Chapter 5 Discrete Probability Distributions Learning Objectives 1.

Understand the concepts of a random variable and a probability distribution.

2.

Be able to distinguish between discrete and continuous random variables.

3.

Be able to compute and interpret the expected value, variance, and standard deviation for a discrete random variable.

4.

Be able to compute and work with probabilities involving a binomial probability distribution.

5.

Be able to compute and work with probabilities involving a Poisson probability distribution.

6.

Know when and how to use the hypergeometric probability distribution.

5-1

Chapter 5

Solutions: 1.

a.

Head, Head (H,H) Head, Tail (H,T) Tail, Head (T,H) Tail, Tail (T,T)

b.

x = number of heads on two coin tosses

c. Outcome (H,H) (H,T) (T,H) (T,T)

2.

Values of x 2 1 1 0

d.

Discrete. It may assume 3 values: 0, 1, and 2.

a.

Let x = time (in minutes) to assemble the product.

b.

It may assume any positive value: x > 0.

c.

Continuous

3.

Let Y = position is offered N = position is not offered a.

S = {(Y,Y,Y), (Y,Y,N), (Y,N,Y), (Y,N,N), (N,Y,Y), (N,Y,N), (N,N,Y), (N,N,N)}

b.

Let N = number of offers made; N is a discrete random variable.

c. Experimental Outcome Value of N 4. 5.

(Y,Y,Y) (Y,Y,N) (Y,N,Y) (Y,N,N) (N,Y,Y) (N,Y,N) (N,N,Y) (N,N,N) 3 2 2 1 2 1 1 0

x = 0, 1, 2, . . ., 12. a.

S = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}

b. Experimental Outcome Number of Steps Required 6.

a.

values: 0,1,2,...,20 discrete

b.

values: 0,1,2,... discrete

c.

values: 0,1,2,...,50 discrete values: 0 ≤ x ≤ 8

d.

(1,1) 2

(1,2) 3

5-2

(1,3) 4

(2,1) 3

(2,2) 4

(2,3) 5

Discrete Probability Distributions continuous

7.

e.

values: x > 0 continuous

a.

f (x) ≥ 0 for all values of x. Σ f (x) = 1 Therefore, it is a proper probability distribution.

8.

b.

Probability x = 30 is f (30) = .25

c.

Probability x ≤ 25 is f (20) + f (25) = .20 + .15 = .35

d.

Probability x > 30 is f (35) = .40

a. x 1 2 3 4

f (x) 3/20 = .15 5/20 = .25 8/20 = .40 4/20 = .20 Total 1.00

b. f (x) .4 .3 .2 .1 x 1

c.

2

f (x) ≥ 0 for x = 1,2,3,4. Σ f (x) = 1

9.

a. Age

Number of Children

f(x)

5-3

3

4

Chapter 5 6 7 8 9 10 11 12 13 14

37,369 87,436 160,840 239,719 286,719 306,533 310,787 302,604 289,168 2,021,175

0.018 0.043 0.080 0.119 0.142 0.152 0.154 0.150 0.143 1.001

b. f(x) .16 .14 .12 .10 .08 .06 .04 .02

x 6

c.

7

8

9

10 11 12 13 14

f(x) ≥ 0 for every x Σ f(x) = 1 Note: Σ f(x) = 1.001 in part (a); difference from 1 is due to rounding values of f(x).

10. a. x 1 2 3 4 5

f(x) 0.05 0.09 0.03 0.42 0.41 1.00

x

f(x)

b.

5-4

Discrete Probability Distributions 1 2 3 4 5

0.04 0.10 0.12 0.46 0.28 1.00

c.

P(4 or 5) = f (4) + f (5) = 0.42 + 0.41 = 0.83

d.

Probability of very satisfied: 0.28

e.

Senior executives appear to be more satisfied than middle managers. 83% of senior executives have a score of 4 or 5 with 41% reporting a 5. Only 28% of middle managers report being very satisfied.

11. a. Duration of Call x 1 2 3 4

f(x) 0.25 0.25 0.25 0.25 1.00

b. f (x) 0.30 0.20

0.10 x 0

1

2

3

4

c.

f (x) ≥ 0 and f (1) + f (2) + f (3) + f (4) = 0.25 + 0.25 + 0.25 + 0.25 = 1.00

d.

f (3) = 0.25

e.

P(overtime) = f (3) + f (4) = 0.25 + 0.25 = 0.50

12. a. b.

Yes; f (x) ≥ 0 for all x and Σ f (x) = .15 + .20 + .30 + .25 + .10 = 1 P(1200 or less) = f (1000) + f (1100) + f (1200) = .15 + .20 + .30 = .65

13. a. b.

Yes, since f (x) ≥ 0 for x = 1,2,3 and Σ f (x) = f (1) + f (2) + f (3) = 1/6 + 2/6 + 3/6 = 1 f (2) = 2/6 = .333

5-5

Chapter 5

c. 14. a.

f (2) + f (3) = 2/6 + 3/6 = .833 f (200) = 1 - f (-100) - f (0) - f (50) - f (100) - f (150) = 1 - .95 = .05 This is the probability MRA will have a $200,000 profit.

b.

P(Profit) = f (50) + f (100) + f (150) + f (200) = .30 + .25 + .10 + .05 = .70

c.

P(at least 100) = f (100) + f (150) + f (200) = .25 + .10 +.05 = .40

15. a. x 3 6 9

f (x) .25 .50 .25 1.00

x f (x) .75 3.00 2.25 6.00

E (x) = µ = 6.00 b. x 3 6 9

x-µ -3 0 3

(x - µ)2 9 0 9

(x - µ)2 f (x) 2.25 0.00 2.25 4.50

f (x) .25 .50 .25

Var (x) = σ2 = 4.50 c.

σ =

4.50 = 2.12

16. a. y 2 4 7 8 E(y) = µ = 5.20

b. y 2

y-µ -3.20

f (y) .20 .30 .40 .10 1.00

(y - µ)2 10.24

y f (y) .40 1.20 2.80 .80 5.20

f (y) .20

5-6

(y - µ)2 f (y) 2.048

Discrete Probability Distributions 4 7 8

-1.20 1.80 2.80

1.44 3.24 7.84

.30 .40 .10

.432 1.296 .784 4.560

Var ( y ) = 4.56

σ = 4.56 = 2.14 17. a/b. x 0 1 2 3 4 5

f (x) .10 .15 .30 .20 .15 .10

x f (x) .00 .15 .60 .60 .60 .50 2.45

x-µ -2.45 -1.45 - .45 .55 1.55 2.55

(x - µ)2 6.0025 2.1025 .2025 .3025 2.4025 6.5025

(x - µ)2 f (x) .600250 .315375 .060750 .060500 .360375 .650250 2.047500

E (x) = µ = 2.45 = 2.0475 = 1.4309

σ2 σ 18. a/b.

x 0 1 2 3 4 Total

f (x) 0.04 0.34 0.41 0.18 0.04 1.00

xf (x) 0.00 0.34 0.82 0.53 0.15 1.84 ↑ E(x)

x-µ -1.84 -0.84 0.16 1.16 2.16

(x - µ)2 3.39 0.71 0.02 1.34 4.66

(x - µ)2 f (x) 0.12 0.24 0.01 0.24 0.17 0.79 ↑ Var(x)

y 0 1 2 3 4 Total

f (y) 0.00 0.03 0.23 0.52 0.22 1.00

yf (y) 0.00 0.03 0.45 1.55 0.90 2.93 ↑ E(y)

y-µ -2.93 -1.93 -0.93 0.07 1.07

(y - µ)2 8.58 3.72 0.86 0.01 1.15

(y - µ)2 f (y) 0.01 0.12 0.20 0.00 0.26 0.59 ↑ Var(y)

c/d.

e.

The number of bedrooms in owner-occupied houses is greater than in renter-occupied houses. The expected number of bedrooms is 1.09 = 2.93 - 1.84 greater. And, the variability in the number of bedrooms is less for the owner-occupied houses.

19. a.

E (x) = Σ x f (x) = 0 (.50) + 2 (.50) = 1.00

b.

E (x) = Σ x f (x) = 0 (.61) + 3 (.39) = 1.17

5-7

Chapter 5

c.

The expected value of a 3 - point shot is higher. So, if these probabilities hold up, the team will make more points in the long run with the 3 - point shot.

20. a. x 0 400 1000 2000 4000 6000

f (x) .90 .04 .03 .01 .01 .01 1.00

x f (x) 0.00 16.00 30.00 20.00 40.00 60.00 166.00

E (x) = 166. If the company charged a premium of $166.00 they would break even. b. Gain to Policy Holder -260.00 140.00 740.00 1,740.00 3,740.00 5,740.00

f (Gain) .90 .04 .03 .01 .01 .01

(Gain) f (Gain) -234.00 5.60 22.20 17.40 37.40 57.40 -94.00

E (gain) = -94.00. The policy holder is more concerned that the big accident will break him than with the expected annual loss of $94.00. 21. a.

E (x) = Σ x f (x) = 0.05(1) + 0.09(2) + 0.03(3) + 0.42(4) + 0.41(5) = 4.05

b.

E (x) = Σ x f (x) = 0.04(1) + 0.10(2) + 0.12(3) + 0.46(4) + 0.28(5) = 3.84

c.

Executives:

σ2 = Σ (x - µ)2 f(x) = 1.2475

Middle Managers: σ2 = Σ (x - µ)2 f(x) = 1.1344 d.

Executives:

σ = 1.1169

Middle Managers: σ = 1.0651 e. 22. a.

The senior executives have a higher average score: 4.05 vs. 3.84 for the middle managers. The executives also have a slightly higher standard deviation. E (x) = Σ x f (x) = 300 (.20) + 400 (.30) + 500 (.35) + 600 (.15) = 445 The monthly order quantity should be 445 units.

b. 23. a.

Cost: 445 @ $50 = $22,250 Revenue: 300 @ $70 = 21,000 $ 1,250 Loss Laptop: E (x) = .47(0) + .45(1) + .06(2) + .02(3) = .63 Desktop: E (x) = .06(0) + .56(1) + .28(2) + .10(3) = 1.42

5-8

Discrete Probability Distributions

b.

Laptop: Var (x) = .47(-.63)2 + .45(.37)2 + .06(1.37)2 + .02(2.37)2 = .4731 Desktop: Var (x) = .06(-1.42)2 + .56(-.42)2 + .28(.58)2 + .10(1.58)2 = .5636

c.

24. a.

From the expected values in part (a), it is clear that the typical subscriber has more desktop computers than laptops. There is not much difference in the variances for the two types of computers. Medium E (x) = Σ x f (x) = 50 (.20) + 150 (.50) + 200 (.30) = 145 Large: E (x) = Σ x f (x) = 0 (.20) + 100 (.50) + 300 (.30) = 140 Medium preferred.

b.

Medium x 50 150 200

f (x) .20 .50 .30

Large y 0 100 300

f (y) .20 .50 .30

x-µ -95 5 55

(x - µ)2 9025 25 3025

y-µ -140 -40 160

(x - µ)2 f (x) 1805.0 12.5 907.5 σ2 = 2725.0

(y - µ)2 19600 1600 25600

(y - µ)2 f (y) 3920 800 7680 σ2 = 12,400

Medium preferred due to less variance. 25. a. S F

S

F

S F

b.

 2 2! f (1) =   (.4)1 (.6)1 = (.4)(.6) = .48 1!1! 1

c.

 2 2! f (0) =   (.4)0 (.6) 2 = (1)(.36) = .36 0!2! 0

5-9

Chapter 5

d.

 2 2! f (2) =   (.4) 2 (.6)0 = (.16)(1) = .16 2!0!  2

e.

P (x ≥ 1) = f (1) + f (2) = .48 + .16 = .64

f.

E (x) = n p = 2 (.4) = .8 Var (x) = n p (1 - p) = 2 (.4) (.6) = .48

σ = .48 = .6928 26. a.

f (0) = .3487

b.

f (2) = .1937

c.

P(x ≤ 2) = f (0) + f (1) + f (2) = .3487 + .3874 + .1937 = .9298

d.

P(x ≥ 1) = 1 - f (0) = 1 - .3487 = .6513

e.

E (x) = n p = 10 (.1) = 1

f.

Var (x) = n p (1 - p) = 10 (.1) (.9) = .9

σ = .9 = .9487 27. a.

f (12) = .1144

b.

f (16) = .1304

c.

P (x ≥ 16) = f (16) + f (17) + f (18) + f (19) + f (20) = .1304 + .0716 + .0278 + .0068 + .0008 = .2374

d.

P (x ≤ 15) = 1 - P (x ≥ 16) = 1 - .2374 = .7626

e.

E (x) = n p = 20(.7) = 14

f.

Var (x) = n p (1 - p) = 20 (.7) (.3) = 4.2

σ =

28. a. b.

4.2 = 2.0494

6 f (2) =   (.33) 2 (.67) 4 = .3292  2

P(at least 2) = 1 - f(0) - f(1)

5 - 10

Discrete Probability Distributions

6 6 = 1 −   (.33)0 (.67)6 −   (.33)1 (.67)5 0 1

= c. 29.

1 - .0905 - .2673 = .6422

 10  f (0) =   (.33)0 (.67)10 = .0182 0

P(At Least 5) = 1 - f (0) - f (1) - f (2) - f (3) - f (4) = 1 - .0000 - .0005 - .0031 - .0123 - .0350 = .9491

30. a. b.

Probability of a defective part being produced must be .03 for each part selected; parts must be selected independently. Let: D = defective G = not defective 1st part

2nd part D

Number Defective

(D, D)

2

(D, G)

1

(G, D)

1

(G, G)

0

G

D

G

Experimental Outcome

D

.

G

c.

2 outcomes result in exactly one defect.

d.

P (no defects) = (.97) (.97) = .9409 P (1 defect) = 2 (.03) (.97) = .0582 P (2 defects) = (.03) (.03) = .0009

31.

Binomial n = 10 and p = .09 f ( x) =

10! (.09) x (.91)10 − x x !(10 − x)!

a.

Yes. Since they are selected randomly, p is the same from trial to trial and the trials are independent.

b. c.

f (2) = .1714 f (0) = .3894

d.

1 - f (0) - f (1) - f (2) = 1 - (.3894 + .3851 + .1714) = .0541

32. a.

.90

5 - 11

Chapter 5

b.

P (at least 1) = f (1) + f (2) f (1) =

2! (.9)1 (.1)1 1! 1!

= 2 (.9) (.1) = .18 f (2) =

2 0 2! (.9) (.1) 2! 0!

= 1 (.81) (1) = .81 ∴ P (at least 1) = .18 + .81 = .99

Alternatively P (at least 1) = 1 - f (0) f (0) =

0 2 2! (.9) (.1) = .01 0! 2!

Therefore, P (at least 1) = 1 - .01 = .99

c.

P (at least 1) = 1 - f (0) f (0) =

3! (.9)0 (.1)3 = .001 0! 3!

Therefore, P (at least 1) = 1 - .001 = .999

d.

33. a.

Yes; P (at least 1) becomes very close to 1 with multiple systems and the inability to detect an attack would be catastrophic. f(12) =

20! (.5)12 (.5)8 12!8!

Using the binomial tables, f(12) = .0708 b.

f(0) + f(1) + f(2) + f(3) + f(4) + f(5) .0000 + .0000 + .0002 + .0011 + .0046 + .0148 = .0207

c.

E(x) = np = 20(.5) = 10

d.

Var (x) = σ2 = np(1 - p) = 20(.5)(.5) = 5

σ = 5 = 2.24

34. a. b.

f (3) = .0634 (from tables) The answer here is the same as part (a). The probability of 12 failures with p = .60 is the same as the probability of 3 successes with p = .40.

5 - 12

Discrete Probability Distributions c.

35. a.

f (3) + f (4) + · · · + f (15) = 1 - f (0) - f (1) - f (2) = 1 - .0005 - .0047 - .0219 = .9729 f (0) + f (1) + f (2) = .0115 + .0576 + .1369 = .2060

b.

f (4) = .2182

c.

1 - [ f (0) + f (1) + f (2) + f (3) ]

d.

µ = n p = 20 (.20) = 4

36. x 0 1 2 3

37.

f (x) .343 .441 .189 .027 1.000

= 1 - .2060 - .2054 = .5886

x-µ -.9 .1 1.1 2.1

(x - µ)2 .81 .01 1.21 4.41

(x - µ)2 f (x) .27783 .00441 .22869 .11907 σ2 = .63000

E(x) = n p = 30(.72) = 21.6 Var(x) = n p (1 - p) = 30(.72)(.28) = 6.048

σ =

6.048 = 2.46

38. a.

f ( x) =

3 x e −3 x!

b.

f (2) =

32 e −3 9(.0498) = = .2241 2! 2

c.

f (1) =

31 e −3 = 3(.0498) = .1494 1!

d. 39. a. b.

P (x ≥ 2) = 1 - f (0) - f (1) = 1 - .0498 - .1494 = .8008 f ( x) =

2 x e −2 x!

µ = 6 for 3 time periods

c.

f ( x) =

6 x e −6 x!

d.

f (2) =

22 e −2 4(.1353) = = .2706 2! 2

e.

f (6) =

6 6 e −6 = .1606 6!

5 - 13

Chapter 5

f. 40. a.

f (5) =

45 e −4 = .1563 5!

µ = 48 (5/60) = 4 3

b.

-4

4 e 3!

f (3) =

(64) (.0183) = .1952 6

µ = 48 (15 / 60) = 12 10

f (10) =

c.

=

-12

12 e 10 !

= .1048

µ = 48 (5 / 60) = 4 I expect 4 callers to be waiting after 5 minutes. 0

f (0) =

-4

4 e 0!

= .0183

The probability none will be waiting after 5 minutes is .0183. d.

µ = 48 (3 / 60) = 2.4 0

f (0) =

2.4 e 0!

-2.4

= .0907

The probability of no interruptions in 3 minutes is .0907. 41. a. b.

30 per hour

µ = 1 (5/2) = 5/2 f (3) =

(5 / 2)3 e − (5 / 2) = .2138 3!

c.

f (0) =

(5 / 2)0 e − (5 / 2) = e − (5 / 2) = .0821 0!

42. a.

f (0) =

70 e −7 = e−7 = .0009 0!

b.

probability = 1 - [f(0) + f(1)] f (1) =

71 e −7 = 7e −7 = .0064 1!

5 - 14

Discrete Probability Distributions probability = 1 - [.0009 + .0064] = .9927 c.

µ = 3.5 f (0) =

3.50 e −3.5 = e −3.5 = .0302 0!

probability = 1 - f(0) = 1 - .0302 = .9698 d.

probability = 1 - [f(0) + f(1) + f(2) + f(3) + f(4)] = 1 - [.0009 + .0064 + .0223 + .0521 + .0912] = .8271 Note: The Poisson tables were used to compute the Poisson probabilities f(0), f(1), f(2), f(3) and f(4) in part (d).

43. a. b.

f (0) =

100 e −10 = e −10 = .000045 0!

f (0) + f (1) + f (2) + f (3) f (0) = .000045 (part a) f (1) =

101 e−10 = .00045 1!

Similarly, f (2) = .00225, f (3) = .0075 and f (0) + f (1) + f (2) + f (3) = .010245 c.

2.5 arrivals / 15 sec. period Use µ = 2.5 f (0) =

d. 44.

2.50 e−2.5 = .0821 0!

1 - f (0) = 1 - .0821 = .9179 Poisson distribution applies

a. b. c. d. 45. a.

µ = 1.25 per month 1.250 e−1.25 = 0.2865 0! 1.251 e−1.25 f (1) = = 0.3581 1!

f (0) =

P (More than 1) = 1 - f (0) - f (1) = 1 - 0.2865 - 0.3581 = 0.3554 average per month =

18 = 1.5 12

5 - 15

Chapter 5

b. c.

f (0) =

1.50 e −1.5 = e−1.5 = .2231 0!

probability = 1 - [f(0) + f(1)] = 1 - [.2231 + .3347] = .4422

46. a.

 3   10 − 3   3!   7!     1   4 − 1   1!2!   3!4!  (3)(35)  = = = .50 f (1) = 10! 210 10    4!6! 4

b.

 3  10 − 3     2 2 − 2  (3)(1) f (2) =    = = .067 45  10    2

c.

 3  10 − 3     0 2 − 0  (1)(21) f (0) =   = = .4667 45 10    2

d.

 3  10 − 3     2 4 − 2  (3)(21) f (2) =    = = .30 210  10    4

47.

 4   15 − 4     3 10 − 3  (4)(330) f (3) =    = = .4396 3003  15     10 

48.

Hypergeometric with N = 10 and r = 6

a.

b.

 6  4     2 1 (15)(4) = .50 f (2) =    = 120  10    3

Must be 0 or 1 prefer Coke Classic. 6 4    1 2 (6)(6) = .30 f (1) =     = 120  10    3

5 - 16

Discrete Probability Distributions

 6  4     0 3 (1)(4) = .0333 f (0) =    = 120  10    3

P (Majority Pepsi) = f (1) + f (0) = .3333 49.

Parts a, b & c involve the hypergeometric distribution with N = 52 and n = 2 a.

r = 20, x = 2  20   32     2 0 (190)(1) = .1433 f (2) =     = 52 1326     2

b.

r = 4, x = 2  4  48     2 0 (6)(1) = .0045 f (2) =    = 52 1326     2

c.

r = 16, x = 2  16   36     2 0 (120)(1) = .0905 f (2) =     = 52 1326     2

d.

Part (a) provides the probability of blackjack plus the probability of 2 aces plus the probability of two 10s. To find the probability of blackjack we subtract the probabilities in (b) and (c) from the probability in (a). P (blackjack) = .1433 - .0045 - .0905 = .0483

50.

N = 60 n = 10 a.

r = 20 x = 0

5 - 17

Chapter 5

f (0)

=

FG 20IJ FG 40IJ b1gFG 40! IJ H 0 K H 10K = H 10!30!K = FG 40! IJ FG 10!50!IJ 60! H 10!30!K H 60! K FG 60IJ 10!50! H 10K

=

40 ⋅ 39 ⋅ 38 ⋅ 37 ⋅ 36 ⋅ 35 ⋅ 34 ⋅ 33 ⋅ 32 ⋅ 31 60 ⋅ 59 ⋅ 58 ⋅ 57 ⋅ 56 ⋅ 55 ⋅ 54 ⋅ 53 ⋅ 52 ⋅ 51

≈ .01

b.

r = 20 x = 1

f (1)

=

FG 20IJ FG 40IJ H 1 K H 9 K = 20FG 40! IJ FG 10!50!IJ H 9!31!K H 60! K FG 60IJ H 10K

≈ .07

c.

1 - f (0) - f (1) = 1 - .08 = .92

d.

Same as the probability one will be from Hawaii. In part b that was found to equal approximately .07.

51. a.

 11 14     2 3 (55)(364) f (2) =     = = .3768 53,130  25    5

b.

 14   11    2 3 (91)(165) f (2) =     = = .2826 53,130  25    5

c.

 14   11    5 0 (2002)(1) f (5) =     = = .0377 25 53,130     5

d.

52.

 14   11    0 5 (1)(462) f (0) =     = = .0087 53,130  25    5 Hypergeometric with N = 10 and r = 2.

Focus on the probability of 0 defectives, then the probability of rejecting the shipment is 1 - f (0).

5 - 18

Discrete Probability Distributions a.

n = 3, x = 0  28    0 3 56 f (0) =     = = .4667 120 10    3

P (Reject) = 1 - .4667 = .5333 b.

n = 4, x = 0  28    0 4 70 f (0) =     = = .3333 10 210     4

P (Reject) = 1 - .3333 = .6667 c.

n = 5, x = 0  28    0 5 56 f (0) =     = = .2222 252 10    5

P (Reject) = 1 - .2222 = .7778 d.

Continue the process. n = 7 would be required with the probability of rejecting = .9333

53. a/b/c. x 1 2 3 4 5 Total

f (x) 0.07 0.21 0.29 0.39 0.04 1.00

xf (x) 0.07 0.42 0.87 1.56 0.20 3.12 ↑ E(x)

x-µ -2.12 -1.12 -0.12 0.88 1.88

(x - µ)2 4.49 1.25 0.01 0.77 3.53

(x - µ)2 f (x) 0.31 0.26 0.00 0.30 0.14 1.03 ↑ Var(x)

σ = 1.03 = 1.01 d.

The expected level of optimism is 3.12. This is a bit above neutral and indicates that investment managers are somewhat optimistic. Their attitudes are centered between neutral and bullish with the consensus being closer to neutral.

54. a/b. x 1 2

f (x) 0.24 0.21

xf (x) 0.24 0.41

x-µ -2.00 -1.00

5 - 19

(x - µ)2 4.00 1.00

(x - µ)2 f (x) 0.97 0.21

Chapter 5 3 4 5 Total

c.

0.10 0.21 0.24 1.00

0.31 0.83 1.21 3.00 ↑ E(x)

0.00 1.00 2.00

For the bond fund categories: E (x) = 1.36 For the stock fund categories: E (x) = 4

0.00 1.00 4.00

0.00 0.21 0.97 2.34 ↑ Var(x)

Var (x) = .23

Var (x) = 1.00

The total risk of the stock funds is much higher than for the bond funds. It makes sense to analyze these separately. When you do the variances for both groups (stocks and bonds), they are reduced. 55. a. x 9 10 11 12 13 b.

E (x)

f (x) .30 .20 .25 .05 .20

= Σ x f (x) = 9 (.30) + 10 (.20) + 11 (.25) + 12 (.05) + 13 (.20) = 10.65

Expected value of expenses: $10.65 million c.

Var (x) = Σ (x - µ)2 f (x) = (9 - 10.65)2 (.30) + (10 - 10.65)2 (.20) + (11 - 10.65)2 (.25) + (12 - 10.65)2 (.05) + (13 - 10.65)2 (.20) = 2.1275

d.

Looks Good: E (Profit) = 12 - 10.65 = 1.35 million However, there is a .20 probability that expenses will equal $13 million and the college will run a deficit.

56. a.

n = 20

and

x = 3

 20  f (3) =   (0.04)3 (0.04)17 = 0.0364 3

b.

n = 20

and

x = 0

 20  f (0) =   (0.04) 0 (0.96) 20 = 0.4420 0

c.

E (x) = n p = 1200 (0.04) = 48 The expected number of appeals is 48.

5 - 20

Discrete Probability Distributions

d.

σ2 = np (1 - p) = 1200 (0.04)(0.96) = 46.08 σ=

57. a.

46.08 = 6.7882

We must have E(x) = np ≥ 10 With p = .4, this leads to: n(.4) ≥ 10 n ≥ 25

b.

With p = .12, this leads to: n(.12) ≥ 10 n ≥ 83.33 So, we must contact 84 people in this age group to have an expected number of internet users of at least 10.

c.

σ = 25(.4)(.6) = 2.45

d.

σ = 84(.12)(.88) = 2.97

58.

Since the shipment is large we can assume that the probabilities do not change from trial to trial and use the binomial probability distribution. a.

n = 5 5 f (0) =   (0.01)0 (0.99)5 = 0.9510 0

b.

 5 f (1) =   (0.01)1 (0.99) 4 = 0.0480 1

c.

1 - f (0) = 1 - .9510 = .0490

d.

No, the probability of finding one or more items in the sample defective when only 1% of the items in the population are defective is small (only .0490). I would consider it likely that more than 1% of the items are defective.

59. a. b.

E(x) = np = 100(.041) = 4.1 Var (x) = np(1 - p) = 100(.041)(.959) = 3.9319

σ = 3.9319 = 1.9829 60. a.

E(x) = 800(.41) = 328

b.

σ = np(1 − p ) = 800(.41)(.59) = 13.91

c.

For this one p = .59 and (1-p) = .41, but the answer is the same as in part (b). For a binomial probability distribution, the variance for the number of successes is the same as the variance for the number of failures. Of course, this also holds true for the standard deviation.

5 - 21

Chapter 5

µ = 15

61.

prob of 20 or more arrivals = f (20) + f (21) + · · · = .0418 + .0299 + .0204 + .0133 + .0083 + .0050 + .0029 + .0016 + .0009 + .0004 + .0002 + .0001 + .0001 = .1249

µ = 1.5

62.

prob of 3 or more breakdowns is 1 - [ f (0) + f (1) + f (2) ]. 1 - [ f (0) + f (1) + f (2) ] = 1 - [ .2231 + .3347 + .2510] = 1 - .8088 = .1912

µ = 10 f (4) = .0189

63. 64. a. b.

f (3) =

3 3 e −3 = 0.2240 3!

f (3) + f (4) + · · · = 1 - [ f (0) + f (1) + f (2) ] 0

f (0) =

3 e 0!

-3

= e

-3

= .0498

Similarly, f (1) = .1494, f (2) = .2240 ∴ 1 - [ .0498 + .1494 + .2241 ] = .5767 65.

Hypergeometric N = 52, n = 5 and r = 4.

a.

FG 4IJ FG 48IJ H 2K H 3 K = 6(17296) =.0399 FG 52IJ 2,598,960 H 5K

b.

FG 4IJ FG 48IJ H 1K H 4 K = 4(194580) =.2995 FG 52IJ 2,598,960 H 5K

c.

FG 4IJ FG 48IJ H 0K H 5 K = 1,712,304 =.6588 FG 52IJ 2,598,960 H 5K

d.

1 - f (0) = 1 - .6588 = .3412

5 - 22

Discrete Probability Distributions

66. a.

 7  3     1 1 (7)(3) f (1) =    = = .4667 45  10    2

b.

 7  3     2 0 (21)(1) f (2) =    = = .4667 45 10    2

c.

7 3    0 2 (1)(3) f (0) =     = = .0667 45  10    2

5 - 23