( x) = (0 − 33.25)2(.525) + (10 − 33.25)2(.0025)

+ (15 − 33.25)2(.0050) + (20 − 33.25)2(.0075) + ...+ (100 − 33.25)2(.0475) = 1471.3125

σ = σ 2  1471.3125 = 38.3577

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Random Variables and Probability Distributions

167

g.

P(x = 0) = .525

h.

Given that the player obtains a 20 on the first spin, the possible values for x (sum of the two spins) are 0 (player spins 85, 90, 95, or 100 on the second spin), 25, 30, ..., 100. In order to get an x of 25, the player would spin a 5 on the second spin. Similarly, the player would have to spin a 10 on the second spin order to get an x of 30, etc. Since all of the outcomes are equally likely on the second spin, the distribution of x is: x

p(x)

0 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

.20 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05

i.

The probability that the players total score will exceed one dollar is the probability that x is zero. P(x = 0) = .20

j.

Given that the player obtains a 65 on the first spin, the possible values for x (sum of the two spins) are 0 (player spins 40, 45, 50, up to 100 on second spin), 70, 75, 80,..., 100. In order to get an x of 70, the player would spin a 5 on the second spin. Similarly, the player would have to spin a 10 on the second spin in order to get an x of 75, etc. Since all of the outcomes are equally likely on the second spin, the distribution of x is: x

p(x)

0 70 75 80 85 90 95 100

.65 .05 .05 .05 .05 .05 .05 .05

The probability that the players total score will exceed one dollar is the probability that x is zero. P(x = 0) = .65.

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168

4.34

Chapter 4

a.

Each point in the system can have one of 2 status levels, “free” or “obstacle”. Define the following events: AF: {Point A is free} BF: {Point B is free} CF: {Point C is free}

AO: {Point A is obstacle} BO: {Point B is obstacle} CO: {Point C is obstacle}

Thus, the sample points for the space are: AFBFCF, AFBFCO, AFBOCF, AFBOCO, AOBFCF, AOBFCO, AOBOCF, AOBOCO b.

Since it is stated that the probability of any point in the system having a “free” status is .5, the probability of any point having an “obstacle” status is also .5, Thus, the probability of each of the sample points above is P(AiBiCi) = .5(.5)(.5) = .125. The values of Y, the number of free links in the system, for each sample point are listed below. A link is free if both the points are free. Thus, a link from A to B is free if A is free and B is free. A link from B to C is free if B is free and C is free. Sample point

Y

Probability

AFBFCF

2

.125

AFBFCO

1

.125

AFBOCF

0

.125

AFBOCO

0

.125

AOBFCF

1

.125

AOBFCO

0

.125

AOBOCF

0

.125

AOBOCO

0

.125

The probability distribution for Y is: Y

Probability

0

.625

1

.250

2

.125

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Random Variables and Probability Distributions

4.35

169

Let x = bookie's earnings per dollar wagered. Then x can take on values $1 (you lose) and $-5 (you win). The only way you win is if you pick 3 winners in 3 games. If the probability of picking 1 winner in 1 game is .5, then P(www) = p(w)p(w)p(w) = .5(.5)(.5) = .125 (assuming games are independent). Thus, the probability distribution for x is: x

p(x)

$1 .875 $-5 .125

4.36

E(x) =

 xp( x) = 1(.875) − 5(.125) = .875 − .625 = $.25

a.

6! 6! 6  5  4  3  2 1    15 2!(6  2)! 2!4! (2  1)(4  3  2  1)

b.

 5 5! 5! 5  4  3  2 1  2   2!(5  2)!  2!3!  (2  1)(3  2  1)  10

c.

7 7! 7! 7  6  5  4  3  2 1  0   0!(7  0)!  0!7!  (1)(7  6  5  4  3  2  1)  1 (Note: 0! = 1)

4.37

d.

6 6! 6! 6  5  4  3  2 1  6   6!(6  6)!  6!0!  (6  5  4  3  2  1)(1)  1

e.

4 4! 4! 4  3  2 1  3   3!(4  3)!  3!1!  (3  2  1)(1)  4

a.

x is discrete. It can take on only six values.

b.

This is a binomial distribution.

c.

 5 5! 5 4 3 2 1 (.7)0(.3)5 = (1)(.00243) = .00243 p(0) =   (.7)0(.3)5-0 = 0 0!5! 1 5 43 21   5 5! (.7)1(.3)4 = .02835 p(1) =   (.7)1(.3)5-1 = 1!4!  1  5 5! (.7)2(.3)3 = .1323 p(2) =   (.7)2(.3)5-2 = 2!3!  2 5 5! (.7)3(.3)2 = .3087 p(3) =   (.7)3(.3)5-3 = 3!2!  3  5 5! (.7)4(.3)1 = .36015 p(4) =   (.7)4(.3)5-4 = 4!1!  4 5 5! (.7)5(.3)0 = .16807 p(5) =   (.7)5(.3)5-5 = 5!0! 5

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170

4.38

Chapter 4

d.

μ = np = 5(.7) = 3.5 σ = npq  5(.7)(.3)  1.0247

e.

μ ± 2σ = 3.5 ± 2(1.0247)  (1.4506, 5.5494)

a.

 3 3! 3  2 1 (.3)0(.7)3 = (1)(.7)3 = .343 p(0) =   (.3)0(.7)3-0 = 0!3! 1 3  2 1 0  3 3! (.3)1(.7)2 = .441 p(1) =   (.3)1(.7)3-1 = 1!2!  1  3 5! (.3)2(.7)1 = .189 p(2) =   (.3)2(.7)3-2 = 2!1!  2  3 5! (.3)3(.7)0 = .027 p(3) =   (.3)3(.7)3-3 = 3!0!  3

4.39

x

p(x)

0 1 2 3

.343 .441 .189 .027

a.

P(x = 1) =

5! 5  4  3  2 1 (.2)1(.8) 4 = (.2)1(.8) 4 = 5(.2)1(.8)4 = .4096 1!4! (1)(4  3  2  1)

b.

P(x = 2) =

4! 4  3  2 1 (.6) 2(.4) 2 = (.6) 2(.4) 2 = 6(.6)2(.4)2 = .3456 2!2! (2  1)(2  1)

c.

P(x = 0) =

3! 3  2 1 (.7) 0(.3) 3 = (.7) 0(.3) 3 = 1(.7)0(.3)3 = .027 0!3! (1)(3  2  1)

d.

P(x = 3) =

5! 5  4  3  2 1 (.1) 3(.9) 2 = (.1) 3(.9) 2 = 10(.1)3(.9)2 = .0081 3!2! (3  2  1)(2  1)

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Random Variables and Probability Distributions

4.40

4.41

e.

P(x = 2) =

4! 4  3  2 1 (.4) 2(.6) 2 = (.4) 2(.6) 2 = 6(.4)2(.6)2 = .3456 2!2! (2  1)(2  1)

f.

P(x = 1) =

3! 3  2 1 (.9)1(.1) 2 = (.9)1(.1) 2 = 3(.9)1(.1)2 = .027 1!2! (1)(2  1)

a.

P(x = 2) = P(x ≤ 2) − P(x ≤ 1) = .167 − .046 = .121 (from Table II, Appendix B)

b.

P(x ≤ 5) = .034

c.

P(x > 1) = 1 − P(x ≤ 1) = 1 − .919 = .081

d.

P(x < 10) = P(x ≤ 9) = 0

e.

P(x ≥ 10) = 1 − P(x ≤ 9) = 1 − .002 = .998

f.

P(x = 2) = P(x ≤ 2) − P(x ≤ 1) = .206 − .069 = .137

a.

μ = np = 25(.5) = 12.5 σ2 = np(1 − p) = 25(.5)(.5) = 6.25 σ = σ 2  6.25  2.5

b.

μ = np = 80(.2) = 16 σ2 = np(1 − p) = 80(.2)(.8) = 12.8 σ = σ 2  12.8  3.578

c.

μ = np = 100(.6) = 60 σ2 = np(1 − p) = 100(.6)(.4) = 24 σ = σ 2  24  4.899

d.

μ = np = 70(.9) = 63 σ2 = np(1 − p) = 70(.9)(.1) = 6.3 σ = σ 2  6.3  2.510

e.

f.

μ = np = 60(.8) = 48 σ2 = np(1 − p) = 60(.8)(.2) = 9.6 σ = σ 2  9.6  3.098 μ = np = 1,000(.04) = 40 σ2 = np(1 − p) = 1,000(.04)(.96) = 38.4 σ = σ 2  38.4  6.197

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171

172

4.42

Chapter 4

x is a binomial random variable with n = 4.

a.

If the probability distribution of x is symmetric, p(0) = p(4) and p(1) = p(3). n Since p(x) =   pxqn-x x = 0, 1, ... , n,  x When n = 4,  4 0 4 4 4 0 4! 0 4 4! 4 0 4 4  0  p q   4  p q  0!4! p q  4!0! p q  q  p  p  q Since p + q = 1, p = .5 Therefore, the probability distribution of x is symmetric when p = .5.

b.

If the probability distribution of x is skewed to the right, then the mean is greater than the median. Therefore, there are more small values in the distribution (0, 1) than large values (3, 4). Therefore, p must be smaller than .5. Let p = .2 and the probability distribution of x will be skewed to the right.

c.

If the probability distribution of x is skewed to the left, then the mean is smaller than the median. Therefore, there are more large values in the distribution (3, 4) than small values (0, 1). Therefore, p must be larger than .5. Let p = .8 and the probability distribution of x will be skewed to the left.

d.

In part a, x is a binomial random variable with n = 4 and p = .5. 4 p(x) =   .5x.54-x  x

x = 0, 1, 2, 3, 4

4 4! 4 .5 = 1(.5)4 = .0625 p(0) =   .50.5 4  0!4! 0 4 4! 4 .5 = 4(.5)4 = .25 p(1) =   .51.53  1!3!  1 4 4! 4 .5 = 6(.5)4 = .375 p(2) =   .5 2.5 2  2!2! 2 p(3) = p(1) = .25 (since the distribution is symmetric) p(4) = p(0) = .0625

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Random Variables and Probability Distributions

173

The probability distribution of x in tabular form is: x

0

1

2

3

4

p(x)

.0625

.25

.375

.25

.0625

μ = np = 4(.5) = 2 The graph of the probability distribution of x when n = 4 and p = .5 is as follows.

In part b, x is a binomial random variable with n = 4 and p = .2.  4 p(x) =   .2 x.8 4  x  x

x = 0, 1, 2, 3, 4

 4 p(0) =   .20.84 = 1(1).84 = .4096  0  4 p(1) =   .21.83 = 4(.2)(.8)3 = .4096  1  4 p(2) =   .22.82 = 6(.2)2(.8)2 = .1536  2  4 p(3) =   .23.81 = 4(.2)3(.8) = .0256  3  4 p(4) =   .24.80 = 1(.2)4(1) = .0016  4 The probability distribution of x in tabular form is: x

0

1

2

3

4

p(x)

.4096

.4096

.1536

.0256

.0016

μ = np = 4(.2) = .8

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174

Chapter 4

The graph of the probability distribution of x when n = 4 and p = .2 is as follows:

In part c, x is a binomial random variable with n = 4 and p = .8.  4 p(x) =   .8 x.2 4- x  x

x = 0, 1, 2, 3, 4

 4 p(0) =   .80.24 = 1(1).24 = .0016  0  4 p(1) =   .81.23 = 4(.8)(.2)3 = .0256  1  4 p(2) =   .82.22 = 6(.8)2(.2)2 = .1536  2  4 p(3) =   .83.21 = 4(.8)3(.2) = .4096  3  4 p(4) =   .84.20 = 1(.8)4(1) = .4096  4 The probability distribution of x in tabular form is: x

0

1

2

3

4

p(x)

.0016

.0256

.1536

.4096

.4096

Note: The distribution of x when n = 4 and p = .2 is the reverse of the distribution of x when n = 4 and p = .8.

μ = np = 4(.8) = 3.2 The graph of the probability distribution of x when n = 4 and p = .8 is as follows:

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Random Variables and Probability Distributions

4.43

175

e.

In general, when p = .5, a binomial distribution will be symmetric regardless of the value of n. When p is less than .5, the binomial distribution will be skewed to the right; and when p is greater than .5, it will be skewed to the left. (Refer to parts a, b, and c.)

a.

We will check the 5 characteristics of a binomial random variable. 1. 2.

3. 4. 5.

The experiment consists of 100 identical trials. There are only 2 possible outcomes for each trial. Let S = internet user goes online at home using a wireless network and F = internet user goes online at home without using a wireless network. The probability of success (S) is the same from trial to trial. For each trial, p = P(S) = .20 and q = 1 – p = 1  .20 = .80. The trials are independent. The binomial random variable x is the number of internet users in 100 trials who go online at home using a wireless connection.

Thus, x is a binomial random variable.

4.44

b.

From the exercise, p = .20. For any internet user who goes online at home, the probability of using a wireless connection is .20.

c.

µ = E(x) = np = 100(.20) = 20. On average, for every 100 internet users who go online at home, 20 will use a wireless connection.

a.

We will check the 5 characteristics of a binomial random variable. 1. 2. 3. 4. 5.

The experiment consists of n = 5 identical trials. We have to assume that the number of bottled water brands is large. There are only 2 possible outcomes for each trial. Let S = brand of bottled water used tap water and F = brand of bottled water did not use tap water. The probability of success (S) is the same from trial to trial. For each trial, p = P(S) = .25 and q = 1 – p = 1 - .25 = .75. The trials are independent. The binomial random variable x is the number of brands in the 5 trials that used tap water.

If the total number of brands of bottled water is large, then the above characteristics will be basically true. Thus, x is a binomial random variable. b.

5  The formula for the probability distribution for x is p( x)    .25x (.75)5 x ,  x for x = 1, 2, 3, 4, 5.

c.

5  5! .252.753  .2637 P( x  2)    .252 (.75)5 2  2!3!  2

d.

5  5 P( x  1)  P ( x  0)  P( x  1)    .250 (.75)5 0    .251 (.75)51 0 1  

5! 5! .250.755  .251.754  .2373  .3955  .6328 0!5! 1!4!

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176

4.45

Chapter 4

a.

Let x = number of small businesses owned by non-Hispanic whites that are female owned in 200 trials. Then x is a binomial random variable with n = 200 and p = .27.

μ  E ( x)  np  200(.27)  54 b.

Let x = number of small businesses owned by non-Hispanic whites that are female owned in 8 trials. Then x is a binomial random variable with n = 8 and p = .27. n 8  P( x  0)    p x (1  p)n  x    .270 (.73)8 0  .738  .0806  x 0 n 8  8! P( x  4)    p x (1  p) n  x    .27 4 (.73)8 4  .27 4.734  .1056 4!(8  4)!  x  4

4.46

a.

We will check the 5 characteristics of a binomial random variable. 1. 2.

The experiment consists of n identical trials. There are only 2 possible outcomes for each trial. Let S = general practice physician in the United States does not recommend medicine as a career and F = general practice physician in the United States does recommend medicine as a career. The probability of success (S) is the same from trial to trial. For each trial, p = P(S) = .60 and q = 1 – p = 1 - .60 = .40. The trials are independent. The binomial random variable x is the number of general practice physicians in the United States in n trials who do not recommend medicine as a career.

3. 4. 5.

Thus, x is a binomial random variable. b.

From the information given, p = .60.

c.

µ = E(x) = np = 25(.60) = 15

σ  npq  25(.60)(.40)  6  2.4495

4.47

d.

From Table II, Appendix B, with n = 25 and p = .60, P(x ≥ 1) = 1 – P(x = 0) = 1 – .000 = 1.000.

a.

We will check the 5 characteristics of a binomial random variable. 1. 2. 3. 4. 5.

The experiment consists of n = 20 identical trials. There are only 2 possible outcomes for each trial. Let S = intruding object is detected and F = intruding object is not detected. The probability of success (S) is the same from trial to trial. For each trial, p = P(S) = .8 and q = 1 – p = 1 − .8 = .2. The trials are independent. The binomial random variable x is the number of intruding objects in the 20 trials that are detected.

Thus, x is a binomial random variable. b.

For this experiment, n = 20 and p = .8.

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Random Variables and Probability Distributions

c.

177

Using Table II, Appendix B, with n = 20 and p = .8, P(x = 15) = P(x ≤ 15) − P(x ≤ 14) = .370 − .196 = .174

d.

Using Table II, Appendix B, with n = 20 and p = .8, P(x ≥ 15) = 1 − P(x ≤ 14) = 1 − .196 = .804

4.48

e.

E(x) = np = 20(.8) = 16. For every 20 intruding objects, SBIRS will detect an average of 16.

a.

Let x = number of commissioners out of 4 who vote in favor of an issue. Then x is a binomial random variable with n = 4 and p = .5 (since they are equally likely to vote for or against an issue). The probability that your vote counts is equal to P(x = 2). P( x  2) 

b.

Let x = number of commissioners out of 2 who vote in favor of an issue. Then x is a binomial random variable with n = 2 and p = .5 (since they are equally likely to vote for or against an issue). The probability that your vote counts is equal to P(x = 1). P( x  1) 

4.49

4! 4  3  2 1 2 .52 (.5) 4  2  .5 (.5) 2  .375 2!(4  2)! 2  1  2 1

2! 2 1 1 1 .51 (.5)2 1  .5 (.5)  .5 1!(2  1)! 1 1

Let x = number of major bridges in Denver that will have a rating of 4 or below in 2020 in 10 trials. Then x has an approximate binomial distribution with n = 10 and p = .09. a.

P( x  3)  1  P( x  2)  1  P( x  0)  P ( x  1)  P( x  2) 10  10  10   1    .09 0 (.91)10  0    .091 (.91)10 1    .092 (.91)10 2 0  1  2  1

b.

4.50

10! 10! 1 9 10! .090.9110  .09 .91  .092.918  1  .389  .385  .171  .055 0!10! 1!9! 2!8!

Since the probability of seeing at least 3 bridges out of 10 with ratings of 4 or less is so small, we can conclude that the forecast of 9% of all major Denver bridges will have ratings of 4 or less in 2020 is too small. There would probably be more than 9%.

Let x = number of packets observed by a network sensor in 150 trials. Then x has an approximate binomial distribution with n = 150 and p = .001. The virus will be detected if at least 1 packets is observed. 150  150! P( x  1)  1  P ( x  0)  1   .0010 (.999)150  0  1  .999150  1  .8606  .1394 0!150!  0 

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178

4.51

Chapter 4

Define the following events: A: {Taxpayer is audited} B: {Taxpayer has income less than $1 million) C: {Taxpayer has income of $1 million or higher} a.

From the information given in the problem, P(A│B) = 1/100 = .01 P(A│C) = 9/100 = .09

b.

Let x = number of taxpayers with incomes under $1 million who are audited. Then x is a binomial random variable with n = 5 and p = .01.

5 5! .011 .99 (4)  .0480 P(x = 1) =   .011 .99 (51)  1!4!  1 P(x > 1) = 1 − [P(x = 0) + P(x = 1)]  5  = 1     .010 .99 (5 0)  .0480 0     5!  = 1  .010 .99 5  .0480 0!5!   = 1 − [.9510 + .0480] = 1 − .9990 = .0010

c.

Let x = number of taxpayers with incomes of $1 million or more who are audited. Then x is a binomial random variable with n = 5 and p = .09. 5 5! .091 .914 = .3086 P(x = 1) =   .091 .91(51)  1!4!  1 P(x > 1) = 1 − [P(x = 0) + P(x = 1)]  5  = 1     .09 0 .91(5 0)  .3086 0   5!  .09 0 .915  .3086 = 1   0!5!  = 1 − [.6240 + .3086] = 1 − .9326= .0674

d.

Let x = number of taxpayers with incomes under $1 million who are audited. Then x is a binomial random variable with n = 2 and p = .01. Let y = number of taxpayers with incomes $1 million or more who are audited. Then y is a binomial random variable with n = 2 and p = .09. 2 2! .010 .99 2 = .9801 P(x = 0) =   .010 .99 (2 0)  0!2! 0 2 2! .09 0 .912 = .8281 P(y = 0) =   .09 0 .91(2 0)  0!2! 0 P(x = 0)P(y = 0) = .9801(.8281) = .8116

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Random Variables and Probability Distributions

e.

4.52

179

We must assume that the variables defined as x and y are binomial random variables. We must assume that the trials are identical, the probability of success is the same from trial to trial, and that the trials are independent.

Let x = number of UWG male students out of 50 who have gambled on sports in the past year. Then x will have an approximate binomial distribution with n = 50 and p = .60. µ = E(x) = np = 50(.60) = 30.

σ  npq  50(.60)(.40)  12  3.464 We would expect that most of the observations would fall within 2 standard deviations of the mean. Thus, a likely range for the number of male students who have gambled on sports in the last year would be: 30 ± 2(3.464)  30 ± 6.928  (23.072, 36.928). The range from 24 to 36 would likely include the actual number of male students in a sample of 50 who gambled on sports in the last year. 4.53

a.

μ  E ( x)  np  800(.70)  560 σ  npq  800(.70)(.30)  168  12.96

b.

Half of the 800 food items would be 400. A value of x = 400 would have a z-score of: z

xμ

σ



400  560  12.35 12.96

Since the z-score associated with 400 items is so small (−12.35), it would be virtually impossible to observe less than half with any pesticides if the 70% value was correct. 4.54

Assuming the supplier's claim is true,

μ = np = 500(.001) = .5 σ = npq  500(.001)(.999)  .4995  .707 If the supplier's claim is true, we would only expect to find .5 defective switches in a sample of size 500. Therefore, it is not likely we would find 4. Based on the sample, the guarantee is probably inaccurate. Note: z 

xμ

σ



4  .5  4.95 .707

This is an unusually large z-score.

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180

4.55

Chapter 4

a.

We must assume that the probability that a specific type of ball meets the requirements is always the same from trial to trial and the trials are independent. To use the binomial probability distribution, we need to know the probability that a specific type of golf ball meets the requirements.

b.

For a binomial distribution,

μ = np σ = npq In this example, n = two dozen = 2 ⋅ 12 = 24. p = .10 (Success here means the golf ball does not meet standards.) q = .90 μ = np = 24(.10) = 2.4 σ = npq  24(.10)(.90) = 1.47

c.

In this situation, p = Probability of success = Probability golf ball does meet standards = .90 q = 1 − .90 = .10 n = 24 E(x) = μ = np = 24(.90) = 21.60 σ = npq  24(.10)(.90) = 1.47 (Note that this is the same as in part b.)

4.56

a.

For this test, n = 20 and p = .10. Then x is a binomial random variable with n = 20 and p = .10. Using Table II, Appendix, with n = 20 and p = .10, P(x ≤ 1) = .392

b.

For the experiment in part a, the level of confidence is 1 − P(x ≤ 1) = 1 − .392 = .608. Since this value is not close to 1, this would not be an acceptable level.

c.

Suppose we increased n from 20 to 25. Using Table II, Appendix B, with n = 25 and p = .10, P(x ≤ 1) = .271. This value is smaller than the value found in part a.

Now, suppose we keep n = 20, but change K to 0 instead of 1. Using Table II, Appendix B, with n = 20 and p = .10, P(x ≤ 0) = .122. This value is again, smaller than the value found in part a.

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Random Variables and Probability Distributions

d.

181

Suppose we let K = 0. Now, we need to find n such that the level of confidence ≥ .95, which means that P(x = 0) ≤ .05. n P( x  0)    .10 (.9) n  0  .05 0

n! n .9  .05 0!n!  .9n  .05 

 ln(.9n )  ln(.05)  nln(.9)  ln(.05) ln(.05) 2.99573 n   28.4 .10536 ln(.9) Thus, if K = 0, then we need a sample size of 28 to get a level of confidence of at least .95. Now, suppose K = 1. Now, we need to find n such that the level of confidence is at least .95, which means that P(x ≤ 1) ≤ .05.

n n P( x  1)  P( x  0)  P( x  1)    .10 (.9) n  0    .11 (.9) n 1  .05 0  1  

n! n n! .9  .11.9n 1  .05 0!n! 1!(n  1)!

 .9n  n.11.9n 1  .05  .9n 1 (.9  .1n)  ln(.05) From here, we will use trial and error. For n = 30, .930-1(.9+.1(30)) = .1837 n

.9n-1(.9+.1n)

30

.930-1(.9+.1(30)) = .1837

40

.940-1(.9+.1(40)) = .0805

45

.945-1(.9+.1(45)) = .0524

46

.946-1(.9+.1(46)) = .0480

Thus, for K = 1, we would need a sample size of 46 to get a level of confidence of at least .95.

4.57

a.

The random variable x is discrete since it can assume a countable number of values (0, 1, 2, ...).

b.

This is a Poisson probability distribution with λ = 3.

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182

Chapter 4

c.

In order to graph the probability distribution, we need to know the probabilities for the possible values of x. Using Table III of Appendix B with λ = 3: p(0) = .050 p(1) = P(x ≤ 1) − P(x = 0) = .199 − .050 = .149 p(2) = P(x ≤ 2) − P(x ≤ 1) = .423 − .199 = .224 p(3) = P(x ≤ 3) − P(x ≤ 2) = .647 − .423 = .224 p(4) = P(x ≤ 4) − P(x ≤ 3) = .815 − .647 = .168 p(5) = P(x ≤ 5) − P(x ≤ 4) = .916 − .815 = .101 p(6) = P(x ≤ 6) − P(x ≤ 5) = .966 − .916 = .050 p(7) = P(x ≤ 7) − P(x ≤ 6) = .988 − .966 = .022 p(8) = P(x ≤ 8) − P(x ≤ 7) = .996 − .988 = .008 p(9) = P(x ≤ 9) − P(x ≤ 8) = .999 − .996 = .003 p(10) ≈ .001 The probability distribution of x in graphical form is:

d.

μ=λ=3 σ2 = λ = 3 σ = 3 = 1.73

e.

The mean of x is the same as the mean of the probability distribution, μ = λ = 3.

The standard deviation of x is the same as the standard deviation of the probability distribution, σ = 1.7321. 4.58

μ = λ = 1.5 Using Table III of Appendix B: a.

P(x ≤ 3) = .934

b.

P(x ≥ 3) = 1 − P(x ≤ 2) = 1 − .809 = .191

c.

P(x = 3) = P(x ≤ 3) − P(x ≤ 2) = .934 − .809 = .125

d.

P(x = 0) = .223

e.

P(x > 0) = 1 − P(x = 0) = 1 − .223 = .777

f.

P(x > 6) = 1 − P(x ≤ 6) = 1 − .999 = .001

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Random Variables and Probability Distributions

4.59

4.60

a.

r   N  r   x   n  x   P(x = 1) = N  n 

 3  5  3  3! 2!  1   3  1  3(1)  1!2! 2!0!  = .3 5! 10 5  3  3!2!

b.

 r   N  r   3  9  3 3! 6!  x   n  x   3  5  3  3!0! 2!4!  1(15) = .119   P(x = 3) = 9! N 9 126      n   5  5!4!

c.

 r   N  r  2  4  2 2! 2!  x   n  x   2   2  2  1(1)   2!0! 0!2!  = .167 P(x = 2) = 4! 6 N 4  n   2  2!2!

d.

 r   N  r  2  4  2 2! 2!  x   n  x   0   2  0  0!2! 2!0!  1(1) = .167   P(x = 0) = 4! 6 N 4  n   2  2!2!

For N = 8, n = 3, and r = 5,

a.

 r   N  r   5  8  5  5! 3!  x   n  x  1   3  1  5(3) P(x = 1) =   1!4! 2!1!  = .268 8! 56 N 8  n   3  3!5!

b.

 r   N  r  5  8  5 5! 3!  x   n  x   0   3  0  0!5! 3!0!  1(1) = .018   P(x = 0) = 8! 56 N 8  n   3  3!5!

c.

 r   N  r  5 8  5 5! 3!  x   n  x   3   3  3  10(1)   3!2! 0!3!  = .179 P(x = 3) = 8! 56 N 8  n   3  3!5!

d.

P(x ≥ 4) = P(x = 4) + P(x = 5) = 0 Since the sample size is only 3, there is no way to get 4 or more successes in only 3 trials.

4.61

a.

For λ = 1, P(x ≤ 2) = .920 (from Table III, Appendix B)

b.

For λ = 2, P(x ≤ 2) = .677

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183

184

4.62

Chapter 4

c.

For λ = 3, P(x ≤ 2) = .423

d.

The probability decreases as λ increases. This is reasonable because λ is equal to the mean. As the mean increases, the probability that x is less than a particular value will decrease.

a.

To graph the Poisson probability distribution with λ = 5, we need to calculate p(x) for x = 0 to 15. Using Table III, Appendix B, p(0) = .007 p(1) = P(x ≤ 1) − P(x ≤ 0) = .040 − .007 = .033 p(2) = P(x ≤ 2) − P(x ≤ 1) = .125 − .040 = .085 p(3) = P(x ≤ 3) − P(x ≤ 2) = .265 − .125 = .140 p(4) = P(x ≤ 4) − P(x ≤ 3) = .440 − .265 = .175 p(5) = P(x ≤ 5) − P(x ≤ 4) = .616 − .440 = .176 p(6) = P(x ≤ 6) − P(x ≤ 5) = .762 − .616 = .146 p(7) = P(x ≤ 7) − P(x ≤ 6) = .867 − .762 = .105 p(8) = P(x ≤ 8) − P(x ≤ 7) = .932 − .867 = .065 p(9) = P(x ≤ 9) − P(x ≤ 8) = .968 − .932 = .036 p(10) = P(x ≤ 10) − P(x ≤ 9) = .986 − .968 = .018 p(11) = P(x ≤ 11) − P(x ≤ 10) = .995 − .986 = .009 p(12) = P(x ≤ 12) − P(x ≤ 11) = .998 − .995 = .003 p(13) = P(x ≤ 13) − P(x ≤ 12) = .999 − .998 = .001 p(14) = P(x ≤ 14) − P(x ≤ 13) = 1.000 − .999 = .001 p(15) = P(x ≤ 15) − P(x ≤ 14) = 1.000 − 1.000 = .000 The graph is shown at right:

b.

μ=λ=5 σ = λ = 5 = 2.2361 μ ± 2σ  5 ± 2(2.2361)  5 ± 4.4722  (.5278, 9.4722)

c.

P(.5278 < x < 9.4722) = P(1 ≤ x ≤ 9) = P(x ≤ 9) − P(x = 0) = .968 − .007 = .961

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Random Variables and Probability Distributions

4.63

4.64

185

For this problem, N = 100, n = 10, and x = 4. a.

If the sample is drawn without replacement, the hypergeometric distribution should be used. The hypergeometric distribution requires that sampling be done without replacement.

b.

If the sample is drawn with replacement, the binomial distribution should be used. The binomial distribution requires that sampling be done with replacement.

With N = 10, n = 5, and r = 7, x can take on values 2, 3, 4, or 5.

a.

 r   N  r   7  10  7  7! 3!  x   n  x   2   5  2  21(1) P(x = 2) =   2!5! 3!0!  = .083 10! 252 N  10   n   5  5!5!  r   N  r   7  10  7  7! 3!  x   n  x   3   5  3  3!4! 2!1!  35(3) = .417 P(x = 3) =   10! 252 N  10   n   5  5!5!  r   N  r   7  10  7  7! 3!  x   n  x   4   5  4  35(3) P(x = 4) =   4!3! 1!2!  = .417 10! 252 N  10   n   5  5!5!  r   N  r   7  10  7  7! 3!  x   n  x   5   5  5  5!2! 0!3!  21(1) = .083 P(x = 5) =   10! 252 N  10   n   5  5!5! The probability distribution of x in tabular form is: x 2 3 4 5

b.

μ= σ2 =

p ( x) .083 .417 .417 .083

nr 5(7)  = 3.5 N 10 r ( N  r ) n( N  n) N ( N  1) 2



7(10  7)5(10  5) 10 (10  1) 2



525 = .583 900

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186

Chapter 4

c.

σ=

.5833 = .764

μ ± 2σ  3.5 ± 2(.764)  3.5 ± 1.528  (1.972, 5.028) The graph of the distribution is:

0.4

p(x)

0.3

0.2

0.1

0.0

2

3

1.972

4.65

4.66

4 x 3.5

5 5.028

d.

P(1.972 < x < 5.028) = P(2 ≤ x ≤ 5) = 1.000

a.

For N = 209, r = 10, and n = 8, E ( x) 

b.

 8   209  8  8! 201!   4   10  4  4!(8  4)! 6!(201  6)! P( x  4)    .0002 209! 209    10  10!(209  10)!

a.

E ( x) = μ = λ = 4

nr 10(8)   .383 N 209

σ  λ  42 xμ

z

c.

Using Table III, Appendix B, with λ = 4,

σ



04  2.00 2

b.

P(x ≤ 10) = 1.000 4.67

a.

With λ = 4.5, P(x = 0) =

b.

P(x = 1) =

c.

μ = E(x) = λ = 4.5

4.50 e 4.5  0.0111 0!

4.51 e 4.5  0.0500 1!

σ  λ  4.5  2.12

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Random Variables and Probability Distributions

4.68

187

Let x = number of male nannies in 10 trials. Then x is a binomial random variable with n = 10 and p = 24 / 4,176 = .00575. 10  P( x  1)  1  P( x  0)  1    .005750.9942510  0  1  .9942510  1  .9440  .0560  0

4.69

Let x = number of “clean” cartridges selected in 5 trials. For this problem, N = 158, r = 122, and n = 5.  r   N  r  122   36  122! 36!  x   n  x   5   0    5!117! 0!36!  .2693 P( x  5)  158! N 158   n   5  5!153!

4.70

4.71

a.

Using Table III, Appendix B, with λ = 1.2, P(x = 0) = .301

b.

Using Table III, Appendix B, with λ = 1.2, P(x ≥ 2) = 1 – P(x ≤ 1) = 1 - .663 = .337

a.

Using Table III, Appendix B, with λ = 5, P(x < 3) = P(x ≤ 2) = .125

4.72

b.

E(x) = λ = 5. The average number of calls blocked during the peak hour of video conferencing call time is 5.

a.

Let x = number of defective items in a sample of size 4. For this problem, x is a hypergeometric random variable with N = 10, n = 4, and r = 1. You will accept the lot if you observe no defectives.  r   N  r  1  10  1 1! 9!  x   n  x   0   4  0  0!1! 4!5!  1(126) = .6 P(x = 0) =   10! 210 N  10   n   4  4!6!

b.

If r = 2,  r   N  r   2  10  2  2! 8!  x   n  x   0   4  0  0!2! 4!4!  1(70) = .333   P(x = 0) = 10! 210 N  10   n   4  4!6!

4.73

Let x = number of spoiled bottles in the sample of 3. Since the sampling will be done without replacement, x is a hypergeometric random variable with N = 12, n = 3, and r = 1.  r   N  r  1 12  1 1! 11!  x   n  x  1  3  1  1!0! 2!9!  55  .25 P(x = 1) =   12! 220 N  12   n   3  3!9!

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188

4.74

Chapter 4

a.

Using Table III and λ = 6.2, P(x = 2) = P(x ≤ 2) − P(x ≤ 1) = .054 − .015 = .039 P(x = 6) = P(x ≤ 6) − P(x ≤ 5) = .574 − .414 = .160 P(x = 10) = P(x ≤ 10) − P(x ≤ 9) = .949 − .902 = .047

b.

The plot of the distribution is:

c.

μ = λ = 6.2, σ = λ = 6.2 = 2.490 μ ± σ  6.2 ± 2.49  (3.71, 8.69) μ ± 2σ  6.2 ± 2(2.49)  6.2 ± 4.98  (1.22, 11.18) μ ± 3σ  6.2 ± 3(2.49)  6.2 ± 7.47  (−1.27, 13.67) See the plot in part b.

d.

First, we need to find the mean number of customers per hour. If the mean number of customers per 10 minutes is 6.2, then the mean number of customers per hour is 6.2(6) = 37.2 = λ.

μ = λ = 37.2 and σ = λ = 37.2 = 6.099 μ ± 3σ  37.2 ± 3(6.099)  37.2 ± 18.297  (18,903, 55.498) Using Chebyshev's Rule, we know at least 8/9 or 88.9% of the observations will fall within 3 standard deviations of the mean. The number 75 is way beyond the 3 standard deviation limit. Thus, it would be very unlikely that more than 75 customers entered the store per hour on Saturdays. 4.75

a.

Using Table III, Appendix B, with λ = 10, P(x = 24) = P(x ≤ 24) – P(x ≤ 23) = 1.000 – 1.000 = .000

b.

Using Table III, Appendix B, with λ = 10, P(x = 23) = P(x ≤ 23) – P(x ≤ 22) = 1.000 – 1.000 = .000

c.

Yes, these probabilities are good approximations for the probability of “fire” and “theft”. The researchers estimated these probabilities to be .0001, indicating that these would be extremely rare events. Our probabilities of .000 are very close to .0001.

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Random Variables and Probability Distributions

4.76

189

μ = λ = 3, using Table III, Appendix B: P(x = 0) = .050

The probability that no bulbs fail in one hour is .050. If we let y = number of one hour intervals out of 8 that have no bulbs fail, then y is a binomial random variable with n = 8 and p = .05. Then, the probability that no bulbs fail in an 8 hour shift is 8  8! .058.950 P(y = 8) =   .058.95(88)  8!(8  8)! 8  =

8  7  6  5  4  3  2 1 .058.950  .058 8  7  6  5  4  3  2 11

We must assume that the 8 one-hour intervals are independent and identical, and that the probability that no bulbs fail is the same for each one-hour interval. 4.77

Let x = number of females promoted in the 72 employees awarded promotion, where x is a hypergeometric random variable. From the problem, N = 302, r = 73, and n = 72. We need to find if observing 5 females who were promoted was fair. E(x) = μ =

nr 72(73)  = 17.40 N 302

If 72 employees are promoted, we would expect that about 17 would be females. V(x) = σ2 =

r ( N  r ) n( N  n) N ( N  1) 2



73(302  73)72(302  72) 3022 (302  1)

= 10.084

σ = 10.084 = 3.176 Using Chebyshev’s Theorem, we know that at least 8/9 of all observations will fall within 3 standard deviations of the mean. The interval from 3 standard deviations below the mean to 3 standard deviations above the mean is:

μ ± 3σ  17.40 ± 3(3.176)  17.40 ± 9.528  (7.872, 26.928) If there is no discrimination in promoting females, then we would expect between 8 and 26 females to be promoted within the group of 72 employees promoted. Since we observed only 5 females promoted, we would infer that females were not promoted fairly. 4.78

If it takes exactly 5 minutes to wash a car and there are 5 cars in line, it will take 5(5) = 25 minutes to wash these 5 cars. Thus, for anyone to be in line at closing time, more than 1 car must arrive in the final ½ hour. In addition, if on average 10 cars arrive per hour, then an average of 5 cars will arrive per ½ hour (30 minutes). If we let x = number of cars to arrive in ½ hour, then x is a Poisson random variable with λ = 5. P(x > 1) = 1 – P(x ≤ 1) = 1 − .04 = .96 (Using Table III, Appendix B) Since this probability is so big, it is very likely that someone will be in line at closing time.

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190

4.79

4.80

Chapter 4

Using Table IV, Appendix B: a.

P(z > 1.46) = .5 − P(0 < z ≤ 1.46) = .5 − .4279 = .0721

b.

P(x 1) = 1 − P(z ≤ 1) = 1 − .8413 = .1587 (Refer to part b.)

Using Table IV, Appendix B: a.

P(z ≥ z0) = .05 A1 = .5 − .05 = .4500 Looking up the area .4500 in Table IV gives z0 = 1.645.

b.

P(z ≥ z0) = .025 A1 = .5 − .025 = .4750 Looking up the area .4750 in Table IV gives z0 = 1.96.

c.

P(z ≤ z0) = .025 A1 = .5 − .025 = .4750 Looking up the area .4750 in Table IV gives z = 1.96. Since z0 is to the left of 0, z0 = −1.96.

d.

P(z ≥ z0) = .10 A1 = .5 − .1 = .4 Looking up the area .4000 in Table IV gives z0 = 1.28.

e.

P(z > z0) = .10 A1 = .5 − .1 = .4 z0 = 1.28 (same as in d)

Using Table IV of Appendix B: a.

P(z ≤ z0) = .2090 A = .5000 − .2090 = .2910 Look up the area .2910 in the body of Table IV; z0 = −.81. (z0 is negative since the graph shows z0 is on the left side of 0.)

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.

Random Variables and Probability Distributions

b.

P(z ≤ z0) = .7090 P(z ≤ z0) = P(z ≤ 0) + P(0 ≤ z ≤ z0) = .5 + P(0 ≤ z ≤ z) = .7090 Therefore, P(0 ≤ z ≤ z0) = .7090 − .5 = .2090 Look up the area .2090 in the body of Table IV; z0 ≈ .55.

c.

P(−z0 ≤ z < z0) = .8472 P(−z0 ≤ z < z0) = 2P(0 ≤ z ≤ z0) 2P(0 ≤ z ≤ z0) = .8472 Therefore, P(0 ≤ z ≤ z0) = .4236. Look up the area .4236 in the body of Table IV; z0 = 1.43.

d.

P(−z0 ≤ z < z0) = .1664 P(−z0 ≤ z ≤ z0) = 2P(0 ≤ z ≤ z0) 2P(0 ≤ z ≤ z0) = .1664 Therefore, P(0 ≤ z ≤ z0) = .0832. Look up the area .0832 in the body of Table IV; z0 = .21.

e.

P(z0 ≤ z ≤ 0) = .4798 P(z0 ≤ z ≤ 0) = P(0 ≤ z ≤ −z0) Look up the area .4798 in the body of Table IV; z0 = −2.05.

f.

P(−1 < z < z0) = .5328 P(−1 < z < z0) = P(−1 < z < 0) + P(0 < z < z0) = .5328 P(0 < z < 1) + P(0 < z < z0) = .5328 Thus, P(0 < z < z0) = .5328 − .3413 = .1915 Look up the area .1915 in the body of Table IV; z0 = .50.

4.86

a.

z=1

b.

z = −1

c.

z=0

d.

z = −2.5

e.

z=3

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193

194

4.87

4.88

Chapter 4

Using Table IV, Appendix B: a.

z=

b.

z=

c.

z=

d.

z=

e.

z=

f.

z=

xμ

σ xμ

σ xμ

σ xμ

σ xμ

σ xμ

σ



20  30 = −2.50 4



30  30 =0 4



27.5  30 = −0.61 4



15  30 = −3.75 4



35  30 = 1.25 4



25  30 = −1.25 4

Using Table IV of Appendix B: a.

To find the probability that x assumes a value more than 2 standard deviations from μ: P(x < μ − 2σ) + P(x > μ + 2σ) = P(z < −2) + P(z > 2) = 2P(z > 2) = 2(.5000 − .4772) = 2(.0228) = .0456 To find the probability that x assumes a value more than 3 standard deviations from μ: P(x < μ − 3σ) + P(x > μ + 3σ) = P(z < −3) + P(z > 3) = 2P(z > 3) = 2(.5000 − .4987) = 2(.0013) = .0026

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Random Variables and Probability Distributions b.

To find the probability that x assumes a value within 1 standard deviation of its mean: P(μ − σ < x < μ + σ) = P(−1 < z < 1) = 2P(0 < z < 1) = 2(.3413) = .6826 To find the probability that x assumes a value within 2 standard deviations of μ: P(μ − 2σ < x < μ + 2σ) = P(−2 < z < 2) = 2P(0 < z < 2) = 2(.4772) = .9544

c.

To find the value of x that represents the 80th percentile, we must first find the value of z that corresponds to the 80th percentile. P(z < z0) = .80. Thus, A1 + A2 = .80. Since A1 = .50, A2 = .80 - .50 = .30. Using the body of Table IV, z0 = .84. To find x, we substitute the values into the z-score formula: z=

xμ

σ

.84 =

x  1000  x = .84(10) + 1000 = 1008.4 10

To find the value of x that represents the 10th percentile, we must first find the value of z that corresponds to the 10th percentile.

P(z < z0) = .10. Thus, A1 = .50 - .10 = .40. Using the body of Table IV, z0 = −1.28. To find x, we substitute the values into the z-score formula: z=

xμ

σ

−1.28 =

x  1000  x = −1.28(10) + 1000 = 987.2 10

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195

196

4.89

Chapter 4

a.

b.

c.

d.

e.

f.

12  11   10  11  z  P(10 ≤ x ≤ 12) = P    2 2  = P(−0.50 ≤ z ≤ 0.50) = A1 + A2 = .1915 + .1915 = .3830 10  11   6  11 P(6 ≤ x ≤ 10) = P  z   2 2  = P(−2.50 ≤ z ≤ −0.50) = P(−2.50 ≤ z ≤ 0) − P(−0.50 ≤ z ≤ 0) = .4938 − .1915 = .3023 16  11   13  11 P(13 ≤ x ≤ 16) = P   z    2 2  = P(1.00 ≤ z ≤ 2.50) = P(0 ≤ z ≤ 2.50) − P(0 ≤ x ≤ 1.00) = .4938 − .3413 = .1525 P(7.8 ≤ x ≤ 12.6) 12.6  11   7.8  11 = P  z    2 2 = P(−1.60 ≤ z ≤ 0.80) = A1 + A2 = .4452 + .2881 = .7333 13.24  11   P(x ≥ 13.24) = P  z    2 = P(z ≥ 1.12) = A2 = .5 − A1 = .5000 − .3686 = .1314 7.62  11   P(x ≥ 7.62) = P  z    2 = P( z ≥ −1.69) = A1 + A2 = .4545 + .5000 = .9545

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Random Variables and Probability Distributions

4.90

The random variable x has a normal distribution with μ = 50 and σ = 3. a.

P(x ≤ x0) = .8413 So, A1 + A2 = .8413 Since A1 = .5, A2 = .8413 − .5 = .3413. Look up the area .3413 in the body of Table IV, Appendix B; z0 = 1.0. To find x0, substitute all the values into the z-score formula: z=

xμ

σ

x  50 1.0 = 0 3 x0 = 50 + 3(1.0) = 53 b.

P(x > x0) = .025 So, A = .5000 − .025 = .4750 Look up the area .4750 in the body of Table IV, Appendix B; z0 = 1.96. To find x0, substitute all the values into the z-score formula: z=

xμ

σ

x  50 1.96 = 0 3 x0 = 50 + 3(1.96) = 55.88 c.

P(x > x0) = .95 So, A1 + A2 = .95. Since A2 = .5, A1 = .95 − .5 = .4500. Look up the area .4500 in the body of Table IV, Appendix B; (since it is exactly between two values, average the z-scores). z0 ≈ −1.645. To find x0, substitute into the z-score formula: z=

xμ

σ

x  50 −1.645 = 0 3 x0 = 50 − 3(1.645) = 45.065

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197

198

Chapter 4

d.

P(41 ≤ x < x0) = .8630 z=

xμ

σ

41  50 = −3 3



A1 = P(41 ≤ x ≤ μ) = P(−3 ≤ z ≤ 0) = P(0 ≤ z ≤ 3) = .4987 A1 + A2 = .8630, since A1 = .4987, A2 = .8630 - .4987 = .3643. Look up .3643 in the body of Table IV, Appendix B; z0 = 1.1. To find x0, substitute into the z-score formula: z=

xμ

σ

x  50 1.1 = 0 3 x0 = 50 + 3(1.1) = 53.3 e.

P(x < x0) = .10 So A = .5000 − .10 = .4000 Look up area .4000 in the body of Table IV, Appendix B; z0 = 1.28. Since z0 is to the left of 0, z0 = −1.28. To find x0, substitute all the values into the z-score formula: z=

xμ

σ

x0  50 3 x0 = 50 − 1.28(3) = 46.16

−1.28 =

f.

P(x > x0) = .01 So A = .5000 − .01 = .4900 Look up area .4900 in the body of Table IV, Appendix B; z0 = 2.33. To find x0, substitute all the values into the z-score formula: z=

xμ

σ

x0  50 3 x0 = 50 + 2.33(3) = 56.99

2.33 =

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Random Variables and Probability Distributions

4.91

Let x = age of a powerful woman. The random variable x has a normal distribution with μ = 50 and σ = 5.3. Using Table IV, Appendix B, a.

b.

c.

d.

4.92

a.

60  50   55  50 P(55  x  60)  P  z   P(.83  z  1.67)  6 6   P(0  z  1.67)  P(0  z  .83)  .4525  .2967  .1558 52  50   48  50 P(48  x  52)  P  z   P(.33  z  .33)  6 6   P(.33  z  0)  P (0  z  .33)  .1293  .1293  .2586 35  50   P( x  35)  P  z    P( z  2.5)  6   .5  P (2.5  z  0)  .5  .4933  .0062 40  50   P( x  40)  P  z    P( z  1.67)  6   .5  P (1.67  z  0)  .5  .4525  .9525 Using Table IV, Appendix B, 0  5.26   P( x  0)  P  z    P( z  0.526)  10   .5  P(0.53  z  0)  .5  .2019  .7019

b.

15  5.26   5  5.26 P(5  x  15)  P  z   P (0.026  z  0.974)  10 10   P(.03  z  0)  P(0  z  .97)  .0120  .3340  .3460

c.

1  5.26   P( x  1)  P  z    P ( z  0.426)  10   .5  P(0.43  z  0)  .5  .1664  .3336

d.

25  5.26   P( x  25)  P  z    P ( z  3.026)  10  .5  P(3.03  z  0)  .5  .4988  .0012 Since the probability of seeing a win percentage of -25% or anything more unusual is so small (p = .0012), we would conclude that the average casino win percentage is not 5.26%.

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199

200

4.93

Chapter 4

a.

Let x = buy-side analyst’s forecast error. Then x has an approximate normal distribution with µ = .85 and σ = 1.93. Using Table IV, Appendix B, 2.00  .85   P( x  2.00)  P  z    P ( z  .60)  .5  .2257  .2743  1.93 

b.

Let y = sell-side analyst’s forecast error. Then y has an approximate normal distribution with µ = -.05 and σ = 85. Using Table IV, Appendix B, 2.00  (.05)   P( y  2.00)  P  z    P( z  2.41)  .5  .4920  .0080  .85

4.94

Let x = driver’s head injury rating. The random variable x has a normal distribution with μ = 605 and σ = 185. Using Table IV, Appendix B, a.

b.

c.

d.

4.95

700  605   500  605 P(500  x  700)  P  z   P (0.57  z  0.51)  185 185   P(0.57  z  0)  P(0  z  0.51)  .2157  .1950  .4107 500  605   400  605 P(400  x  500)  P  z   P(1.11  z  0.57)  185 185   P(1.11  z  0)  P(0.57  z  0)  .3665  .2157  .1508 850  605   P( x  850)  P  z    P( z  1.32)  .5  P (0  z  1.32)  185   .5  .4066  .9066 1, 000  605   P( x  1, 000)  P  z    P( z  2.14)  .5  P (0  z  2.14)  185  .5  .4838  .0162

Let x = transmission delay. The random variable x has a normal distribution with μ = 48.5 and σ = 8.5. Using Table IV, Appendix B, a.

b.

57  48.5   P( x  57)  P  z    P( z  1.00)  8.5   .5  P (0  z  1)  .5  .3413  .8413 60  48.5   40  48.5 P(40  x  60)  P  z   P (1  z  1.35)  8.5 8.5   P(1  z  0)  P(0  z  1.35)  .3413  .4115  .7528

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Random Variables and Probability Distributions

4.96

201

Let x = number of defects per million. Then x has an approximate normal distribution with µ = 3. Using Table IV, Appendix B, 3  1.5σ  3   3  1.5σ  3 P(3  1.5σ  x  3  1.5σ )  P  z   P(1.5  z  1.5)  .4332  .4332  .8664  σ σ It is fairly likely that the goal will b met. Since the probability is .8664, the goal would be met approximately 86.64% of the time.

4.97

a.

Let x = rating. Then x has a normal distribution with µ = 50 and σ = 15. Using Table IV, Appendix B, P(x > xo) = .10. Find xo. x  50   P( x  xo )  P  z  o  P ( z  zo )  .10  15  A1 = .5 − .10 = .4000 Looking up area .4000 in Table IV, zo = 1.28 zo 

b.

xo  50 x  50  1.28  o  xo  50  1.28(15)  69.2 15 15

P(x > xo) = .10 + .20 + .40 = .70. Find xo. x  50   P( x  xo )  P  z  o  P ( z  zo )  .70  15  A1 = .70 − .5 = .2000 Looking up area .2000 in Table IV, zo = −.52 zo 

4.98

a.

xo  50 x  50  .52  o  xo  50  .52(15)  42.2 15 15

Let x = crop yield. The random variable x has a normal distribution with μ = 1,500 and σ = 250. 1, 600 -1,500   P(x < 1,600) = P  z   = P(z < .4) = .5 + .1554 = .6554  250 (Using Table IV)

b.

Let x1 = crop yield in first year and x2 = crop yield in second year. If x1 and x2 are independent, then the probability that the farm will lose money for two straight years is: 1, 600  1,500   1, 600  1,500   P(x1 < 1,600) P(x2 < 1,600) = P  z1   P  z2    250 250 = P(z1 < .4) P(z2 < .4) = (.5 + .1554)(.5 + .1554) = .6554(.6554) = .4295 (Using Table IV)

c.

[1,500  2σ ]  1,500   [1,500  2σ ]  1,500 z P(1,500 − 2σ ≤ x ≤ 1,500 + 2σ) = P    σ σ (Using Table IV) = P(−2 ≤ z ≤ 2) = 2P(0 ≤ z ≤ 2) = 2(.4772) = .9544

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202

4.99

Chapter 4

a.

Using Table IV, Appendix B, and μ = 75 and σ = 7.5, 80  75   P(x > 80) = P  z   = P(z > .67) = .5 - .2486 = .2514  7.5  Thus, 25.14% of the scores exceeded 80.

b.

P(x ≤ x0) = .98. Find x0.  x  75  P(x ≤ x0) = P  z  0  = P(z ≤ z0) = .98  7.5  A1 = .98 - .5 = .4800 Looking up area .4800 in Table IV, z0 = 2.05. z0 =

4.100

x 0  75 x  75  2.05 = 0  x0 = 90.375 7.5 7.5

Let x = wage rate. The random variable x is normally distributed with μ = 17 and σ = 1.25. Using Table IV, Appendix B, a.

18.30  17   P( x  18.30)  P  z    P( z  1.04)  1.25   .5  P(0  z  1.04)  .5  .3508  .1492

b.

18.30  17   P( x  18.30)  P  z    P( z  1.04)  1.25   .5  P(0  z  1.04)  .5  .3508  .1492

c.

P(x ≤ η) = P(x ≥ η) = .5 Thus, μ = η = 17. (Recall from section 2.4 that in a symmetric distribution, the mean equals the median.)

4.101

a.

The contract will be profitable if total cost, x, is less than $1,000,000.  1, 000, 000  850, 000  P(x < 1,000,000) = P  z   = P(z < .88) = .5 + .3106 = .8106  170, 000

b.

The contract will result in a loss if total cost, x, exceeds 1,000,000. P(x > 1,000,000) = 1 − P(x < 1,000,000) = 1 − .8106 = .1894

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Random Variables and Probability Distributions

c.

203

P(x < R) = .99. Find R.  R  850, 000  = P(z < z0) = .99 P(x < R) = P  z   170, 000  A1 = .99 − .5 = .4900 Looking up the area .4900 in Table IV, z0 = 2.33

R  850, 000 R  850, 000  2.33 = 170, 000 170, 000  R = 2.33(170,000) + 850,000 = $1,246,100

z0 =

4.102

a.

Let x = quantity injected per container. The random variable x has a normal distribution with μ = 10 and σ = .2. 10  10   P(x < 10) = P  z   = P(z < 0.0) = .5  .2  10  10   P(x ≥ 10) = P  z   = P(z ≥ 0.0) = .5  .2 

4.103

b.

Since the container needed to be reprocessed, it cost $10. Upon refilling, it contained 10.60 units with a cost of 10.60($20) = $212. Thus, the total cost for filling this container is $10 + $212 = $222. Since the container sells for $230, the profit is $230 − $222 = $8.

c.

Let x = quantity injected per container. The random variable x has a normal distribution with μ = 10.10 and σ = .2. The expected value of x is E(x) = μ = 10.10. The cost of a container with 10.10 units is 10.10($20) = $202. Thus, the expected profit would be the selling price minus the cost or $230 − $202 = $28.

Let x = load. Then x has a normal distribution with µ = 20. We are given P(10 < x < 30) = .95. We want to find σ. P(10 < x < 30) = .95  P(z1 < z < z2) = .95  P(z1 < z < 0) = P(0 < z < z2) = .95/2 = .4750 Looking up area .4750 in Table IV, Appendix B, z2 = 1.96 and z1 = −1.96. z2 

4.104

a.

x  30

σ



30  20

σ

 1.96  σ 

10  5.1 1.96

If z is a standard normal random variable, QL = zL is the value of the standard normal distribution which has 25% of the data to the left and 75% to the right. Find zL such that P(z < zL) = .25 A1 = .50 − .25 = .25.

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204

Chapter 4

Look up the area A1 = .25 in the body of Table IV of Appendix B; zL = −.67 (taking the closest value). If interpolation is used, −.675 would be obtained. QU = zU is the value of the standard normal distribution which has 75% of the data to the left and 25% to the right. Find zU such that P(z < zU) = .75 A1 + A2 = P(z ≤ 0) + P(0 ≤ z ≤ zU) = .5 + P(0 ≤ z ≤ zU) = .75 Therefore, P(0 ≤ z ≤ zU) = .25. Look up the area .25 in the body of Table IV of Appendix B; zU = .67 (taking the closest value). b.

Recall that the inner fences of a box plot are located 1.5(QU − QL) outside the hinges (QL and QU). To find the lower inner fence, QL − 1.5(QU − QL) = −.67 − 1.5(.67 − (−.67)) = −.67 − 1.5(1.34) = −2.68 (−2.70 if zL = −.675 and zU = +.675) The upper inner fence is: QU + 1.5(QU − QL) = .67 + 1.5(.67 − (−.67)) = .67 + 1.5(1.34) = 2.68 (+2.70 if zL = −.675 and zU = +.675)

c.

Recall that the outer fences of a box plot are located 3(QU − QL) outside the hinges (QL and QU). To find the lower outer fence, QL − 3(QU − QL) = −.67 − 3(.67 − (−.67)) = −.67 − 3(1.34) = −4.69 (−4.725 if zL = −.675 and zU = +.675) The upper outer fence is: QU + 3(QU − QL) = .67 + 3(.67 − (−.67)) = .67 + 3(1.34) = 4.69 (4.725 if zL = −.675 and zU = +.675)

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Random Variables and Probability Distributions

d.

205

P(z < −2.68) + P(z > 2.68) = 2P(z > 2.68) = 2(.5000 − .4963) (Table IV, Appendix B) = 2(.0037) = .0074 (or 2(.5000 − .4965) = .0070 if −2.70 and 2.70 are used) P(z < −4.69) + P(z > 4.69) = 2P(z > 4.69) ≈ 2(.5000 − .5000) ≈ 0

4.105

4.106

e.

In a normal probability distribution, the probability of an observation being beyond the inner fences is only .0074 and the probability of an observation being beyond the outer fences is approximately zero. Since the probability is so small, there should not be any observations beyond the inner and outer fences. Therefore, they are probably outliers.

a.

The proportion of measurements that one would expect to fall in the interval μ ± σ is about .68.

b.

The proportion of measurements that one would expect to fall in the interval μ ± 2σ is about .95.

c.

The proportion of measurements that one would expect to fall in the interval μ ± 3σ is about 1.00.

a.

IQR = QU − QL = 195 − 72 = 123

b.

IQR/s = 123/95 = 1.295

c.

Yes. Since IQR is approximately 1.3, this implies that the data are approximately normal.

4.107

If the data are normally distributed, then the normal probability plot should be an approximate straight line. Of the three plots, only plot c implies that the data are normally distributed. The data points in plot c form an approximately straight line. In both plots a and b, the plots of the data points do not form a straight line.

4.108

a.

Using MINITAB, the stem-and-leaf display is: Stem-and-leaf of X N = 28 Leaf Unit = 0.10 5 6 8 11 14 14 10 7 2

2 3 4 5 6 7 8

11266 1 35 035 039 3457 346 24469 47

Since the data do not form a mound-shape, it indicates that the data may not be normally distributed.

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206

Chapter 4

b.

Using MINITAB, the descriptive statistics are: Variable

N

Mean

Median

TrMean

StDev

SE Mean

X

28

5.511

6.100

5.519

2.765

0.5230

Variable X

Minimum

Maximum

Q1

Q3

1.100

9.700

3.350

8.050

The standard deviation is 2.765. c.

Using the printout from MINITAB in part b, QL = 3.35, and QU = 8.05. The IQR = QU − QL = 8.05 − 3.35 = 4.7. If the data are normally distributed, then IQR/s ≈ 1.3. For this data, IQR/s = 4.7/2.765 = 1.70. This is a fair amount larger than 1.3, which indicates that the data may not be normally distributed.

d.

Using MINITAB, the normal probability plot is:

The data at the extremes are not particularly on a straight line. This indicates that the data are not normally distributed. 4.109

a.

IQR = QU – QL = 54 – 47 = 7 This agrees with IQR from the printout.

b.

From the printout, s = 6.444

c.

If the data are approximately normal, then

IQR  1.3 . For this problem, s

IQR 7   1.086  1.3 . Thus, the distribution is approximately normal. s 6.444

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Random Variables and Probability Distributions

d.

207

From Exercise 2.48 d, the histogram is: Histogram of AGE 14 12

Frequency

10 8 6 4 2 0

32

40

48

56

64

AGE

From the histogram, the data appear to be approximately mound-shaped. Thus, the data are approximately normal. 4.110

Based on the normal probability plot, it appears that the data are not approximately normal. If the data are normal, then the probability plot should reflect a straight line. In this graph, the plot of the data is not a straight line.

4.111

The information given in the problem states that x = 4.71, s = 6.09, QL = 1, and QU = 6. To be normal, the data have to be symmetric. If the data are symmetric, then the mean would equal the median and would be half way between the lower and upper quartile. Half way between the upper and lower quartiles is 3.5. The sample mean is 4.71, which is much larger than 3.5. This implies that the data may not be normal. In addition, the interquartile range divided by the standard deviation will be approximately 1.3 if the data are normal. For this data,

IQR QU  QL 6  1    .82 s 6.09 s The value of .82 is much smaller than the necessary 1.3 to be normal. Again, this is an indication that the data are not normal. Finally, the standard deviation is larger than the mean. Since one cannot have values of the variable in this case less than 0, a standard deviation larger than the mean indicates that the data are skewed to the right. This implies that the data are not normal.

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208

We will look at the 4 methods or determining if the data are normal. First, we will look at a histogram of the data. Using MINITAB, the histogram of the failure times of the 50 used panels is: Histogram of Fail Normal 12

Mean StDev N

10

1.935 0.9287 50

8 Frequency

4.112

Chapter 4

6 4 2 0

0

1

2 Fail

3

4

From the histogram, the data appear to have a somewhat normal distribution. Next, we look at the intervals x  s, x  2s, x  3s . If the proportions of observations falling in each interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB, the summary statistics are: Descriptive Statistics: Fail Variable Fail

N 50

Mean 1.935

StDev 0.929

Q1 1.218

Median 1.835

Q3 2.645

x  s  1.935  .929  (1.006, 2.864) 33 of the 50 values fall in this interval. The proportion is 33/50 = .66. This is fairly close to the .68 we would expect if the data were normal. x  2s  1.935  2(.929)  1.935  1.858  (0.077, 3.793) 49 of the 50 values fall in this interval. The proportion is 49/50 = .98. This is a fair amount above the .95 we would expect if the data were normal. x  3s  1.935  3(.929)  1.935  2.787  (0.852, 4.722) 50 of the 50 values fall in this interval. The proportion is 50/50 =1.00. This is equal to the 1.00 we would expect if the data were normal. From this method, it appears that the data may be normal. Next, we look at the ratio of the IQR to s. IQR = QU – QL = 2.645 – 1.218 = 1.427. IQR 1.427   1.54 . This is somewhat larger than the 1.3 we would expect if the data were normal. This s .929 method indicates the data may be normal.

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Random Variables and Probability Distributions

209

Finally, using MINTAB, the normal probability plot is: Probability Plot of Fail Normal - 95% CI 99

Mean StDev N AD P-Value

95 90

1.935 0.9287 50 0.305 0.557

Percent

80 70 60 50 40 30 20 10 5

1

-1

0

1

2 Fail

3

4

5

Since the data form a fairly straight line, the data may be normal. From the 4 different methods, all indications are that the failure times are approximately normal. We will look at the 4 methods for determining if the data are normal. First, we will look at a histogram of the data. Using MINITAB, the histogram of the driver’s head injury rating is:

20

Frequency

4.113

10

0 200

300 400

500 600

700

800 900 1000 1100 1200

DrivHead

From the histogram, the data appear to be somewhat skewed to the right, but is fairly mound-shaped. This indicates that the data are normal.

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210

Chapter 4 Next, we look at the intervals x  s, x  2s, x  3s . If the proportions of observations falling in each interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB, the summary statistics are: Descriptive Statistics: DrivHead Variable DrivHead

N 98

Mean 603.7

Median 605.0

TrMean 600.3

Variable DrivHead

Minimum 216.0

Maximum 1240.0

Q1 475.0

Q3 724.3

StDev 185.4

SE Mean 18.7

x  s  603.7  185.4  (418.3, 789.1) 68 of the 98 values fall in this interval. The proportion is .69. This is very close to the .68 we would expect if the data were normal. x  2s  603.7  2(185.4)  603.7  370.8  (232.9, 974.5) 96 of the 98 values fall in this interval. The proportion is .98. This is a fair amount larger than the .95 we would expect if the data were normal. x  3s  603.7  3(185.4)  603.7  556.2  (47.5, 1,159.9) 97 of the 98 values fall in this interval. The proportion is .99. This is fairly close to the 1.00 we would expect if the data were normal. From this method, it appears that the data may be normal. Next, we look at the ratio of the IQR to s. IQR = QU – QL = 724.3 – 475 = 249.3. IQR 249.3   1.3 This is equal to the 1.3 we would expect if the data were normal. This method 185.4 s indicates the data may be normal. Finally, using MINITAB, the normal probability plot is:

Since the data form a fairly straight line, the data may be normal. From the 4 different methods, all indications are that the driver’s head injury rating data are normal.

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.

Random Variables and Probability Distributions

We will look at the 4 methods for determining if the data are normal. First, we will look at a histogram of the data. Using MINITAB, the histogram of the sanitation scores is: Histogram of Score 60 50 40 Frequency

4.114

211

30 20 10 0

60.0

67.5

75.0 Score

82.5

90.0

97.5

From the histogram, the data appear to be skewed to the left. This indicates that the data are not normal. Next, we look at the intervals x  s, x  2s, x  3s . If the proportions of observations falling in each interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB, the summary statistics are: Descriptive Statistics: DDT Variable Score

N 182

Mean 95.044

StDev 5.391

Minimum 56.000

Q1 94.000

Median 96.500

Q3 98.000

Maximum 100.000

x  s  95.044  5.391  (89.653, 100.435) 164 of the 182 values fall in this interval. The proportion is .90. This is much larger than the .68 we would expect if the data were normal. x  2s  95.044  2(5.391)  95.044  10.782  (84.262, 105.826) 174 of the 182 values fall in this interval. The proportion is .96. This is slightly larger than the .95 we would expect if the data were normal. x  3s  95.044  3(5.391)  95.044  16.173  (78.871, 111.271) 178 of the 182 values fall in this interval. The proportion is .978. This is somewhat smaller than the 1.00 we would expect if the data were normal. From this method, it appears that the data are not normal. Next, we look at the ratio of the IQR to s. IQR = QU – QL = 98 – 94 = 4. IQR 4   .742 This is much smaller than the 1.3 we would expect if the data were normal. This s 5.391 method indicates the data are not normal.

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212

Chapter 4

Finally, using MINITAB, the normal probability plot is: Probability Plot of Score Normal - 95% CI 99.9

Mean StDev N AD P-Value

99

Percent

95 90

95.04 5.391 182 10.307 .03) = .5 − .0120 = .4880  15.811 (from Table IV, Appendix B)

b.

(500  .5)  500   (490  .5)  500 z P(490 ≤ x < 500) ≈ P    15.811 15.811 = P(−.66 ≤ z < −.03) = .2454 − .0120 = .2334 (from Table IV, Appendix B)

c.

4.121

(550  .5)  500   P(x > 550) ≈ P  z   = P(z > 3.19) ≈ .5 − .5 = 0  15.811 (from Table IV, Appendix B)

x is a binomial random variable with n = 100 and p = .4. μ ± 3σ  np ± 3 npq  100(.4) ± 3 100(.4)(1  .4)  40 ± 3(4.8990)  (25.303, 54.697)

Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the probabilities. a.

b.

c.

(35  .5)  40   P(x ≤ 35) ≈ P  z    4.899 = P(z ≤ −.92) = .5000 −.3212 = .1788 (Using Table IV in Appendix B.) P(40 ≤ x ≤ 50) (50  .5)  40   (40  .5)  40 z ≈ P   4.899 4.899 = P(−.10 ≤ z ≤ 2.14) = P(−.10 ≤ z ≤ 0) + P(0 ≤ z ≤ 2.14) = .0398 + .4838 = .5236 (Using Table IV in Appendix B.)

(38  .5)  40   P(x ≥ 38) ≈ P  z    4.899 = P(z ≥ −.51) = .5000 + .1950 = .6950 (Using Table IV in Appendix B.)

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Random Variables and Probability Distributions

4.122

a.

x is a binomial random variable with n = 100 and p = .2.

b.

µ = E(x) = np = 100(.20) = 20 σ  npq  100(.2)(.8)  16  4

c.

z

d.

4.123

xμ

σ



27.5  20  1.875 4

25  .5  20   P( x  25)  P  z    P( z  1.38)  .5  .4162  .9162  4 (Using Table IV, Appendix B)

Let x = number of patients who undergo laser surgery who have serious post-laser vision problems in 100,000 trials. Then x is a binomial random variable with n = 100,000 and p = .01. E(x) = μ = np = 100,000(.01) = 1,000.

σ  σ 2  npq  100, 000(.01)(.99)  990  31.464 To see if the normal approximation is appropriate, we use:

μ  3σ  1, 000  3(31.464)  1, 000  94.392  (905.608, 1, 094.392) Since the interval lies in the range of 0 to 100,000, the normal approximation is appropriate. 949.5  1000   P( x  950)  P  z    P ( z  1.61)  .5  .4463  .0537  31.464  (Using Table IV, Appendix B) 4.124

a.

E(x) = μ = np = 350(.27) = 94.5.

b.

σ  σ 2  npq  350(.27)(.73)  68.985  8.306

c.

z

d.

To see if the normal approximation is appropriate, we use:

xμ

σ



99.5  94.5  0.60 8.306

μ  3σ  94.5  3(8.306)  94.5  24.918  (69.582, 119.418) Since the interval lies in the range of 0 to 350, the normal approximation is appropriate. P( x  100)  P( z  0.60)  .5  .2257  .2743 (Using Table IV, Appendix B)

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221

222

4.125

Chapter 4

a.

x is a binomial random variable with n = 1,000 and p = .28. µ = E(x) = np = 1000(.28) = 280

b.

σ  npq  1000(.28)(.72)  201.6  14.2 μ  3σ  280  3(14.2)  280  42.6  (237.4, 322.6) Since the interval lies in the range 0 to 1000, we can use the normal approximation to approximate the probability. 750  .5  280   P( x  750)  P  z    P ( z  33.1)  .5  .5  0  14.2

4.126

a.

For n = 100 and p = .01: μ ± 3σ  np ± 3 npq  100(.01) ± 3 100(.01)(.99)  1 ± 3(.995)  1 ± 2.985  (−1.985, 3.985) Since the interval does not lie in the range 0 to 100, we cannot use the normal approximation to approximate the probabilities.

b.

For n = 100 and p = .5: μ ± 3σ  np ± 3 npq  100(.5) ± 3 100(.5)(.5)  50 ± 3(5)  50 ± 15  (35, 65)

Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the probabilities. c.

For n = 100 and p = .9: μ ± 3σ  np ± 3 npq  100(.9) ± 3 100(.9)(.1)  90 ± 3(3)  90 ± 9  (81, 99)

Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the probabilities. 4.127

Let x = number of bottles (brands) selected in 65 trials that contain tap water. Then x is binomial random variable with n = 65 and p = .25. E(x) = μ = np = 65(.25) = 16.25

σ  σ 2  npq  65(.25)(.75)  12.1875  3.49 To see if the normal approximation is appropriate, we use:

μ ± 3σ => 16.25 ± 3(3.49) => 16.25 ± 10.47 => (5.78, 26.72) Since this interval lies in the range from 0 to 65, the normal approximation is appropriate. (20  .5)  16.25   P( x  20)  P  z    P ( z  .93)  .5  P (0  z  .93)  .5  .3238  .1762  3.49 (Using Table IV, Appendix B)

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Random Variables and Probability Distributions

4.128

b.

223

Let v = number of credit card users out of 100 who carry Visa. Then v is a binomial random variable with n = 100 and pv = .46. E(v) = npv = 100(.46) = 46. Let d = number of credit card users out of 100 who carry Discover. Then d is a binomial random variable with n = 100 and pd = .06. E(d) = npd = 100(.06) = 6.

c.

To see if the normal approximation is valid, we use:

μ  3σ  npv  3 npv qv  100(.46)  3 100(.46)(.54)  46  3(4.984)  46  14.952  (31.048, 60.952) Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the probability. (50  .5)  46   P(v  50)  P  z    P( z  .70)  .5  .2580  .2420  4.984 Let a = number of credit card users out of 100 who carry American Express. Then a is a binomial random variable with n = 100 and pa = .12. To see if the normal approximation is valid, we use:

μ  3σ  npa  3 npa qa  100(.12)  3 100(.12)(.88)  12  3(3.250)  12  9.75  (2.25, 21.75) Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the probability. (50  .5)  12   P(a  50)  P  z    P( z  11.54)  .5  .5  0  3.25 d.

4.129

In order for the normal approximation to be valid, μ ± 3σ must lie in the interval (0, n). This check was done in part c for both portions of the question. In both cases, the normal approximation was justified.

Let x = number of defective CDs in n = 1,600 trials. Then x is a binomial random variable with n = 1,600 and p = .006. E(x) = μ = np = 1,600(.006) = 9.6.

σ  σ 2  npq  1, 600(.006)(.994)  9.5424  3.089 To see if the normal approximation is appropriate, we use:

μ  3σ  9.6  3(3.089)  9.6  9.267  (0.333, 18.867)

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224

Chapter 4

Since the interval lies in the range of 0 to 1,600, the normal approximation is appropriate. 11.5  9.6   P( x  12)  P  z    P ( z  0.62)  .5  .2324  .2676  3.089  (Using Table IV, Appendix B) Since this probability is fairly large, it would not be unusual to see 12 or more defectives in a sample of 1,600 if 99.4% were defect-free. Thus, there would be no evidence to cast doubt on the manufacturer’s claim. 4.130

a.

There are well over a million college students in 4-year public and private institutions. In order to collect a truly random sample, each of the college students must have an equal chance of being selected. This would be extremely hard to do.

b.

Suppose your institution was a 4-year public institution. Let x = number of students receiving financial aid in 100 trials. The random variable x has a binomial distribution with n = 100 and p = .45. To determine if the normal approximation is appropriate, we check: μ ± 3σ  np ± 3 npq  100(.45) ± 3 100(.45)(.55)  45 ± 3(4.9749)  45 ± 14.9247  (30.0753, 59.9247)

Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the probabilities. (50  .5)  45   P(x ≥ 50) ≈ P  z   = P(z ≥ .90) = .5 − .3159 = .1841  4.9749  (25  .5)  45   P(x < 25) ≈ P  z   = P(z < −4.12) = .5 − .5 = 0  4.9749  Suppose your institution was a 4-year private institution. Let x = number of students receiving financial aid in 100 trials. The random variable x has a binomial distribution with n = 100 and p = .52. To determine if the normal approximation is appropriate, we check:

μ ± 3σ  np ± 3 npq  100(.52) ± 3 100(.52)(.48)  52 ± 3(4.996)  52 ± 14.988  (37.012, 66.988)

Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the probabilities. (50  .5)  52   P(x ≥ 50) ≈ P  z   = P(z ≥ −.50) = .5 + .1915 = .6915  4.996 (25  .5)  52   P(x < 25) ≈ P  z   = P(z < −5.50) ≈ .5 − .5 = 0  4.996 c.

In order for the normal approximation to be appropriate, the interval μ ± 3σ should lie in the interval 0 to n. This assumption was checked in part b.

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Random Variables and Probability Distributions

4.131

225

Let x = number patients out of 150 who wait more than 30 minutes to see a doctor in a typical U.S. emergency room. Then x is a binomial random variable with n = 150 and p = .5. µ = np = 150(.5) = 75

σ  npq  150(.5)(.5)  37.5  6.124

To see if the normal approximation is valid, we use:

μ  3σ  75  3(6.124)  75  18.372  (56.628, 93.372) Since the interval lies in the range 0 to 150, we can use the normal approximation to approximate the probability. a.

b.

75  .5  75   P( x  75)  P  z    P( z  .08)  .5  .0319  .4681  6.124  (Using Table IV, Appendix B) 85  .5  75   P( x  85)  P  z    P( z  1.71)  .5  .4564  .0436  6.124  (Using Table IV, Appendix B)

c.

90  .5  75   60  .5  75 P(60  x  90)  P  z   P(2.37  z  2.37)  6.124 6.124   P(2.37  z  0)  P(0  z  2.37)  .4911  .4911  .9822 (Using Table IV, Appendix B)

4.132

a.

Let x = number of passengers in 1500 who will be detained for luggage inspection. Then x is a binomial random variable with n = 1500 and p = .20. The expected number of passengers detained will be: E(x) = np = 1,500(.2) = 300

4.133

b.

For n = 4,000, E(x) = np = 4,000(.2) = 800

c.

 (600  .5)  800  P(x > 600) ≈ P  z   = P(z > −7.89) = .5 + .5 = 1.0 4000(.2)(.8)  

a.

f(x) =

1 (c ≤ x ≤ d) d c 1 1 1   = .04 d  c 45  20 25

 .04 (20  x  45) So, f(x) =   0 otherwise

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226

Chapter 4

b.

c  d 20  45 65   = 32.5 2 2 2 d  c 45  20  = 7.22 σ= 12 12

μ=

c.

μ ± 2σ  32.5 ± 2(7.22)  (18.06, 46.94)

P(18.06 < x < 46.94) = P(20 < x < 45) = (45 − 20).04 = 1 4.134

 .04 (20  x  45) From Exercise 4.133, f(x) =   0 otherwise a.

P(20 ≤ x ≤ 30) = (30 − 20)(.04) = .4

b.

P(20 < x < 30) = (30 − 20)(.04) = .4

c.

P(x ≥ 30) = (45 − 30)(.04) = .6

d.

P(x ≥ 45) = (45 − 45)(.04) = 0

e.

P(x ≤ 40) = (40 − 20)(.04) = .8

f.

P(x < 40) = (40 − 20)(.04) = .8

g.

P(15 ≤ x ≤ 35) = (35 − 20)(.04) = .6

h.

P(21.5 ≤ x ≤ 31.5) = (31.5 − 21.5)(.04) = .4

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Random Variables and Probability Distributions

4.135

a.

1 (c ≤ x ≤ d) d c 1 1 1   d c 73 4 1  (3  x  7) f(x) =  4  0 otherwise f(x) =

c  d 3  7 10   =5 2 2 2 d c 73 4   = 1.155 σ= 12 12 12

b.

μ=

c.

μ ± σ  5 ± 1.155  (3.845, 6.155) P(μ − σ ≤x ≤ μ + σ) = P(3.845 ≤ x ≤ 6.155) =

4.136

4.137

4.138

4.139

227

a.

If θ = 1, a = 1, then e−a/θ = e−1 = .367879.

b.

If θ = 1, a = 2.5, then e−a/θ = e−2.5 = .082085

c.

If θ = .4, a = 3, then e−a/θ = e−7.5 = .000553

d.

If θ = .2, a = .3, then e−a/θ = e−1.5 = .223130

b  a 6.155  3.845 2.31  = d c 73 4 = .5775

P(x ≥ a) = e−a/θ = e−a/1. Using Table V, Appendix B: a.

P(x > 1) = e−1/1 = e−1 = .367879

b.

P(x ≤ 3) = 1 − P(x > 3) = 1 − e−3/1 = 1 − e−3 = 1 − .049787 = .950213

c.

P(x > 1.5) = e−1.5/1 = e−1.5 = .223130

d.

P(x ≤ 5) = 1 − P(x > 5) = 1 − e−5/1 = 1 − e−5 = 1 − .006738 = .993262

a.

P(x ≤ 4) = 1 − P(x > 4) = 1 − e−4/2.5 = 1 − e−1.6 = 1 − .201897 = .798103 (Using Table V, Appendix B)

b.

P(x > 5) = e−5/2.5 = e−2 = .135335

c.

P(x ≤ 2) = 1 − P(x > 2) = 1 − e−2/2.5 = 1 − e −.8 = 1 − .449329 = .550671

d.

P(x > 3) = e−3/2.5 = e−1.2 = .301194

1 1 1   = .01 d  c 200  100 100  .01 (100  x  200) f(x) =  0 otherwise  f(x) =

c  d 100  200 300   = 150 2 2 2 d  c 200  100 100 σ=   = 28.8675 12 12 12

μ=

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228

Chapter 4

a.

μ ± 2σ  150 ± 2(28.8675)  150 ± 57.735  (92.265, 207.735) P(x < 92.265) + P(x > 207.735) = P(x < 100) + P(x > 200) = 0 + 0 =0

b.

μ ± 3σ  150 ± 3(28.8675)  150 ± 86.6025  (63.3975, 236.6025) P(63.3975 < x < 236.6025) = P(100 < x < 200) = (200 − 100)(.01) = 1

c.

From a, μ ± 2σ  (92.265, 207.735). P(92.265 < x < 207.735) = P(100 < x < 200) = (200 − 100)(.01) = 1

4.140

With θ = 2, f(x)=

1 −x/2 (x > 0) e 2

μ=σ=θ=2 a.

μ ± 3σ  2 ± 3(2)  (−4, 8) Since μ − 3σ lies below 0, find the probability that x is more than μ + 3σ = 8. P(x > 8) = e−8/2 = e−4 = .018316

b.

(using Table V, Appendix B)

μ ± 2σ  2 ± 2(2)  (−2, 6) Since μ ± 2σ lies below 0, find the probability that x is between 0 and 6. P(x < 6) = 1 − P(x ≥ 6) = 1 − e−6/2 = 1 − e−3 = 1 − .049787 = .950213 (using Table V, Appendix B)

c.

μ ± .5σ  2 ± .5(2)  (1, 3) P(1 < x < 3) = P(x > 1) − P(x > 3) = e−1/2 − e−3/2 = e−.5 − e−1.5 = .606531 − .223130 = .383401 (using Table V in Appendix B)

4.141

a.

Let x = temperature with no bolt-on trace elements. Then x has a uniform distribution. f ( x) 

1 d c

(c  x  d )

1 1 1   d  c 290  260 30

1  Therefore, f ( x)   30  0

(260  x  290) otherwise

P(280  x  284)  (284  280)

1  1   4    .133  30  30

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Random Variables and Probability Distributions

229

Let y = temperature with bolt-on trace elements. Then y has a uniform distribution. f ( y) 

1 d c

(c  y  d )

1 1 1   d  c 285  278 7 1 (278  y  285)  Therefore, f ( y )   7  0 otherwise P(280  y  284)  (284  280)

b.

P( x  268)  (268  260)

1 1  4    .571 7 7

1  1   8    .267  30  30

P( y  268)  (268  260)(0)  0 4.142 a.

Let x = number of anthrax spores. Then x has an approximate uniform distribution. f ( x) 

1 d c

(c ≤ x ≤ d)

1 1 1    .1 d  c 10  0 10 .1 Therefore, f ( x)   0

(0  x  10) otherwise

P(x ≤ 8) = (8 – 0)(.1) = .8 b. 4.143

4.144

P(2 ≤ x ≤ 5) = (5 – 2)(.1) = .3

a.

P(x > 2) = e−2/2.5 = e−.8 = .449329

b.

P(x < 5) = 1 – P(x ≥ 5) = 1 – e−5/2.5 = 1 – e−2 = 1 − .135335 = .864665 (using Table V, Appendix B)

a.

Let x = time until the first critical part failure. Then x has an exponential distribution with θ = .1.

(using Table V, Appendix B)

P( x  1)  e 1/.1  e 10  .0000454 (using Table V, Appendix B) b.

30 minutes = .5 hours. P( x  .5)  1  P ( x  .5)  1  e .5/.1  1  e5  1  .0067  .9933 (using table V, Appendix B)

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230

4.145

Chapter 4

To construct a relative frequency histogram for the data, we can use 7 measurement classes. Interval width =

Largest number - smallest number 98.0716  .7434 = = 13.9 Number of classes 7

We will use an interval width of 14 and a starting value of .74335. The measurement classes, frequencies, and relative frequencies are given in the table below. Class

Measurement Class

Class Frequency

1 2 3 4 5 6 7

.74335 − 14.74335 14.74335 − 28.74335 28.74335 − 42.74335 42.74335 − 56.74335 56.74335 − 70.74335 70.74335 − 84.74335 84.74335 − 98.74335

6 4 6 6 5 4 9 40

Class Relative Frequency

6/40 = .15 .10 .15 .15 .125 .10 .225 1.000

The histogram looks like the data could be from a uniform distribution. The last class (84.74335 − 98.74335) has a few more observations in it than we would expect. However, we cannot expect a perfect graph from a sample of only 40 observations.

4.146

a.

For layer 2, let x = amount loss. Since the amount of loss is random between .01 and .05 million dollars, the uniform distribution for x is: f(x) =

1 d c

(c ≤ x ≤ d)

1 1 1   = 25 d  c .05  .01 .04  25 (.01  x  .05) Therefore, f(x) =   0 otherwise

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Random Variables and Probability Distributions

A graph of the distribution looks like the following:

μ=

σ=

c+d .01 + .05 = = .03 2 2

d c 12



.05  .01 12

= .0115, σ2 = (.0115)2 = .00013

The mean loss for layer 2 is .03 million dollars and the variance of the loss for layer 2 is .00013 million dollars squared. b.

For layer 6, let x = amount loss. Since the amount of loss is random between .50 and 1.00 million dollars, the uniform distribution for x is: f(x) =

1 d c

(c ≤ x ≤ d )

1 1 1   =2 d  c 1.00  .50 .50  2 (.50  x  1.00) Therefore, f(x) =   0 otherwise A graph of the distribution looks like the following:

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231

232

Chapter 4 μ=

σ=

c  d .50  1.00  = .75 2 2 d c 12



1.00  .50 12

= .1443, σ2 = (.1443)2 = .0208

The mean loss for layer 6 is .75 million dollars and the variance of the loss for layer 6 is .0208 million dollars squared. c.

A loss of $10,000 corresponds to x = .01. P(x > .01) = 1 A loss of $25,000 corresponds to x = .025. 1  1    P(x < .025) = (Base)(Height) = (x − c)  = (.025 − .01)   d  c  .05  .01 

d.

= .015(25) = .375 A loss of $750,000 corresponds to x = .75. A loss of $1,000,000 corresponds to x = 1. 1  1    P(.75 < x < 1) = (Base)(Height) = (d - x)  = (1.00 - .75)   d c  1.00  .50  = .25(2) = .5 A loss of $900,000 corresponds to x = .90. 1  1    P(x > .9) = (Base)(Height) = (d − x)  = (1.00 − .90)   d  c   1.00  .50  = .10(2) = .20 P(x = .9) = 0

4.147

a.

The amount dispensed by the beverage machine is a continuous random variable since it can take on any value between 6.5 and 7.5 ounces.

b.

Since the amount dispensed is random between 6.5 and 7.5 ounces, x is a uniform random variable. f ( x) 

1 (c ≤ x ≤ d) d c

1 1 1   =1 d  c 7.5  6.5 1  1 (6.5  x  7.5) Therefore, f ( x)    0 otherwise

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Random Variables and Probability Distributions

The graph is as follows:

c.

μ

σ

c  d 6.5  7.5 14   7 2 2 2 d c 12



7.5  6.5 12

 .2887

μ ± 2σ  7 ± 2(.2887)  7 ± .5774  (6.422, 7.577) d.

P(x ≥ 7) = (7.5 − 7)(1) = .5

e.

P(x < 6) = 0

f.

P(6.5 ≤ x ≤ 7.25) = (7.25 − 6.5)(1) = .75

g.

The probability that the next bottle filled will contain more than 7.25 ounces is: P(x > 7.25) = (7.5 − 7.25)(1) = .25 The probability that the next 6 bottles filled will contain more than 7.25 ounces is: P[(x > 7.25) ∩ (x > 7.25) ∩ (x > 7.25) ∩ (x > 7.25) ∩ (x > 7.25) ∩ (x > 7.25)] = [P(x >7.25)]6 = .256 = .0002

4.148

a.

Let x = length of time elapsed before the winning goal is scored. Then x has an exponential distribution with μ = 9.15. P(x ≤ 3) = 1 − P(x > 3) = 1 − e−3/9.15 = 1 − e−.327869 = 1 − .720457 = .279543

b.

P(x > 20) = e−20/9.15 = e−2.185792 = .1123887

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233

234

a.

For μ = 17 = θ. To graph the distribution, we will pick several values of x and find the value of f(x), where x = time between arrivals of the smaller craft at the pier. f(x) =

1

θ

e x /θ 

1  x /17 e 17

1 1/17 = .0555 e 17 1 f(3) = e3/17 = .0493 17 1 5 /17 = .0438 f(5) = e 17 1 7 /17 = .0390 f(7) = e 17 1 10 /17 = .0327 f(10) = e 17 1 15/17 = .0243 f(15) = e 17 1 20 /17 = .0181 f(20) = e 17 1 25/17 = .0135 f(25) = e 17 f(1) =

The graph is: 0.06

0.05

0.04 f(x)

4.149

Chapter 4

0.03

0.02

0.01 0

5

10

15

20

25

x

b.

We want to find the probability that the time between arrivals is less than 15 minutes. P(x < 15) = 1 − P(x ≥ 15) = 1 − e−15/17 = 1 − .4138 = .5862

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Random Variables and Probability Distributions 4.150

Let x = cycle availability, where x has a uniform distribution on the interval from 0 to 1. Mean = μ 

c  d 0 1   .5 2 2

Standard deviation = σ 

d c 12



1 0 12

 .289

The 10th percentile is that value of x such that 10% of all observations are below it. Let K1 = 10th percentile. P(x ≤ K1) = (K1 − 0)(1 − 0) = K1 = .10

The lower quartile is that value of x such that 25% of all observations are below it. Let K2 = 25th percentile. P(x ≤ K2) = (K2 − 0)(1 − 0) = K2 = .25

The UPPER quartile is that value of x such that 75% of all observations are below it. Let K3 = 75th percentile. P(x ≤ K3) = (K3 − 0)(1 − 0) = K3 = .75

4.151

a.

Let x = product’s lifetime at the end of its lifetime. Then x has an exponential distribution with μ = 500,000. P(x < 700,000) = 1 – P(x ≥ 700,000) = 1 – e-700000/5000000 = 1 – e-1.4 = 1 − .246597 = .753403

(Using Table V, Appendix B) b.

Let y = product’s lifetime during its normal life. Then y has a uniform distribution. f ( y) 

1 d c

(c ≤ y ≤ d)

1 1 1   d  c 1, 000, 000  100, 000 900, 000

 1  Therefore, f ( y )   900,000 0 

(100,000  y  1,000,000) otherwise

 1  P( y  700, 000)  (700, 000  100, 000)   .667  900, 000  c.

P(x < 830,000) = 1 – P(x ≥ 830,000) = 1 – e-830000/5000000 = 1 – e-1.66 = 1 − .190139 = .809861 (Using a calculator)  1  P( y  830, 000)  (830, 000  100, 000)   .811  900, 000 

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235

236

4.152

Chapter 4

a.

Let x = life length of CD-ROM. Then x has an exponential distribution with θ = 25,000. R(t )  P ( x  t )  e  t / 25,000

b.

R(8,760)  P ( x  8, 760)  e8,760 / 25,000  e.3504  .7044

c.

S(t) = probability that at least one of two drives has a length exceeding t hours = 1 – probability that neither has a length exceeding t hours = 1 – P(x1 ≤ t)P(x2 ≤ t) = 1 – [1 – P(x1 > t)][1 – P(x2> t)] = 1 – [1 – e-t/25,000][1 – e-t/25,000] =1 – [1 – 2e-t/25,000 + e-t/12,500] = 2e-t/25,000 – e-t/12,500

4.153

d.

S (8, 760)  2e 8,760 / 25,000  e8,760 /12,500  2(.7044)  .4962  1.4088  .4962  .9126

e.

The probability in part d is greater than that in part b. We would expect this. The probability that at least one of the systems lasts longer than 8,760 hours would be greater than the probability that only one system lasts longer than 8,760 hours.

Let x = number of inches a gouge is from one end of the spindle. Then x has a uniform distribution with f(x) as follows: 1 1  1    f(x) =  d  c 18  0 18 0

0  x  18 otherwise

In order to get at least 14 consecutive inches without a gouge, the gouge must be within 4 inches of either end. Thus, we must find: P(x < 4) + P(x > 14) = (4 − 0)(1/18) + (18 − 14)(1/18) = 4/18 + 4/18 = 8/18 = .4444 4.154

a.

b.

c.

c  d 0 1  = .5 2 2 d  c 1 0  = .289 σ= 12 12 μ=

σ2 = .2892 = .083

P(p > .95) = (1 − .95)(1) = .05 P(p < .95) = (.95 − 0)(1) = .95

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Random Variables and Probability Distributions

d.

237

The analyst should use a uniform probability distribution with c = .90 and d = .95.

1 1  1 = = = 20 (.90 ≤ p ≤ .95)  f(p) =  d − c .95 − .90 .05  0 otherwise 4.155

a.

For θ = 250, P(x > a) = e−a/250 For a = 300 and b = 200, show P(x > a + b) ≥ P(x > a)P(x > b) P(x > 300 + 200) = P(x > 500) = e−500/250 = e−2 = .1353 P(x > 300) P(x > 200) = e−300/250 e−200/250 = e−1.2 e−.8 = .3012(.4493) = .1353

Since P(x > 300 + 200) = P(x > 300) P(x > 200), then P(x > 300 + 200) ≥ P(x > 300) P(x > 200) Also, show P(x > 300 + 200) ≤P(x > 300) P(x > 200). Since we already showed that P(x > 300 + 200) = P(x > 300) P(x > 200), then P(x > 300 + 200) ≤ P(x > 300) P(x > 200). b.

Let a = 50 and b = 100. Show P(x > a + b) ≥ P(x > a) P(x > b) P(x > 50 + 100) = P(x > 150) = e−150/250 = e−.6 = .5488 P(x > 50) P(x > 100) = e−50/250 e−100/250 = e−.2e−.4 = .8187(.6703) = .5488

Since P(x > 50 + 100) = P(x > 50) P(x > 100), then P(x > 50 + 100) ≥ P(x > 50) P(x > 100) Also, show P(x > 50 + 100) ≤ P(x > 50) P(x > 100). Since we already showed that P(x > 50 + 100) = P(x > 50) P(x > 100), then P(x > 50 + 100) ≤ P(x > 50) P(x > 100). c.

4.156

Show P(x > a + b) ≥ P(x > a) P(x > b) P(x > a + b) = e−(a+b)/250 = e−a/250 e−b/250 = P(x > a) P(x > b)

a–b. The different samples of n = 2 with replacement and their means are: Possible Samples 0, 0 0, 2 0, 4 0, 6 2, 0 2, 2 2, 4 2, 6

c.

x 0 1 2 3 1 2 3 4

Possible Samples 4, 0 4, 2 4, 4 4, 6 6, 0 6, 2 6, 4 6, 6

x 2 3 4 5 3 4 5 6

Since each sample is equally likely, the probability of any 1 being selected is

1 1 1   4  4  16

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238

Chapter 4

d.

P( x = 0) = P( x = 1) = P( x = 2) = P( x = 3) = P( x = 4) = P( x = 5) = P( x = 6) =

1 16 1 1 2   16 16 16 1 1 1 3    16 16 16 16 1 1 1 1 4     16 16 16 16 16 1 1 1 3    16 16 16 16 1 1 2   16 16 16 1 16

x

p( x )

0 1 2 3 4 5 6

1/16 2/16 3/16 4/16 3/16 2/16 1/16

e.

4.157

Solution will vary. See page 1037 for Guided Solutions.

4.158

If the observations are independent of each other, then P(1, 1) = p(1)p(1) = .2(.2) = .04 P(1, 2) = p(1)p(2) = .2(.3) = .06 P(1, 3) = p(1)p(3) = .2(.2) = .04 etc. a. Possible Sample

1, 1 1, 2 1, 3 1, 4 1, 5 2, 1 2, 2 2, 3 2, 4 2, 5 3, 1 3, 2 3, 3

x

1 1.5 2 2.5 3 1.5 2 2.5 3 3.5 2 2.5 3

p( x )

Possible Samples

.04 .06 .04 .04 .02 .06 .09 .06 .06 .03 .04 .06 .04

3, 4 3, 5 4, 1 4, 2 4, 3 4, 4 4, 5 5, 1 5, 2 5, 3 5, 4 5, 5

x

3.5 4 2.5 3 3.5 4 4.5 3 3.5 4 4.5 5

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.

p( x )

.04 .02 .04 .06 .04 .04 .02 .02 .03 .02 .02 .01

Random Variables and Probability Distributions

Summing the probabilities, the probability distribution of is: x

p( x )

1 1.5 2 2.5 3 3.5 4 4.5 5

.04 .12 .17 .20 .20 .14 .08 .04 .01

b.

4.159

c.

P( x ≥ 4.5) = .04 + .01 = .05

d.

No. The probability of observing = 4.5 or larger is small (.05).

E ( x) = μ =

 xp( x) = 1(.2) + 2(.3) + 3(.2) + 4(.2) + 5(.1) = .2 + .6 + .6 + .8 + .5 = 2.7

E( x ) =

 xp( x ) = 1.0(.04) + 1.5(.12) + 2.0(.17) + 2.5(.20) + 3.0(.20) + 3.5(.14) + 4.0(.08) + 4.5(.04) + 5.0(.01) = .04 + .18 + .34 + .50 + .60 + .49 + .32 + .18 + .05 = 2.7

4.160

a.

For a sample of size n = 2, the sample mean and sample median are exactly the same. Thus, the sampling distribution of the sample median is the same as that for the sample mean (see Exercise 4.158a).

b.

The probability histogram for the sample median is identical to that for the sample mean (see Exercise 4.158b).

4.161

Solution will vary. See page 1037 for Guided Solutions.

4.162

Solution will vary. See page 1037 for Guided Solutions.

4.163

The sampling distribution is approximately normal only if the sample size is sufficiently large or if the population being sampled from is normal.

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239

240

4.164

4.165

4.166

Chapter 4

a.

μ x = μ = 10, σ x = σ / n  3 / 25 = 0.6

b.

μ x = μ = 100, σ x = σ / n  25 / 25 = 5

c.

μ x = μ = 20, σ x = σ / n  40 / 25 = 8

d.

μ x = μ = 10, σ x = σ / n  100 / 25 = 20

a.

μ x  μ  100, σ x 

b.

μ x  μ  100, σ x 

c.

μ x  μ  100, σ x 

d.

μ x  μ  100, σ x 

e.

μ x  μ  100, σ x 

f.

μ x  μ  100, σ x 

a.

μ x  μ  20, σ x  σ / n  16 / 64  2

b.

By the Central Limit Theorem, the distribution of is approximately normal. In order for the Central Limit Theorem to apply, n must be sufficiently large. For this problem, n = 64 is sufficiently large.

c.

z=

d.

z=

x  μx

σx x  μx

σx

σ n

σ n

σ n

σ n

σ n

σ n



100



100



100



100



100



100

4

25

100

50

5

2

1

 1.414

500

 .447

1000



15.5  20   2.25 2



23  20  1.50 2

 .316

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Random Variables and Probability Distributions

4.167

In Exercise 4.166, it was determined that the mean and standard deviation of the sampling distribution of the sample mean are 20 and 2 respectively. Using Table IV, Appendix B: a.

16  20   P( x < 16) = P  z   = P(z < −2) = .5 − .4772 = .0228  2 

b.

23  20   P( x > 23) = P  z   = P(z > 1.50) = .5 − .4332 = .0668  2 

c.

25  20   P( x > 25) = P  z   = P(z > 2.5) = .5 − .4938 = .0062  2 

d.

e. 4.168

241

22  20   16  20 P(16 < x < 22) = P  z  = P(−2 < z < 1)  2 2  = .4772 + .3413 = .8185 14  20   P( x < 14) = P  z   = P(z < −3) = .5 − .4987 = .0013  2 

For this population and sample size, E ( x ) = μ = 100, σ x = σ / n  10 / 900 = 1/3 a.

Approximately 95% of the time, will be within two standard deviations of the mean, i.e., μ ± 2σ  2 1 100 ± 2    100 ±  (99.33, 100.67). Almost all of the time, the sample mean will be within 3 3 1 three standard deviations of the mean, i.e., μ ± 3σ  100 ± 3    100 ± 1  (99, 101). 3

b. c.

4.169

1 No more than three standard deviations, i.e., 3   = 1 3 No, the previous answer only depended on the standard deviation of the sampling distribution of the sample mean, not the mean itself.

By the Central Limit Theorem, the sampling distribution of is approximately normal with μ x = μ = 30 and σ x  σ / n  16 / 100 = 1.6. Using Table IV, Appendix B: a.

28  30   P( ≥ 28) = P  z   = P(z ≥ −1.25) = .5 + .3944 = .8944  1.6 

b.

26.8  30   22.1  30 z P(22.1 ≤ x ≤ 26.8) = P   = P(−4.94 ≤ z ≤ −2) = .5 − .4772 = .0228  1.6 1.6 

c.

28.2  30   P( x ≤ 28.2) = P  z   = P(z ≤ −1.13) = .5 − .3708 = .1292  1.6 

d.

27.0  30   P( x ≥ 27.0) = P  z   = P(z ≥ −1.88) = .5 + .4699 = .9699  1.6 

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242

Chapter 4

4.170

Solution will vary. See page 1037 for Guided Solutions.

4.171

a.

μ x  μ  141

b.

σx 

c.

By the Central Limit Theorem, the sampling distribution of x is approximately normal.

d.

z

e.

P( x  142)  P( z  0.56)  .5  .2123  .2877 (Using Table IV, Appendix B)

a.

μ x  μ  97,300

b.

σx 

4.172

c.

4.173

σ n

x  μx

σx

σ n

18





100



 1.8

142  141  0.56 1.8

30, 000 50

 4, 242.6407

By the Central Limit Theorem, the sampling distribution of x is approximately normal. x  μx

z

e.

P( x  89,500)  P ( z  1.84)  .5  .4671  .9671 (Using Table IV, Appendix B)

σx



89,500  97,300  1.84 4, 242.6407

d.

By the Central Limit Theorem, the sampling distribution of x is approximately normal with μ x  μ  19 and σ x 

σ n



65 100

 6.5 .

Using Table IV, Appendix B, 10  19   P( x  10)  P  z    P ( z  1.38)  .5  .4162  .0838  6.5  4.174

a. b. c.

4.175

a.

b.

σ

.10   .0141 n 50 Since n > 30, the sampling distribution of x is approximately normal by the Central Limit Theorem. E ( x )  μ x  μ  .10

σx 

.13  .10   P( x  .13)  P  z    P ( z  2.13)  .5  .4834  .0166  .0141  (Using Table IV, Appendix B) By the Central Limit Theorem, the sampling distribution of x is approximately normal with a mean σ .193   .0273 . μ x  μ  .53 and standard deviation σ x  n 50 .58  .53   P( x  .58)  P  z    P ( z  1.83)  .5  .4664  .0336  .0273 

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Random Variables and Probability Distributions

c.

243

If Before Tensioning: μ x  μ  .53 .59  .53   P( x  .59)  P  z    P( z  2.20)  .5  .4861  .0139  .0273  If After Tensioning: μ x  μ  .58 .59  .58   P( x  .59)  P  z    P( z  0.37)  .5  .1443  .3557  .0273  Since the probability of getting a maximum differential of .59 or more Before Tensioning is so small, it would be very unlikely that the measurements were obtained before tensioning. However, since the probability of getting a maximum differential of .59 or more After Tensioning is not small, it would not be unusual that the measurements were obtained after tensioning. Thus, most likely, the measurements were obtained After Tensioning.

4.176

a.

Since the sample size is small, we also have to assume that the distribution from which the sample σ .5 was drawn is normal. μ x  μ  1.8 , σ x    .1118 n 20 1.85  1.8   P( x  1.85)  P  z    P ( z  0.45)  .5  .1736  .3264  .1118  (using Table IV, Appendix B)

b.

Using MINITAB, the descriptive statistics are:

Descriptive Statistics: Rough Variable Rough

N 20

N* 0

Mean 1.881

SE Mean 0.117

StDev 0.524

Minimum 1.060

Q1 1.303

Median 2.040

Q3 2.293

Maximum 2.640

From this output, the value of x is 1.881. c.

For x = 1.881: 1.881  1.8   P( x  1.881)  P  z    P( z  0.72)  .5  .1736  .3264  .1118  Since this probability is so high, observing a sample mean of x = 1.881, is not unusual. The assumptions in part a appear to be valid.

4.177

a.

By the Central Limit Theorem, the sampling distribution of x is approximately normal with a mean σ 10 μ x  μ  6 and standard deviation σ x    .5538 . n 326 7.5  6   P( x  7.5)  P  z    P( z  2.71)  .5  .4966  .0034  .5538  (Using Table IV, Appendix B)

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244

Chapter 4

b.

We first need to find the probability of observing the current data or anything more unusual if the true mean is 6. 300  6   P( x  300)  P  z    P( z  530.88)  .5  .5  0  .5538  Since the probability of observing a sample mean of 300 ppb or higher is essentially 0 if the true mean is 6 ppb, we would infer that the true mean PFOA concentration for the population of people who live near DuPont’s Teflon facility is not 6 ppb but higher than 6 ppb.

4.178

a.

By the Central Limit Theorem, the sampling distribution of x is approximately normal with μ x  μ and σ x  σ / n  σ / 100 .

b.

The mean of the x distribution is equal to the mean of the distribution of the fleet or the fleet mean score.

c.

μ x  μ  30 and σ x  σ / n  σ / 100  60 / 100  6 .

d.

4.179

45  30   P( x  45)  P  z    P  z  2.5  .5  .4938  .0062 (Using Table IV, Appendix B)  6  The sample mean of 45 tends to refute the claim. If the true fleet mean was as high as 30, observing a sample mean of 45 or higher would be extremely unlikely (probability = .0062). Thus, we would infer that the true mean is actually not 30 but something higher. Thus, we would refute the company’s claim that the mean “couldn’t possibly be as large as 30.”

a.

By the Central Limit Theorem, the sampling distribution of is approximately normal with μ x = μ and σ x = σ / n = σ / 50 .

b.

μ x = μ = 40 and σ x = σ / 50 = 12 / 50 = 1.6971. 44  40   P( x ≥ 44) = P  z   = P(z ≥ 2.36) = .5 − .4909 = .0091  1.6971  (using Table IV, Appendix B)

c.

μ ± 2σ / n  40 ± 2(1.6971)  40 ± 3.3942  (36.6058, 43.3942) 43.3942  40   36.6058  40 P(36.6058 ≤ x ≤ 43.3942) = P  z   1.6971 1.6971  = P(−2 ≤ z ≤ 2) = 2(.4772) = .9544 (using Table IV, Appendix B)

4.180

For n = 36, μ x = μ = 406 and σ x = σ / n = 10.1/ 36 = 1.6833. By the Central Limit Theorem, the sampling distribution is approximately normal (n is large). 400.8  406   P( x ≤ 400.8) = P  z   = P(z ≤ −3.09) = .5 − .4990 = .0010  1.6833  (using Table IV, Appendix B) The first. If the true value of μ is 406, it would be extremely unlikely to observe an as small as 400.8 or smaller (probability .0010). Thus, we would infer that the true value of μ is less than 406.

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Random Variables and Probability Distributions

4.181

245

For n = 50, we can use the Central Limit Theorem to decide the shape of the distribution of the sample mean bacterial counts. For the handrubbing sample, the sampling distribution of x is approximately σ 59   8.344 . For the handwashing sample, normal with a mean of μ = 35 and standard deviation n 50 the sampling distribution of x is approximately normal with a mean of μ = 69 and standard deviation σ 106   14.991 . n 50 For Handrubbing:

30  35   P( x  30 | μ  35)  P  z    P ( z  .60)  .5  .2257  .2743  8.344  (using Table IV, Appendix B) For Handwashing:

30  69   P( x  30 | μ  69)  P  z    P( z  2.60)  .5  .4953  .0047  14.991  (using Table IV, Appendix B) Since the probability of getting a sample mean of less than 30 for the handrubbing is not small compared with that for the handwashing, the sample of workers probably came from the handrubbing group. 4.182

a.

This experiment consists of 100 trials. Each trial results in one of two outcomes: chip is defective or not defective. If the number of chips produced in one hour is much larger than 100, then we can assume the probability of a defective chip is the same on each trial and that the trials are independent. Thus, x is a binomial. If, however, the number of chips produced in an hour is not much larger than 100, the trials would not be independent. Then x would not be a binomial random variable.

b.

This experiment consists of two trials. Each trial results in one of two outcomes: applicant qualified or not qualified. However, the trials are not independent. The probability of selecting a qualified applicant on the first trial is 3 out of 5. The probability of selecting a qualified applicant on the second trial depends on what happened on the first trial. Thus, x is not a binomial random variable. It is a hypergeometric random variable.

c.

The number of trials is not a specified number in this experiment, thus x is not a binomial random variable. In this experiment, x is counting the number of calls received.

d.

The number of trials in this experiment is 1000. Each trial can result in one of two outcomes: favor state income tax or not favor state income tax. Since 1000 is small compared to the number of registered voters in Florida, the probability of selecting a voter in favor of the state income tax is the same from trial to trial, and the trials are independent of each other. Thus, x is a binomial random variable.

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246

4.183

4.184

Chapter 4

n p(x) =   p x q n - x x = 0, 1, 2, ... , n  x

a.

7 7! 3 4 P(x = 3) = p(3) =   .53.5 4  .5 .5 = 35(.125)(.0625) = .2734 3!4!  3

b.

4 4! 3 1 P(x = 3) = p(3) =   .83.21  .8 .2 = 4(.512)(.2) = .4096 3!1!  3

c.

15  15! 1 14 .1 .9 = 15(.1)(.228768) = .3432 P(x = 1) = p(1) =   .11.914  1!14!  1

a.

μ= σ2 =

 xp( x) = 10(.2) + 12(.3) + 18(.1) + 20(.4) = 15.4  (x  μ )

2

p( x)

= (10 − 15.4) (.2) + (12 − 15.4)2(.3) + (18 − 15.4)2(.1) + (20 − 15.4)2(.4) = 18.44 σ = 18.44 ≈ 4.294

4.185

2

b

P(x < 15) = p(10) + p(12) = .2 + .3 = .5

c.

μ ± 2σ = 15.4 ± 2(4.294)  (6.812, 23.988)

d.

P(6.812 < x < 23.988) = .2 + .3 + .1 + .4 = 1.0

From Table II, Appendix B: a.

P(x = 14) = P(x ≤ 14) − P(x ≤ 13) = .584 − .392 = .192

b.

P(x ≤ 12) = .228

c.

P(x > 12) = 1 − P(x ≤ 12) = 1 − .228 = .772

d.

P(9 ≤ x ≤ 18) = P(x ≤ 18) − P(x ≤ 8) = .992 − .005 = .987

e.

P(8 < x < 18) = P(x ≤ 17) − P(x ≤ 8) = .965 − .005 = .960

f.

μ = np = 20(.7) = 14 σ2 = npq = 20(.7)(.3) = 4.2, σ =

g.

4.2 = 2.049

μ ± 2σ  14 ± 2(2.049)  14 ± 4.098  (9.902, 18.098) P(9.902 < x < 18.098) = P(10 ≤ x ≤ 18) = P(x ≤ 18) − P(x ≤ 9) = .992 − .017 = .975

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Random Variables and Probability Distributions

4.186

4.187

4.188

4.189

4.190

Using Table III, Appendix B, a.

When λ = 2, p(3) = P(x ≤ 3) − P(x ≤ 2) = .857 − .677 = .180

b.

When λ = 1, p(4) = P(x ≤ 4) − P(x ≤ 3) = .996 − .981 = .015

c.

When λ = .5, p(2) = P(x ≤ 2) − P(x ≤ 1) = .986 − .910 = .076

a.

Poisson

b.

Binomial

c.

Binomial

a.

 r   N  r  3   8  3 3! 5!  x   n  x   2   5  2  2!1! 3!2!  3(10) = .536   P(x = 2) = 8! 56 N 8   n   5  5!3!

b.

r   N  r  2 6  2 2! 4!  x   n  x   2   2  2  1(1)   2!0! 0!4!  = .067 P(x = 2) = 6! 15 N 6  n   2  2!4!

c.

 r   N  r   4  5  4 4! 1!  x   n  x   3   4  3  3!1! 1!0!  4(1) = .8   P(x = 3) = 5! 5 N 5  n   3  4!1!

a.

Discrete - The number of damaged inventory items is countable.

b.

Continuous - The average monthly sales can take on any value within an acceptable limit.

c.

Continuous - The number of square feet can take on any positive value.

d.

Continuous - The length of time we must wait can take on any positive value.

a.

1 1  1   ,10  x  90  f(x) =  d  c 90  10 80  0 otherwise

b.

μ=

cd 10  90 = = 50 2 2 d c 90  10 = = 23.094011 σ= 12 12

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247

248

4.191

Chapter 4

c.

The interval μ ± 2σ  50 ± 2(23.094)  50 ± 46.188  (3.812, 96.188) is indicated on the graph.

d.

P(x ≤ 60) = Base(height) = (60 − 10)

e.

P(x ≥ 90) = 0

f.

P(x ≤ 80) = Base(height) = (80 − 10)

g.

P(μ −σ ≤ x ≤ μ + σ) = P(50 − 23.094 ≤ x ≤ 50 + 23.094) = P(26.906 ≤ x ≤ 73.094) = Base(height)  1  46.188 = (73.094 − 26.906)    = .577  80  80

h.

P(x > 75) = Base(height) = (90 − 75)

a.

P(z ≤ 2.1) = A1 + A2 = .5 + .4821 = .9821

b.

P(z ≥ 2.1) = A2 = .5 − A1 = .5 − .4821 = .0179

c.

P(z ≥ −1.65) = A1 + A2 = .4505 + .5000 = .9505

d.

P(−2.13 ≤ z ≤ −.41) = P(−2.13 ≤ z ≤ 0) − P(−.41 ≤ z ≤ 0) = .4834 − .1591 = .3243

1 5  = .625 80 8

1 7  = .875 80 8

1 15  = .1875 80 80

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Random Variables and Probability Distributions

4.192

249

e.

P(−1.45 ≤ z ≤ 2.15) = A1 + A2 = .4265 + .4842 = .9107

f.

P(z ≤ −1.43) = A1 = .5 − A2 = .5000 − .4236 = .0764

a.

P(z ≤ z0) = .5080  P(0 ≤ z ≤ z0) = .5080 − .5 = .0080 Looking up the area .0080 in Table IV,  z0 = .02

b.

P(z ≥ z0) = .5517  P(z0 ≤ z ≤ 0) = .5517 − .5 = .0517 Looking up the area .0517 in Table IV, z0 = −.13.

c.

P(z ≥ z0) = .1492  P(0 ≤ z ≤ z0) = .5 − .1492 = .3508 Looking up the area .3508 in Table IV,  z0 = 1.04

d.

P(z0 ≤ z ≤ .59) = .4773  P(z0 ≤ z ≤ 0) + P(0 ≤ z ≤ .59) = .4773 P(0 ≤ z ≤ .59) = .2224 Thus, P(z0 ≤ z ≤ 0) = .4773 − .2224 = .2549 Looking up the area .2549 in Table IV, z0 = -.69

4.193

x

7

a.

For the probability density function, f ( x) 

e

b.

For the probability density function, f ( x) 

1 , 5  x  25 , x is a uniform random variable. 20

c.

For the probability function, f ( x) 

7

e .5[( x 10) / 5]

, x  0 , x is an exponential random variable.

2

5 2π

, x is a normal random variable.

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250

4.194

4.195

Chapter 4

a.

P(x ≤ 1) = 1 − P(x > 1) = 1 − e−1/3 = 1 − .716531 = .283469 (using calculator)

b.

P(x > 1) = e −1/3 = .716531

c.

P(x = 1) = 0

d.

P(x ≤ 6) = 1 − P(x > 6) = 1 − e−6/3 = 1 − e−2 = 1 − .135335 = .864665 (using Table V, Appendix B)

e.

P(2 ≤ x ≤ 10) = P(x ≥ 2) − P(x > 10) = e−2/3 − e−10/3 = .513417 − .035674 = .477743 (using calculator)

a.

b.

c.

(x is a continuous random variable. There is no probability associated with a single point.)

80  75   P(x ≤ 80) = P  z   = P(z ≤ .5)  10  = .5000 + .1915 = .6915 (Table IV, Appendix B)

85  75   P(x ≥ 85) = P  z   = P(z ≥ 1)  10  = .5000 - .3413 = .1587 (Table IV, Appendix B)

75  75   70  75 z P(70 ≤ x ≤ 75) = P    10 10  = P(−.5 ≤ z ≤ 0) = P(0 ≤ z ≤ .5) = .1915

d.

P(x > 80) = 1 − P(x ≤ 80) = 1 − .6915 = .3085 (Refer to part a.)

e.

P(x = 78) = 0, since a single point does not have an area.

f.

110  75   P(x ≤ 110) = P  z   = P(z ≤ 3.5)  10  ≈ .5000 + .5000 = 1.0 (Table IV, Appendix B)

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Random Variables and Probability Distributions

4.196

4.174 μ = np = 100(.5) = 50, σ =

a.

b.

c.

npq  100(.5)(.5) = 5

(48  .5)  50   P(x ≤ 48) = P  z    5 = P(z ≤ −.30) = .5 − .1179 = .3821 P(50 ≤ x ≤ 65) (65  .5)  50   (50  .5)  50 = P  z    5 5 = P(−.10 ≤ z ≤ 3.10) = .0398 + .5000 = .5398 (70  .5)  50   P(x ≥ 70) = P  z    5 = P(z ≥ 3.90) = .5 − .5 = 0

d.

P(55 ≤ x ≤ 58) (58  .5)  50   (55  .5)  50 = P  z    5 5 = P(.90 ≤ z ≤ 1.70) = P(0 ≤ z ≤ 1.70) − P(0 ≤ z ≤ .90) = .4554 − .3159 = .1395

e.

P(x = 62) (62  .5)  50   (62  .5)  50 = P  z    5 5 = P(2.30 ≤ z ≤ 2.50) = P(0 ≤ z ≤ 2.50) − (0 ≤ z ≤ 2.30) = .4938 − .4893 = .0045

f.

P(x ≤ 49 or x ≥ 72) (49  .5)  50    Pz    5 (72  .5)  50    Pz    5 = P(z ≤ −.10) + P(z ≥ 4.30) = (.5 − .0398) + (.5 − .5) = .4602

4.197

x is normal random variable with μ = 40, σ2 = 36, and σ = 6. a.

P(x ≥ x0) = .10 So, A = .5000 − .1000 = .4000. (See part a.) z0 = 1.28 To find x0, substitute the values into the z-score formula: x μ x  40 z0 = 0  1.28 = 0  x0 = 1.28(6) + 40 = 47.68 6 σ

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251

252

Chapter 4

b.

P(μ ≤ x ≤ x0) = .40 Look up the area .4000 in the body of Table IV, Appendix B; (take the closest value) z0 = 1.28. To find x0, substitute the values into the z-score formula: z0 =

c.

x0  μ

σ

 1.28 =

x0  40  x0 = 40 + 6(1.28) = 47.68 6

P(x < x0) = .05 So, A = .5000 − .0500 = .4500. Look up the area .4500 in the body of Table IV, Appendix B; z0 = −1.645. (.45 is halfway between .4495 and .4505; therefore, we average the z-scores 1.64  1.65 = 1.645 2 z0 is negative since the graph shows z0 is on the left side of 0. To find x0, substitute the values into the z-score formula: z0 =

d.

x0  μ

σ

 −1.645 =

x0  40  x0 = −1.645(6) + 40 = 30.13 6

P(x > x0) = .40 So, A = .5000 − .4000 = .1000. Look up the area .1000 in the body of Table IV, Appendix B; (take the closest value) z0 = .25. To find x0, substitute the values into the z-score formula: z0 =

e.

x0  μ

σ

 .25 =

x0  40  x0 = 40 + 6(.25) = 41.5 6

P(x0 ≤ x < μ) = .45 Look up the area .4500 in the body of Table IV, Appendix B; z0 = −1.645. (.45 is halfway between .4495 and .4505; therefore, we average the z-scores 1.64  1.65 = 1.645 2 z0 is negative since the graph shows z0 is on the left side of 0.

4.198

a.

To find x0, substitute the values into the z-score formula: x μ x  40 z0 = 0  −1.645 = 0  x0 = 40 − 6(1.645) = 30.13 6 σ First we must compute μ and σ. The probability distribution for x is:

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Random Variables and Probability Distributions

x 1 2 3 4

μ = E(x) = σ2 = E

253

p(x) .3 .2 .2 .3

 xp( x) = 1(.3) + 2(.2) + 3(.2) + 4(.3) = 2.5

 (x  μ)

2

=

 (x  μ )

2

p( x)

= (1 − 2.5) (.3) + (2 − 2.5)2(.2) + (3 − 2.5)2(.2)+ (4 − 2.52)(.3) = 1.45 σ 1.45 μ x = μ = 2.5, σ x =  = .1904 n 40 b.

4.199

2

By the Central Limit Theorem, the distribution of is approximately normal. The sample size, n = 40, is sufficiently large. Our answer does depend on n. If n is not sufficiently large, the Central Limit Theorem would not apply.

By the Central Limit Theorem, the sampling distribution of is approximately normal.

μ x = μ = 19.6, σ x 

a.

b.

c.

d.

3.2 68

= .388

19.6  19.6   P( x ≤ 19.6) = P  z   = P(z ≤ 0) = .5  .388 

(using Table IV, Appendix B)

19  19.6   P( x ≤ 19) = P  z   = P(z ≤ −1.55) = .5 − .4394 = .0606  .388  (using Table IV, Appendix B) 20.1  19.6   P( x ≥ 20.1) = P  z   = P(z ≥ 1.29) = .5 − .4015 = .0985  .388  (using Table IV, Appendix B) 20.6  19.6   19.2  19.6 P(19.2 ≤ x ≤ 20.6) = P  z   .388 .388  = P(−1.03 ≤ z ≤ 2.58) = .3485 + .4951 = .8436 (using Table IV, Appendix B)

4.200

Solution will vary. See page 1037 for Guided Solutions.

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254

4.201

Chapter 4

Given: μ = 100 and σ = 10 n

σ n

1

5

10

20

30

40

50

10 4.472 3.162 2.236 1.826 1.581 1.414

The graph of σ / n against n is given here:

4.202

a.

In order for x to be a binomial random variable, the n trials must be identical. We can assume that the process of selecting of a worker is identical from trial to trial. There are two possible outcomes a worker missed work due to a back injury or not. The probability of success must be the same from trial to trial. We can assume that the probability of missing work due to a back injury is constant. The trials must be independent of each other. We can assume that the outcome of one trials will not affect the outcome of any other. Thus, x is a binomial random variable.

b.

From the information given in the problem, the estimate of p is .40.

c.

The mean is μ = E(x) = np = 10(.40) = 4. The standard deviation is σ =

d.

np(1  p)  10(.40)(.60)  2.4 1 = 1.549

Using Table II, Appendix B, with n = 10 and p = .40, P(x = 1) = P(x ≤ 1) − P(x ≤ 0) = .046 − .006 = .040 P(x > 1) = 1 − P(x ≤ 1) = 1 − .046 = .954 6

4.203

a.

 p( x )  p(0)  p(1)  p(2)  p(3)  p(4)  p(5) i

i 1

 .0102  .0768  .2304  .3456  .2592  .0778  1.0000 b.

P(x = 4) = .2592

c.

P(x < 2) = P(x = 0) + P(x = 1) = .0102 + .0768 = .0870

d.

P(x ≥ 3) = P(x = 3) + P(x = 4) + P(x = 5) = .3456 + .2592 + .0778 = .6826

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Random Variables and Probability Distributions

255

6

e.

 x p( x )  0(.0102)  1(.0768)  2(.2304)  3(.3456)  4(.2592)  5(.0778)

μ  E ( x) 

i

i

i 1

 0  .0768  .4608  1.0368  1.0368  .3890  3.0002 On the average, 3 out of every 5 dentists will use nitrous oxide. 4.204

a.

Using Table IV, Appendix B, with μ = 8.72 and σ = 1.10, 6  8.72   P(x < 6) = P  z   = P(z < −2.47) = .5 − .4932 = .0068  1.10  Thus, approximately .68% of the games would result in fewer than 6 hits.

4.205

b.

The probability of observing fewer than 6 hits in a game is p = .0068. The probability of observing 0 hits would be even smaller. Thus, it would be extremely unusual to observe a no hitter.

a.

For this problem, c = 0 and d = 1. 1  1  (0  x  1)  f(x) =  d  c 1  0  0 otherwise

cd 0 1 = = .5 2 2 (d  c) 2 (1  0) 2 1 = = = .0833 σ2 = 12 12 12 σ = .0833 = .289

μ=

4.206

b.

P(.2 < x < .4) = (.4 − .2)(1) = .2

c.

P(x > .995) = (1 − .995)(1) = .005. Since the probability of observing a trajectory greater than .995 is so small, we would not expect to see a trajectory exceeding .995.

a.

In the problem, it is stated that E(x) = .03. This is also the value of λ.

σ2 = λ = .03 b.

c.

The experiment consists of counting the number of deaths or missing persons in a three- year interval. We must assume that the probability of a death or missing person in a three-year period is the same for any three-year period. We must also assume that the number of deaths or missing persons in any three-year period is independent of the number of deaths or missing persons in any other three-year period. λ 1e -λ .031e -.03 P(x = 1) =  = .0291 1! 1!

P(x = 0) =

λ 0e - λ 0!



.030e -.03 = .9704 0!

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256

4.207

Chapter 4

a.

μ x = μ = 89.34; σ x =

σ n

=

7.74 35

= 1.31

b.

c.

d.

4.208

a.

88  89.34   P( x > 88) = P  z   = P(z > −1.02) = .5 + .3461 = .8461  1.31  (using Table IV, Appendix B) 87  89.34   P( x < 87) = P  z   = P(z < −1.79) = .5 − .4633 = .0367  1.31  (using Table IV, Appendix B) Let x = number of trees infected with the Dutch elm disease in the two trees purchased. For this problem, x is a hypergeometric random variable with N = 10, n = 2, and r = 3. The probability that both trees will be healthy is:

 r   N  r   3  10  3  3! 7!  x   n  x   0   2  0  0!3! 2!5!  1(21) = .467   P(x = 0) = 10! 45 N 10   n   2  2!8! b.

The probability that at least one tree will be infected is: P(x ≥ 1) = 1 − P(x = 0) = 1 − .467 = .533.

4.209

Let x = interarrival time between patients. Then x is an exponential random variable with a mean of 4 minutes. a.

P(x < 1) = 1 − P(x ≥ 1) = 1 − e−1/4 = 1 − e−.25 = 1 − .778801 = .221199

b.

Assuming that the interarrival times are independent, P(next 4 interarrival times are all less than 1 minute) = {P(x < 1)}4 = .2211994 = .002394 P(x > 10) = e−10/4 = e−2.5 = .082085

c.

4.210

a.

We must assume that the trials are identical, the probability of success is constant from trial to trial, and the trials are independent of each other.

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Random Variables and Probability Distributions

b.

257

From the problem, we estimate p to be .20. Using Table II, Appendix B, with n = 25 and p = .20, P(x ≤ 10) = .994

c.

E(x) = np = 25(.20) = 5

σ=

np(1  p)  25(.20)(.80)  4 = 2

d.

μ ± 2σ  5 ± 2(2)  5 ± 4  (1, 9)

e.

Using Table II, Appendix B, with n = 25 and p = .20, P(1 < x < 9) = P(x ≤ 8) − P(x ≤ 1) = .953 − .027 = .926

4.211

a.

Let x1 = repair time for machine 1. Then x1 has an exponential distribution with μ1 = 1 hour. P(x1 > 1) = e−1/1 = e−1 = .367879 (using Table V, Appendix B)

b.

Let x2 = repair time for machine 2. Then x2 has an exponential distribution with μ2 = 2 hours. P(x2 > 1) = e−1/2 = e−.5 = .606531 (using Table V, Appendix B)

c.

Let x3 = repair time for machine 3. Then x3 has an exponential distribution with μ3 = .5 hours. P(x3 > 1) = e−1/.5 = e−2 = .135335 (using Table V, Appendix B) Since the mean repair time for machine 4 is the same as for machine 3, P(x4 > 1) = P(x3 > 1) = .135335.

d.

The only way that the repair time for the entire system will not exceed 1 hour is if all four machines are repaired in less than 1 hour. Thus, the probability that the repair time for the entire system exceeds 1 hour is: P(Repair time entire system exceeds 1 hour) = 1 − P((x1 ≤ 1) ∩ (x2 ≤ 1) ∩ (x3 ≤ 1) ∩ (x4 ≤ 1)) = 1 − P(x1 ≤ 1)P(x2 ≤ 1)P(x3 ≤ 1)P(x4 ≤ 1) = 1 − (1 − .367879)(1 − .606531)(1 − .135335)(1 − .135335) = 1 − (.632121)(.393469)(.864665)(.864665) = 1 − .185954 = .814046

4.212

Let x = demand for white bread. Then x is a normal random variable with μ = 7200 and σ = 300: a.

P(x ≤ x0) = .94. Find x0.  x  7200  P(x ≤ x0) = P  z  0   300  = P(z ≤ z0) = .94 A1 = .94 − .50 = .4400 Using Table IV and area .4400, z0 = 1.555. x 0  7200 x  7200  1.555  0  x0 = 7666.5 ≈ 7667 300 300 If the company produces 7,667 loaves, the company will be left with more than 500 loaves if the demand is less than 7,667 - 500 = 7167. 7167  7200   P(x < 7167) = P  z   = P(z < −.11)  300 = .5 − .0438 = .4562 (from Table IV, Appendix B) z0 =

b.

Thus, on 45.62% of the days the company will be left with more than 500 loaves.

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258

4.213

Chapter 4

a.

In order for the number of deaths to follow a Poisson distribution, we must assume that the probability of a death is the same for any week. We must also assume that the number of deaths in any week is independent of any other week. The first assumption may not be valid. The probability of a death may not be the same for every week. The number of passengers varies from week to week, so the probability of a death may change. Also, things such as weather, which varies from week to week may increase or decrease the chance of derailment.

b.

E(x) = λ = 20 σ = λ = 20 = 4.47

c.

The z-score corresponding to x = 4 is: 4  20 z= = −3.58 4.47 Since this z-score is more than 3 standard deviations from the mean, it would be very unlikely that only 4 or fewer deaths occur next week.

d.

Using Table III, Appendix B with λ = 20, P(x ≤ 4) = 0.000 This probability is consistent with the answer in part c. The probability of 4 or fewer deaths is essentially zero, which is very unlikely.

4.214

Let x = number of inches a gouge is from one end of the spindle. Then x has a uniform distribution with f(x) as follows:

1 1  1    f ( x)   d  c 18  0 18  0

0  x  18 otherwise

In order to get at least 14 consecutive inches without a gouge, the gouge must be within 4 inches of either end. Thus, we must find: P(x < 4) + P(x > 14) = (4 − 0)(1/18) + (18 − 14)(1/18) = 4/18 + 4/18 = 8/18 = .4444

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Random Variables and Probability Distributions

4.215

259

Using MINITAB, the stem-and-leaf display is: Stem-and-leaf of Time Leaf Unit = 0.10 (26) 23 15 9 4 2 2 1 1 1

1 2 3 4 5 6 7 8 9 10

N

= 49

00001122222344444445555679 11446799 002899 11125 24 8

1

The data are skewed to the right, and do not appear to be normally distributed. Using MINITAB, the descriptive statistics are: Variable Time Variable Time

N 49

Mean 2.549 Minimum 1.000

Median 1.700 Maximum 10.100

TrMean 2.333 Q1 1.350

StDev 1.828

SE Mean 0.261

Q3 3.500

x ± s  2.549 ± 1.828  (0.721, 4.377) x ± 2s  2.549 ± 2(1.828)  2.549 ± 3.656  (±1.107, 6.205) x ± 3s  2.549 ± 3(1.828)  2.549 ± 5.484  (−2.935, 8.033)

Of the 49 measurements, 44 are in the interval (0.721, 4.377). The proportion is 44/49 = .898. This is much larger than the proportion (.68) stated by the Empirical Rule. Of the 49 measurements, 47 are in the interval (−1.107, 6.205). The proportion is 47/49 = .959. This is close to the proportion (.95) stated by the Empirical Rule. Of the 49 measurements, 48 are in the interval (−2.935, 8.033). The proportion is 48/49 = .980. This is smaller than the proportion (1.00) stated by the Empirical Rule. This would imply that the data are not normal. IQR = QU − QL = 3.500 − 1.350 = 2.15. IQR/s = 2.15/1.828 = 1.176. If the data are normally distributed, this ratio should be close to 1.3. Since 1.176 is smaller than 1.3, this indicates that the data may not be normal.

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260

Chapter 4

Using MINITAB, the normal probability plot is:

Since this plot is not a straight line, the data are not normal. All four checks indicate that the data are not normal. 4.216

a.

By the Central Limit Theorem, the sampling distribution of is approximately normal with μ x = μ and σ x  σ / n .

b.

Let μ = 18.5. Since we do not know σ we will estimate it with s = 6.  19.1  18.5  P( x  19.1)  P  z   = P(z ≥ 1.85) = .5 − .4678 = .0322  6 / 344 

c.

Let μ = 19.5. Since we do not know σ we will estimate it with s = 6.  19.1  19.5  P( x  19.1)  P  z   = P(z ≥ −1.24) = .5 + .3925 = .8925  6 / 344 

d.

If P( x ≥ 19.1) = .5, then the population mean must be equal to 19.1. (For a normal distribution, half of the distribution is above the mean and half is below the mean.)

e.

If P( x ≥ 19.1) = .2, then the population mean is less than 19.1. We know the probability that is greater than the mean is .5. Since P(x ≥ 19.1) = .2 which is less than .5, we know that 19.1 must be to the right of the mean. Thus, the population mean must be less than 19.1.

Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.

Random Variables and Probability Distributions

4.217

261

Let x equal the difference between the actual weight and recorded weight (the error of measurement). The random variable x is normally distributed with μ = 592 and σ = 628. a.

We want to find the probability that the weigh-in-motion equipment understates the actual weight of the truck. This would be true if the error of measurement is positive.

0  592   P(x > 0) = P  z    628  = P(z > −.94) = .5000 + .3264 = .8264 b.

P(overstate the weight) = 1 − P(understate the weight) = 1 − .8264 = .1736 (Refer to part a.) For 100 measurements, approximately 100(.1736) = 17.36 or 17 times the weight would be overstated.

c.

d.

400  592   P(x > 400) = P  z    628  = P(z > −.31) = .5000 + .1217 = .6217 We want P(understate the weight) = .5 To understate the weight, x > 0. Thus, we want to find μ so that P(x > 0) = .5 0 μ   P(x > 0) = P  z   = .5  628 

μ

From Table IV, Appendix B, z0 = 0. To find μ, substitute into the z-score formula: x μ 0μ z0 = 0 0= μ=0 628 σ Thus, the mean error should be set at 0. We want P(understate the weight) = .4 To understate the weight, x > 0. Thus, we want to find μ so that P(x > 0) = .4

μ

A = .5 − .40 = .1. Look up the area .1000 in the body of Table IV, Appendix B, z0 = .25. To find μ, substitute into the z-score formula: z0 =

x0  μ

σ

 .25 =

0μ  μ = 0 − (.25)628 = −157 628

Thus, the mean error should be set at −157.

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262

4.218

Chapter 4

a.

Let p1 = probability of an error = 1/100 = .01 and p2 = probability of an error resulting in a significant problem = 1/500 = .002. Let x = number of errors in 60,000 trials. Then E(x) = μ1 = np1 = 60,000(.01) = 600. Let y = number of significant errors in 60,000 trials. Then E(y) = μ2 = np2 = 60,000(.002) = 120.

b.

σ = np2q2 = 60,000(.002)(.998) = 119.76 σ = 119.76 = 10.94 μ2 ± 3σ  120 ± 3(10.94)  120 ± 32.82  (87.18, 152.82)

Using Chebyshev's Rule, at least 88.9% of the observations will fall within 3 standard deviations of the mean. We would expect the number of significant errors to fall between 87 and 153.

4.219

c.

We must assume that the trials are independent and that the probability of a significant error is constant from trial to trial.

a.

μ = n ⋅ p = 25(.05) = 1.25 σ = npq  25(.05)(.95) = 1.09 Since μ is not an integer, x could not equal its mean.

b.

The event is (x ≥ 5). From Table II with n = 25 and p = .05: P(x ≥ 5) = 1 − P(x ≤ 4) = 1 − .993 = .007

4.220

c.

Since the probability obtained in part b is so small, it is unlikely that 5% applies to this agency. The percentage is probably greater than 5%.

a.

By the Central Limit Theorem, the sampling distribution of x is approximately normal since n > 30 and σ 15 σx  μ x = μ = 840  = 2.1213 n 50

b.

830  840   P( x ≤ 830) = P  z   = P(z ≤ −4.71) ≈ .5 − .5 = 0  2.1213 

c.

Since the probability of observing a mean of 830 or less is extremely small (≈0) if the true mean is 840, we would tend to believe that the mean is not 840, but something less.

d.

By the Central Limit Theorem, the sampling distribution of is approximately normal since n > 30 and σ 45 σx  μ x = μ = 840  = 6.3640 n 50 830  840   P( x ≤ 830) = P  z   = P(z ≤ −1.57) ≈ .5 − .4418 = .0582  6.3640 

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Random Variables and Probability Distributions

4.221

263

Let x = number of grants awarded to the north side in 140 trials. The random variable x has a hypergeometric distribution with N = 743, n = 140, and r = 601. a.

μ = E(x) =

nr 140(601)  = 113.24 N 743

r ( N  r ) n( N  n)

σ2 =

N ( N  1) 2



601(743  601)140(743  140) 7432 (743  1)

= 17.5884

σ = 17.5884 = 4.194 b.

If the grants were awarded at random, we would expect approximately 113 to be awarded to the north side. We observed 140. The z-score associated with 140 is: z=

xμ

σ



140  113.24 = 6.38 4.194

Because this z-score is so large, it would be extremely unlikely to observe all 140 grants to the north side if they are randomly selected. Thus, we would conclude that the grants were not randomly selected. 4.222

Let x = length of time a bus is late. Then x is a uniform random variable with probability distribution: 1 (0  x  20)  f(x) =  20  0 otherwise

μ=

b.

 1  1 P(x ≥ 19) = (20 − 19) ⋅    = .05  20  20

c.

4.223

0  20 = 10 2

a.

It would be doubtful that the director’s claim is true, since the probability of the being more than 19 minutes late is so small.

We know from the Empirical Rule that almost all the observations are larger than μ − 2σ. (≈ 95% are between μ − 2σ and μ + 2σ). Thus μ − 2σ > 100. For the binomial, μ = np = n(.4) and σ =

npq  n(.4)(.6) =

.24n

μ − 2σ > 100  .4n − 2 .24n > 100  .4n − .98 n − 100 > 0 Solving for n

n , we get: .98  .98 2  4(.4)(100) .98  12.687  2(.4) .8 

n = 17.084  n = 17.0842 = 291.9 ≈ 292

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264

4.224

Chapter 4

Let x = tensile strength of a particular metal part. Then x is a normal random variable with mean = μ = 25 and standard deviation = σ = 2. The tolerance limits are 21 and 30. 21  25   P( x  21)  P  z    P( z  2)  .5  .4772  .0228 (Using Table IV, Appendix B).  2  30  25   21  25 P(21  x  30)  P  z   P (2  z  2.5)  .4772  .4938  .9710  2 2  30  25   P( x  30)  P  z    P ( z  2.5)  .5  .4938  .0062  2  E(Profit) = −$2(.0228) + $10(.9710) − $1(.0062) = −$.0456 + $9.71 − $.0062 = $9.66

4.225

Even though the number of flaws per piece of siding has a Poisson distribution, the Central Limit Theorem implies that the distribution of the sample mean will be approximately normal with μ x = μ = 2.5 and

σx 

σ n



2.5 35

 .2673 . Therefore,

 2.1  2.5  P( x > 2.1) + P  z   = P(z > −1.50) = .5 + .4332 = .9332 (using Table IV, Appendix B)  2.5 / 35  4.226

Let x = number of packets out of 4 that contain cocaine. Then x is an approximate binomial random variable with n = 4 and p= 331/496 = .667.  4 P( x  4)    .667 4 (.333)4  4  .6674  .197  4 Let y = number of packets out of 2 not containing cocaine, given the first 4 contained cocaine. Then y is an approximate binomial random variable with n = 2 and p = 165/492 = .335.  2 P( y  2)    .3352 (.665)2  2  .3352  .112  2 The probability that the first 4 packets contained cocaine and the next 2 did not is: P( x  4) P ( y  2)  .197(.112)  .022 Thus, it is very unlikely that the first 4 packets contained cocaine and the last 2 did not.

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Random Variables and Probability Distributions

4.227

265

Using MINITAB, the descriptive statistics are: Descriptive Statistics: INSOMNIA Variable INSOMNIA

N 40

Mean 5.935

StDev 5.392

Minimum 1.300

Q1 2.250

Median 3.650

Q3 7.675

Maximum 22.800

To see if the mean time to fall asleep for those taking melatonin is different from those taking a placebo, we will see how unusual it is to observe a sample mean of 5.935 if the true mean is 15.   5.935  15   P( x  5.935)  P z   P ( z  10.63)  .5  .5  0 (Using Table IV, Appendix B.)   5.392  40 

Since this probability is so small, we would infer that the mean time to fall asleep for those taking melatonin is not 15 minutes, but something less than 15 minutes.

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