Chapter 6: Random Variables and the Normal Distribution. 6.1 Discrete Random Variables. 6.2 Binomial Probability Distribution

Chapter 6: Random Variables and the Normal Distribution  6.1 Discrete Random Variables  6.2 Binomial Probability Distribution  6.3 Continuous ...
46 downloads 0 Views 5MB Size
Chapter 6: Random Variables and the Normal Distribution 

6.1 Discrete Random Variables



6.2 Binomial Probability Distribution



6.3 Continuous Random Variables and the Normal Probability Distribution



6.4 Standard Normal Distribution



6.5 Applications of the Normal Distribution

6.1 Discrete Random Variables Objectives: By the end of this section, I will be able to… 1) 2)

3)

Identify random variables. Explain what a discrete probability distribution is and construct probability distribution tables and graphs. Calculate the mean, variance, and standard deviation of a discrete random variable.

Random Variables 

A variable whose values are determined by chance



Chance in the definition of a random variable is crucial

Example 6.2 - Notation for random variables Suppose our experiment is to toss a single fair die, and we are interested in the number rolled. We define our random variable X to be the outcome of a single die roll. a. Why is the variable X a random variable? b. What are the possible values that the random variable X can take? c. What is the notation used for rolling a 5? d. Use random variable notation to express the probability of rolling a 5.

Example 6.2 continued Solution a)

We don’t know the value of X before we toss the die, which introduces an element of chance into the experiment

b)

Possible values for X: 1, 2, 3, 4, 5, and 6.

c)

When a 5 is rolled, then X equals the outcome 5, or X = 5.

d)

Probability of rolling a 5 for a fair die is 1/6, thus P(X = 5) = 1/6.

Types of Random Variables 

Discrete random variable - either a finite number of values or countable number of values, where “countable” refers to the fact that there might be infinitely many values, but they result from a counting process



Continuous random variable infinitely many values, and those values can be associated with measurements on a continuous scale without gaps or interruptions

Example Identify each as a discrete or continuous random variable. (a)

Total amount in ounces of soft drinks you consumed in the past year.

(b)

The number of cans of soft drinks that you consumed in the past year.

Example ANSWER: (a)

continuous

(b)

discrete

Example Identify each as a discrete or continuous random variable. (a) The number of movies currently playing in U.S. theaters. (b) The running time of a randomly selected movie

(c) The cost of making a randomly selected movie.

Example ANSWER

(a) discrete (b) continuous

(c) continuous

Discrete Probability Distributions 

Provides all the possible values that the random variable can assume



Together with the probability associated with each value



Can take the form of a table, graph, or formula



Describe populations, not samples

Example Table 6.2 in your textbook

The probability distribution table of the number of heads observed when tossing a fair coin twice

Probability Distribution of a Discrete Random Variable 

The sum of the probabilities of all the possible values of a discrete random variable must equal 1.



That is, ΣP(X) = 1.



The probability of each value of X must be between 0 and 1, inclusive.



That is, 0 ≤ P(X ) ≤ 1.

Example Let the random variable x represent the number of girls in a family of four children. Construct a table describing the probability distribution.

Example Determine the outcomes with a tree diagram:

Example 

Total number of outcomes is 16



Total number of ways to have 0 girls is 1 P(0 girls) 1 / 16 0.0625



Total number of ways to have 1 girl is 4 P(1 girl) 4 / 16 0.2500



Total number of ways to have 2 girls is 6 P(2 girls) 6 / 16 0.375

Example 

Total number of ways to have 3 girls is 4 P(3 girls) 4 / 16 0.2500



Total number of ways to have 4 girls is 1 P(4 girls) 1 / 16 0.0625

Example Distribution is:

NOTE:

P(x) 1

x

P(x)

0

0.0625

1

0.2500

2

0.3750

3

0.2500

4

0.0625

Mean of a Discrete Random Variable 

The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times



Also called the expected value or expectation of the random variable X.



Denoted as E(X )



Holds for discrete and continuous random variables

Finding the Mean of a Discrete Random Variable 

Multiply each possible value of X by its probability.



Add the resulting products.

X P X

Variability of a Discrete Random Variable Formulas for the Variance and Standard Deviation of a Discrete Random Variable Definition Formulas 2

X X

2

2

P X P X

Computational Formulas 2

X2

P X

X2

P X

2

2

Example

x

P(x)

0

x P(x)

x2

x 2 P( x)

0.0625

0

0

0

1

0.2500

0.25

1

0.2500

2

0.3750

0.75

4

1.5000

3

0.2500

0.75

9

2.2500

4

0.0625

0.25

16

1.0000

xP(x) 2.0

Example

2

x

P(x)

0

x P(x)

x2

x 2 P( x)

0.0625

0

0

0

1

0.2500

0.25

1

0.2500

2

0.3750

0.75

4

1.5000

3

0.2500

0.75

9

2.2500

4

0.0625

0.25

16

1.0000

x 2 P( x)

2

5.0000 4.0000 1.0000

1.0000 1.0

Discrete Probability Distribution as a Graph 

Graphs show all the information contained in probability distribution tables



Identify patterns more quickly

FIGURE 6.1 Graph of probability distribution for Kristin’s financial gain.

Example 

Page 270

Example 



Probability distribution (table) x

P(x)

0

0.25

1

0.35

2

0.25

3

0.15

Omit graph

Example 

Page 270

Example 

ANSWER X

number of goals scored

(a) Probability X is fewer than 3 P( X 0 X 1 X 2) P( X 0) P( X 1) P( X 0.25 0.35 0.25 0.85

2)

Example ANSWER (b) The most likely number of goals is the expected value (or mean) of X 

x

P(x)

x P(x)

0

0.25

0

1

0.35

0.35

2

0.25

0.50

3

0.15

0.45

xP(x) 1.3 1

She will most likely score one goal

Example ANSWER (c) Probability X is at least one 

P( X 1 X 2 X 3) P( X 1) P( X 2) P( X 0.35 0.25 0.15 0.75

3)

Summary 

Section 6.1 introduces the idea of random variables, a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text.



Random variables are variables whose value is determined at least partly by chance.



Discrete random variables take values that are either finite or countable and may be put in a list.



Continuous random variables take an infinite number of possible values, represented by an interval on the number line.

Summary 

Discrete random variables can be described using a probability distribution, which specifies the probability of observing each value of the random variable.



Such a distribution can take the form of a table, graph or formula.



Probability distributions describe populations, not samples.



We can find the mean μ, standard deviation σ, and variance σ2 of a discrete random variable using formulas.



6.2 Binomial Probability Distribution

6.2 Binomial Probability Distribution Objectives: By the end of this section, I will be able to… 1) 2) 3) 4)

Explain what constitutes a binomial experiment. Compute probabilities using the binomial probability formula. Find probabilities using the binomial tables. Calculate and interpret the mean, variance, and standard deviation of the binomial random variable.

Factorial symbol 

For any integer n ≥ 0, the factorial symbol n! is defined as follows:



0! = 1



1! = 1



n! = n(n - 1)(n - 2) · · · 3 · 2 · 1

Example Find each of the following 1. 4!

2. 7!

Example ANSWER

1. 4! 4 3 2 1 24 2. 7! 7 6 5 4 3 2 1 5040

Factorial on Calculator Calculator 7

MATH

PRB

4:!

which is

7! Enter gives the result 5040

Combinations An arrangement of items in which  r items are chosen from n distinct items.  repetition of items is not allowed (each item is distinct).  the order of the items is not important.

Example of a Combination The number of different possible 5 card poker hands. Verify this is a combination by checking each of the three properties. Identify r and n.

Example 

Five cards will be drawn at random from a deck of cards is depicted below

Example An arrangement of items in which  5 cards are chosen from 52 distinct items.  repetition of cards is not allowed (each card is distinct).  the order of the cards is not important.

Combination Formula The number of combinations of r items chosen from n different items is denoted as nCr and given by the formula:

n

Cr

n! r! n r !

Example Find the value of

C 7 4

Example ANSWER:

7

C4

7! 4! (7 4)! 7! 4! 3! 7 6 5 4 3 21 ( 4 3 2 1) (3 2 1) 7 6 5 3 21 7 5 35

Combinations on Calculator Calculator 7

MATH

To get:

7

PRB

C4

Then Enter gives 35

3:nCr

4

Example of a Combination Determine the number of different possible 5 card poker hands.

Example ANSWER: 52

C5

2,598,960

Motivational Example Genetics •

In mice an allele A for agouti (gray-brown, grizzled fur) is dominant over the allele a, which determines a non-agouti color. Suppose each parent has the genotype Aa and 4 offspring are produced. What is the probability that exactly 3 of these have agouti fur?

Motivational Example •

A single offspring has genotypes: A

a

A

AA

Aa

a

aA

aa

Sample Space { AA, Aa, aA, aa}

Motivational Example •

Agouti genotype is dominant

• Event that offspring is agouti:

{ AA, Aa, aA} •

Therefore, for any one birth: P(agouti genotype) 3 / 4 P(not agouti genotype) 1 / 4

Motivational Example •



Let G represent the event of an agouti offspring and N represent the event of a non-agouti Exactly three agouti offspring may occur in four different ways (in order of birth): NGGG, GNGG, GGNG, GGGN

Motivational Example •

Consecutive events (birth of a mouse) are independent and using multiplication rule: P( N

P(G

G

N

G

G

G)

G)

P( N ) P(G ) P(G ) P(G ) 1 3 3 3 27 4 4 4 4 256 P(G ) P( N ) P(G ) P(G ) 3 1 3 3 27 4 4 4 4 256

Motivational Example P(G

P(G

G

G

N

G

G)

N)

P(G ) P(G ) P( N ) P(G ) 3 3 1 3 27 4 4 4 4 256 P(G ) P(G ) P(G ) P( N ) 3 3 3 1 27 4 4 4 4 256

Motivational Example •

P(exactly 3 offspring has agouti fur) P(first birth N OR second birth N OR third birth N OR fourth birth N) 1 3 3 3 3 1 3 3 3 3 1 3 3 3 3 1 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4

4

3 4

3

1 4

108 0.422 256

Binomial Experiment Agouti fur example may be considered a binomial experiment

Binomial Experiment Four Requirements: 1)

Each trial of the experiment has only two possible outcomes (success or failure)

2)

Fixed number of trials

3)

Experimental outcomes are independent of each other

4)

Probability of observing a success remains the same from trial to trial

Binomial Experiment Agouti fur example may be considered a binomial experiment 1)

Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)

2)

Fixed number of trials (4 births)

3)

Experimental outcomes are independent of each other

4)

Probability of observing a success remains the same from trial to trial (¾)

Binomial Probability Distribution When a binomial experiment is performed, the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution.

Binomial Probability Distribution Formula For a binomial experiment, the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is

P( X ) where n

CX

X n C p ( 1 p ) n X

n! X! n X !

X

Binomial Probability Distribution Formula Let q=1-p

P( X )

X n C p q n X

X

Rationale for the Binomial Probability Formula P(x) =

n! • (n – x )!x!

The number of outcomes with exactly x successes among n trials

px •

n-x q

Binomial Probability Formula P(x) =

n! • (n – x )!x!

Number of outcomes with exactly x successes among n trials

px •

n-x q

The probability of x successes among n trials for any one particular order

Agouti Fur Genotype Example X

event of a birth with agouti fur

P( X )

4! (4 3)! 3! 4

3 4

3

3 4

3

1 4

27 1 4 64 4 108 0.422 256

1

1 4

1

Binomial Probability Distribution Formula: Calculator

2ND VARS

A:binompdf(4, .75, 3) n, p, x

Enter gives the result 0.421875

Binomial Distribution Tables   

n is the number of trials X is the number of successes p is the probability of observing a success

See Example 6.16 on page 278 for more information FIGURE 6.7 Excerpt from the binomial tables.

Example Page 284

Example ANSWER X

number of heads

P( X

5) 5

20

C5 (0.5) (0.5) 5

20 5 15

15,504 (0.5) (0.5) 0.0148

Binomial Mean, Variance, and Standard Deviation 

Mean (or expected value): μ = n · p



Variance:

2

np(1 p)

Use q 1 p, then 

2

npq

Standard deviation:

np(1 p)

npq

Example 20 coin tosses The expected number of heads:

np (20)(0.50) 10 Variance and standard deviation: 2

npq (20)(0.50)(0.50) 5.0 5

2.24

Example Page 284

Is this a Binomial Distribution? Four Requirements: 1)

Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)

2)

Fixed number of trials (50)

3)

Experimental outcomes are independent of each other

4)

Probability of observing a success remains the same from trial to trial (assumed to be 58.4%=0.584)

Example ANSWER (a) X number of baskets

P( X

25) 50

C25 (0.584 )

0.0549

25

(0.416 )

25

Example ANSWER (b) The expected value of X

np (50)(0.584)

29.2

The most likely number of baskets is 29

Example ANSWER (c) In a random sample of 50 of O’Neal’s shots he is expected to make 29.2 of them.

Example Page 285

Is this a Binomial Distribution? Four Requirements: 1)

Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)

2)

Fixed number of trials (120)

3)

Experimental outcomes are independent of each other

4)

Probability of observing a success remains the same from trial to trial (assumed to be 11%=0.11)

Example ANSWER (a) X=number of white males who contracted AIDS through injected drug use

P( X

10) 10

120

C10 (0.11)

0.0816

110

(0.89)

Example ANSWER (b) At most 3 men is the same as less than or equal to 3 men: P( X

3)

P( X

0) P( X

1) P( X

Why do probabilities add?

2) P( X

3)

Example Use TI-83+ calculator 2ND VARS to get A:binompdf(n, p, X) P( X

0) P( X

1) P( X

2) P( X

3)

= binompdf(120, .11, 0) + binompdf(120, .11, 1)

+ binompdf(120, .11, 2) + binompdf(120, .11,3)

5.535172385 E 4 0.000554

Example ANSWER (c) Most likely number of white males is the expected value of X

np (120)(0.11) 13.2

Example ANSWER (d) In a random sample of 120 white males with AIDS, it is expected that approximately 13 of them will have contracted AIDS by injected drug use

Example Page 286

Example ANSWER (a) 2

npq (120)(0.11)(0.89) 11.748

11.748 3.43

RECALL: Outliers and z Scores Data values are not unusual if

2

z - score

2

Otherwise, they are moderately unusual or an outlier (see page 131)

Z score Formulas Sample z

x x s

Population z

x

Example Z-score for 20 white males who contracted AIDS through injected drug use:

z

20 13.2 1.98 3.43

It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS.

Summary 

The most important discrete distribution is the binomial distribution, where there are two possible outcomes, each with probability of success p, and n independent trials.



The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula.

Summary 

Binomial probabilities can also be found using the binomial tables or using technology.



There are formulas for finding the mean, variance, and standard deviation of a binomial random variable.

6.3 Continuous Random Variables and the Normal Probability Distribution Objectives: By the end of this section, I will be able to… 1)

Identify a continuous probability distribution.

2)

Explain the properties of the normal probability distribution.

FIGURE 6.15- Histograms (a) Relatively small sample (n = 100) with large class widths (0.5 lb).

(b) Large sample (n = 200) with smaller class widths (0.2 lb).

Figure 6.15 continued (c) Very large sample (n = 400) with very small class widths (0.1 lb). (d) Eventually, theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small.

Continuous Probability Distributions 

A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take



Density curve is drawn above the horizontal axis



Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

Requirements for Probability Distribution of a Continuous Random Variable 1)

The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables).

2)

The vertical height of the density curve can never be negative. That is, the density curve never goes below the horizontal axis.

Probability for a Continuous Random Variable 

Probability for Continuous Distributions is represented by area under the curve above an interval.

The Normal Probability Distribution 

Most important probability distribution in the world



Population is said to be normally distributed, the data values follow a normal probability distribution



population mean is μ



population standard deviation is σ



μ and σ are parameters of the normal distribution

FIGURE 6.19 

The normal distribution is symmetric about its mean μ (bell-shaped).

Properties of the Normal Density Curve (Normal Curve) 1)

It is symmetric about the mean μ.

2)

The highest point occurs at X = μ, because symmetry implies that the mean equals the median, which equals the mode of the distribution.

3)

It has inflection points at μ-σ and μ+σ.

4)

The total area under the curve equals 1.

Properties of the Normal Density Curve (Normal Curve) continued 5)

Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 0.5 (Figure 6.19).

6)

The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions. As X moves farther from the mean, the density curve approaches but never quite touches the horizontal axis.

The Empirical Rule For data sets having a distribution that is approximately bell shaped, the following properties apply:

 About 68% of all values fall within 1 standard deviation of the mean.  About 95% of all values fall within 2 standard deviations of the mean.

 About 99.7% of all values fall within 3 standard deviations of the mean.

FIGURE 6.23 The Empirical Rule.

Drawing a Graph to Solve Normal Probability Problems 1.

Draw a “generic” bell-shaped curve, with a horizontal number line under it that is labeled as the random variable X.



Insert the mean μ in the center of the number line.

Steps in Drawing a Graph to Help You Solve Normal Probability Problems Mark on the number line the value of X indicated in the problem. Shade in the desired area under the normal curve. This part will depend on what values of X the problem is asking about. 3) Proceed to find the desired area or probability using the empirical rule. 2)

Example Page 295

Example ANSWER

Example Page 296

Example ANSWER

Example Page 296

Example ANSWER

Summary 

Continuous random variables assume infinitely many possible values, with no gap between the values.



Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve.

Summary 

The normal distribution is the most important continuous probability distribution.



It is symmetric about its mean μ and has standard deviation σ.



One should always sketch a picture of a normal probability problem to help solve it.

6.4 Standard Normal Distribution Objectives: By the end of this section, I will be able to… 1)

Find areas under the standard normal curve, given a Z-value.

2)

Find the standard normal Z-value, given an area.

The Standard Normal (Z) Distribution 

A normal distribution with

mean μ = 0 and standard deviation σ = 1. When the normal curve is plotted, we plot Z on the horizontal axis

The Empirical Rule For Standard Normal Distribution

-3

-2

-1

0

1

2

3

Example Find the area that lies under the standard normal curve between Z=-1 and Z=2

Example ANSWER: 0.34+0.34+0.135=0.815

Example Find the area that lies under the standard normal curve that lies above Z=3

Example ANSWER: Since 100%-99.7%=0.3%=0.003, we take half of 0.003 to get:

0.0015

The Standard Normal (Z) Distribution 

In fact, we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z



Table C in Appendix on pages T-9 and T-10

Table C: Standard Normal Distribution 1.

It is designed only for the standard normal distribution, which has a mean of 0 and a standard deviation of 1.

2.

It is on two pages, with one page for negative z-scores and the other page for positive z-scores.

3.

Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score.

Table continues to -0.0 in the appendix

Table continues to 3.4 in the appendix

Table C: Standard Normal Distribution 4. When working with a graph, avoid confusion between z-scores and areas: z Score Distance along horizontal scale of the standard normal distribution; refer to the leftmost column and top row of Table C. Area Region under the curve; refer to the values in the body of Table C. 5. The part of the z-score denoting hundredths is found across the top.

Table 6.7 Steps for finding areas under the standard normal curve

Example Page 307

Do problems 18 and 20

Example ANSWER (18)

Use Table C for Z=2.00

Example ANSWER (20)

Use Table C for Z=0.500

Example Page 307

Do problems 26 and 28

Example

Example ANSWER 26(c) Use Table C to find that the area to the left of Z=2.12 is 0.9830. We want the area to the right of Z=2.12 which will be 1-0.9830=0.017

Example

Example ANSWER 28(c) Use Table C to find that the area to the left of Z=-0.69 is 0.2451. We want the area to the right of Z=-0.69 which will be 1-0.2451=0.7549

Example Page 308

Do problems 32 and 36

Example

Example Page 308

Do problems 32 and 38

Example

Example ANSWER 32(c) Use Table C to find that the area to the left of Z=1.28 is 0.8997. Use Table C to find that the area to the left of Z=1.96 is 0.9750. We want the area to the between Z=1.28 and Z=1.96 which will be

0.9750-0.8997=0.0753

Example

Example ANSWER 38(c) Use Table C to find that the area to the left of Z=-2.01 is 0.0222. Use Table C to find that the area to the left of Z=2.37 is 0.9911. We want the area to the between Z=-2.01 and Z=2.37 which will be

0.9911-0.0222=0.9689

Example Page 308

Do problem 42

Example

Example Page 308

Do problem 52

Example

Example Page 308

Do problem 58

Example

Note: this answer is the Z value in the table that corresponds to: 1 - 0.5120 = 0.4880

Summary 

The standard normal distribution has mean μ=0 and standard deviation σ=1.



This distribution is often called the Z distribution.



The Z table and technology can be used to find areas under the standard normal curve.

Summary 

In the Z table, the numbers on the outside are values of Z, and the numbers inside are areas to the left of values of Z.



The Z table and technology can also be used to find a value of Z, given a probability or an area under the curve.

6.5 Applications of the Normal Distribution Objectives: By the end of this section, I will be able to… 1)

Compute probabilities for a given value of any normal random variable.

2)

Find the appropriate value of any normal random variable, given an area or probability.

3)

Calculate binomial probabilities using the normal approximation to the binomial distribution. (OMIT)

Standardizing a Normal Random Variable 

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

Z

x

Finding Probabilities for Any Normal Distribution Step 1  Determine the random variable X, the mean μ, and the standard deviation σ. 

Draw the normal curve for X, and shade the desired area.

Step 2  Standardize by using the formula Z = (X - μ )/σ to find the values of Z corresponding to the X-values.

Finding Probabilities for Any Normal Distribution Step 3  Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X. Step4  Find the area under the standard normal curve using either (a) the Z table or (b) technology.  This area is equal to the area under the normal curve for X drawn in Step 1.

Example Page 323

Do problems 6,8,10,12

Example Solutions: 6) 0.1587 8) 0.0062 10) 0.8413 12) 0.8400

Calculator Use

Calculator Use

Do problem 12 again with calculator

Calculator Use

Or use 10^99 as larger value and -10^99 as the smaller value

Do problems 6,8 again with calculator

Finding Normal Data Values for Specified Probabilities Step 1 

Determine X, μ, and σ, and draw the normal curve for X.



Shade the desired area.



Mark the position of X1, the unknown value of X.

Finding Normal Data Values for Specified Probabilities Step 2 

Find the Z-value corresponding to the desired area.



Look up the area you identified in Step 1 on the inside of the Z table.



If you do not find the exact value of your area, by convention choose the area that is closest.

Finding Normal Data Values for Specified Probabilities Step 3 

Transform this value of Z into a value of X, which is the solution.



Use the formula X1 = Zσ + μ.

Example Page 323

Do problems 18,20,22

Example Solutions: 18) X=57.2 20) X=53.55 and X=86.45 22) X=44.2 and X=95.8

Calculator Use

Do problem 18 and 22 again with calculator

Calculator Use

Do problem 18 and 22 again with calculator

Example 6.38 - Finding the Xvalues that mark the Boundaries of the middle 95% of X-values Edmunds.com reported that the average amount that people were paying for a 2007 Toyota Camry XLE was $23,400. Let X = price, and assume that price follows a normal distribution with μ = $23,400 and σ = $1000. Find the prices that separate the middle 95% of 2007 Toyota Camry XLE prices from the bottom 2.5% and the top 2.5%.

Example 6.38 continued Solution

Step 1  Determine X, μ, and σ, and draw the normal curve for X.  Let X = price, μ = $23,400, and σ = $1000.  The middle 95% of prices are delimited by X1 and X2, as shown in Figure 6.49.

FIGURE 6.49

Example 6.38 continued Solution Step 2 

Find the Z-values corresponding to the desired area.



The area to the left of X1 equals 0.025, and the area to the left of X2 equals 0.975.



Looking up area 0.025 on the inside of the Z table gives us Z1 = –1.96.



Looking up area 0.975 on the inside of the Z table gives us Z2 = 1.96.

Example 6.38 continued Solution Step 3  Transform using the formula X1 = Zσ + μ.  X1 = Z1σ + μ =(–1.96)(1000) + 23,400 = 21,440 X2 =Z2σ + μ =(1.96)(1000) + 23,400 = 25,360  The prices that separate the middle 95% of 2007 Toyota Camry XLE prices from the bottom 2.5% of prices and the top 2.5% of prices are $21,440 and $25,360.

Example Page 324

Example Solution: Let X be the random variable that represents how many points McGrady scores in a randomly chosen game. Find: P(X

30)

Example Solution: P( X

30) 1 P( X

30)

1 0.5987 0.4013

Example Page 324

Example Solution: P(10 X

20) 0.1464

Example Page 324

Example Solution: Find X 1 so that P( X

X1)

0.05

that is, the 5th percentile (only 5% of his games did he have a lower score)

Example Solution: P( X

X1)

0.05

Gives the value: X 1 14.84 points

Example Page 324

Example Solution: The z-score for X=40 points is 1.50. Since this z-score is more than -2 and less than 2, this is not unusual.

Summary 

Section 6.5 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 6.4.



Methods for finding probabilities for a given value of the normal random variable X were discussed.

Summary 

For any normal probability distribution, values of X can be found for given probabilities using the “backward” methods of Section 6.4 and the formula X1 = Z1σ + μ.

Suggest Documents