© 2010 Pearson Prentice Hall. All rights reserved Chapter The Normal Probability Distribution © 2010 Pearson Prentice Hall. All rights reserved 3 7.

© 2010 Pearson Prentice Hall. All rights reserved

Chapter

The Normal Probability Distribution


37

© 2010 Pearson Prentice Hall. All rights reserved 7-2

Section 7.1 Properties of the Normal Distribution



Suppose that United Parcel Service is supposed to deliver a package to your front door and the arrival time is somewhere between 10 am and 11 am. Let the random variable X represent the time from 10 am when the delivery is supposed to take place. The delivery could be at 10 am (x = 0) or at 11 am (x = 60) with all 1-minute interval of times between x = 0 and x = 60 equally likely. That is to say your package is just as likely to arrive between 10:15 and 10:16 as it is to arrive between 10:40 and 10:41. The random variable X can be any value in the interval from 0 to 60, that is, 0 < X < 60. Because any two intervals of equal length between 0 and 60, inclusive, are equally likely, the random variable X is said to follow a uniform probability distribution.

EXAMPLE Illustrating the Uniform DistributionEXAMPLE Illustrating the Uniform Distribution

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The graph below illustrates the properties for the “time” example. Notice the area of the rectangle is one and the graph is greater than or equal to zero for all x between 0 and 60, inclusive.

Because the area of a rectangle is height times width, and the width of the rectangle is 60, the height must be 1/60.

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Values of the random variable X less than 0 or greater than 60 are impossible, thus the equation must be zero for X less than 0 or greater than 60.

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The area under the graph of the density function over an interval represents the probability of observing a value of the random variable in that interval.

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The probability of choosing a time that is between 15 and 30 seconds after the minute is the area under the uniform density function.

15 30

Area = P(15 < x < 30)

= 15/60 = 0.25

EXAMPLE Area as a ProbabilityEXAMPLE Area as a Probability

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Relative frequency histograms that are symmetric and bell-shaped are said to have the shape of a normal curve.

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If a continuous random variable is normally distributed, or has a normal probability distribution, then a relative frequency histogram of the random variable has the shape of a normal curve (bell-shaped and symmetric).

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Copyright © 2010 Pearson Education, Inc.

Use the graph of the given normal distribution to identify μ and σ.

A. μ = 80 and σ = 36

B. μ = 80 and σ = 6

C. μ = 62 and σ = 98

D. μ = 6 and σ = 80

62 8668 74 80 92 98 X

Slide 7- 17


Use the graph of the given normal distribution to identify μ and σ.

A. μ = 80 and σ = 36

B. μ = 80 and σ = 6

C. μ = 62 and σ = 98

D. μ = 6 and σ = 80

62 8668 74 80 92 98 X

Slide 7- 18




EXAMPLE Interpreting the Area Under a Normal CurveEXAMPLE Interpreting the Area Under a Normal Curve

The weights of giraffes are approximately normally distributed with mean μ = 2200 pounds and standard deviation σ = 200 pounds.

(a)Draw a normal curve with the parameters labeled.(b)Shade the area under the normal curve to the left of x = 2100 pounds.(c)Suppose that the area under the normal curve to the left of x = 2100 pounds is 0.3085. Provide two interpretations of this result.

(a), (b)(c) • The proportion of giraffes whose weight is less than 2100 pounds is 0.3085• The probability that a randomly selected giraffe weighs less than 2100 pounds is 0.3085.

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Section 7.2 The Standard Normal Distribution

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The table gives the area under the standard normal curve for values to the left of a specified Z-score, zo, as shown in the figure.

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Find the area under the standard normal curve to the left of z = -0.38.

EXAMPLE Finding the Area Under the Standard Normal CurveEXAMPLE Finding the Area Under the Standard Normal Curve

Area left of z = -0.38 is 0.3520.

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Area under the normal curve to the right of zo = 1 – Area to the left of zo

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Find the area under the standard normal curve to the right of Z = 1.25.

Area right of 1.25 = 1 – area left of 1.25= 1 – 0.8944= 0.1056 7-32


Find the area under the standard normal curve between z = -1.02 and z = 2.94.


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Find the z-score such that the area to the left of the z-score is 0.7157.

EXAMPLE Finding a z-score from a Specified Area to the LeftEXAMPLE Finding a z-score from a Specified Area to the Left

The z-score such that the area to the left of the z-score is 0.7157 is z = 0.57.

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EXAMPLE Finding a z-score from a Specified Area to the RightEXAMPLE Finding a z-score from a Specified Area to the Right

Find the z-score such that the area to the right of the z-score is 0.3021.

The area left of the z-score is 1 – 0.3021 = 0.6979.

The approximate z-score that corresponds to an area of 0.6979 to the left (0.3021 to the right) is 0.52. Therefore, z = 0.52.

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Find the z-scores that separate the middle 80% of the area under the normal curve from the 20% in the tails.

EXAMPLE Finding a z-score from a Specified AreaEXAMPLE Finding a z-score from a Specified Area

Area = 0.8

Area = 0.1Area = 0.1

z1 is the z-score such that the area left is 0.1, so z1 = -1.28.

z2 is the z-score such that the area left is 0.9, so z2 = 1.28.

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Notation for the Probability of a Standard Normal Random Variable

P(a < Z < b) represents the probability a standard normal random variable is between

a and b

P(Z > a) represents the probability a standard normal random variable is greater than a.

P(Z < a) represents the probability a standard normal random variable is less than a.

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Find the probability of the standard normal random variable Z.

P(Z < 1.49)

A. 0.9319

B. 0.0681

C. 0.6879

D. 0.3121

Slide 7- 41



P(Z < 1.49)

A. 0.9319

B. 0.0681

C. 0.6879

D. 0.3121

Slide 7- 42


Find each of the following probabilities:

(a) P(Z < -0.23)

(b) P(Z > 1.93)

(c) P(0.65 < Z < 2.10)

EXAMPLE Finding Probabilities of Standard Normal Random VariablesEXAMPLE Finding Probabilities of Standard Normal Random Variables

(a) P(Z < -0.23) = 0.4090

(b) P(Z > 1.93) = 0.0268

(c) P(0.65 < Z < 2.10) = 0.2399

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P(Z ≥ –2.31)

A. 0.0104

B. 0.0087

C. 0.9896

D. 0.9913

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P(Z ≥ –2.31)

A. 0.0104

B. 0.0087

C. 0.9896

D. 0.9913

Slide 7- 45



P(–2.14 < Z < 0.95)

A. 0.1170

B. 0.0681

C. 0.1873

D. 0.8127

Slide 7- 46



P(–2.14 < Z < 0.95)

A. 0.1170

B. 0.0681

C. 0.1873

D. 0.8127

Slide 7- 47


Find the Z-score such that the area under the standard normal curve to the right is 0.10.

A. 1.28

B. -1.28

C. 0.5398

D. 0.8159

Slide 7- 48


Find the Z-score such that the area under the standard normal curve to the right is 0.10.

A. 1.28

B. -1.28

C. 0.5398

D. 0.8159

Slide 7- 49


For any continuous random variable, the probability of observing a specific value of the random variable is 0. For example, for a standard normal random variable, P(a) = 0 for any value of a. This is because there is no area under the standard normal curve associated with a single value, so the probability must be 0. Therefore, the following probabilities are equivalent:

P(a < Z < b) = P(a < Z < b) = P(a < Z < b) = P(a < Z < b)

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IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability a randomly selected person has an IQ score greater than 120.

A. 0.9082

B. 0.6293

C. 0.0918

D. 0.3707

Slide 7- 51


IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability a randomly selected person has an IQ score greater than 120.

A. 0.9082

B. 0.6293

C. 0.0918

D. 0.3707

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Section 7.4 Assessing Normality

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Suppose that we obtain a simple random sample from a population whose distribution is unknown. Many of the statistical tests that we perform on small data sets (sample size less than 30) require that the population from which the sample is drawn be normally distributed. Up to this point, we have said that a random variable X is normally distributed, or at least approximately normal, provided the histogram of the data is symmetric and bell-shaped. This method works well for large data sets, but the shape of a histogram drawn from a small sample of observations does not always accurately represent the shape of the population. For this reason, we need additional methods for assessing the normality of a random variable X when we are looking at sample data.

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The idea behind finding the expected Z-score is that if the data comes from a population that is normally distributed, we should be able to predict the area left of each of the data values. The value of fi represents the expected area left of the ith data value assuming the data comes from a population that is normally distributed. For example, f1 is the expected area left of the smallest data value, f2 is the expected area left of the second smallest data value, and so on.

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A discrete random variable is given. Assume the probability will be approximated using a normal distribution. Describe the area under the normal curve that will be computed.

The probability of getting at least 80 successes

A. Right of X = 80.5

B. Right of X = 79.5

C. Left of X = 80.5

D. Left of X = 79.5

Slide 7- 56


A discrete random variable is given. Assume the probability will be approximated using a normal distribution. Describe the area under the normal curve that will be computed.

The probability of getting at least 80 successes

A. Right of X = 80.5

B. Right of X = 79.5

C. Left of X = 80.5

D. Left of X = 79.5

Slide 7- 57


Section 7.5 The Normal Approximation to the Binomial Probability Distribution

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Criteria for a Binomial Probability Experiment

An experiment is said to be a binomial experiment provided:

1. The experiment is performed n independent times. Each repetition of the experiment is called a trial. Independence means that the outcome of one trial will not affect the outcome of the other trials.

2. For each trial, there are two mutually exclusive outcomes - success or failure.

3. The probability of success, p, is the same for each trial of the experiment.

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For a fixed p, as the number of trials n in a binomial experiment increases, the probability distribution of the random variable X becomes more nearly symmetric and bell-shaped. As a general rule of thumb, if np(1 – p) > 10, the probability distribution will be approximately symmetric and bell-shaped.

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P(X = 18) ≈ P(17.5 < X < 18.5)

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P(X < 18) ≈ P(X < 18.5)

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EXAMPLE Using the Binomial Probability Distribution FunctionEXAMPLE Using the Binomial Probability Distribution Function

According to the Experian Automotive, 35% of all car-owning households have three or more cars.

(a)In a random sample of 400 car-owning households, what is the probability that fewer than 150 have three or more cars?

(b) In a random sample of 400 car-owning households, what is the probability that at least 160 have three or more cars?

(1 ) 400(0.35)(1 0.35)

91 10

np p

400(0.35)

140X

400(0.35)(1 0.35)

9.54X

( 150) ( 149)

( 149.5)

149.5 140

9.54

0.8413

P X P X

P X

P Z

( 160) ( 159.5)

159.5 140

9.54

0.0207

P X P X

P Z

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Forty-three percent of marriages end in divorce. You randomly select 150 married couples. Find the probability that fewer than 70 of the marriages will end in divorce.

A. 0.8389

B. 0.8186

C. 0.8251

D. 0.7967

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Forty-three percent of marriages end in divorce. You randomly select 150 married couples. Find the probability that fewer than 70 of the marriages will end in divorce.

A. 0.8389

B. 0.8186

C. 0.8251

D. 0.7967

Slide 7- 68

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