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The Normal Distribution

Chapter 5

Properties of a Normal Distribution

- The mean, median, and mode are equal

- Bell shaped and symmetric about the mean

- The total area under the curve is one (1) or 100

- The curve approaches but never touches the x-

axis as it - extends farther and farther away from the mean in

both directions.

- The points at which the curvature changes are

called inflection points.

Means and Standard Deviations

Curves with different means, same standard

deviation

Curves with different means different standard

deviations

Empirical Rule

Determining Intervals

An instruction manual claims that the assembly

time for a product is normally distributed with

a mean of 4.2 hours and standard deviation 0.3

hours. Determine the interval in which 95 of

the assembly times fall.

95 of the data will fall within 2 standard

deviations of the mean. 4.2 - 2 (0.3) 3.6

and 4.2 2 (0.3) 4.8. 95 of the assembly

times will be between 3.6 and 4.8 hrs.

The Standard Score

The standard score, or z-score, represents the

number of standard deviations a random variable x

falls from the mean.

The test scores for a civil service exam are

normally distributed with a mean of 152 and

standard deviation of 7. Find the standard

z-score for a person with a score of (a) 161

(b) 148 (c) 152

From z-Scores to Raw Scores

The test scores for a civil service exam are

normally distributed with a mean of 152 and

standard deviation of 7. Find the test score for

a person with a standard score of (a) 2.33

(b) -1.75 (c) 0

(a) x 152 (2.33)(7) 168.31

(b) x 152 ( -1.75)(7) 139.75

(c) x 152 (0)(7) 152

The Standard Normal Distribution

The standard normal distribution has a mean of 0

and a standard deviation of 1.

Using z- scores any normal distribution can be

transformed into the standard normal distribution.

z

Cumulative Areas

The total area under the curve is one.

- The cumulative area is close to 0 for z-scores

close to -3.49.

- The cumulative area for z 0 is 0.5000

- The cumulative area is close to 1 for z scores

close to 3.49.

Cumulative Areas

Find the cumulative area for a z-score of -1.25.

Read down the z column on the left to z -1.2

and across to the column under .05. The value in

the cell is 0.1056, the cumulative area.

The probability that z is at most -1.25 is

0.1056. P ( z ? -1.25) 0.1056

From Areas to z-scores

Find the z-score corresponding to a cumulative

area of 0.9803.

0.9803

0.9803

z

Locate 0.9803 in the area portion of the table.

Read the values at the beginning of the

corresponding row and at the top of the column.

The z-score is 2.06.

z 2.06 is roughly the 98th percentile.

Finding Probabilities

To find the probability that z is less than a

given value, read the cumulative area in the

table corresponding to that z-score.

Find P( z lt -1.24)

Read down the z-column to -1.2 and across to .04.

The cumulative area is 0.1075.

P ( z lt 1.24) 0.1075

Finding Probabilities

To find the probability that z is greater than a

given value, subtract the cumulative area in the

table from 1.

Find P( z gt -1.24)

0.1075

The cumulative area (area to the left) is 0.1075.

So the area to the right is 1 - 0.1075 0.8925.

P( z gt -1.24) 0.8925

Finding Probabilities

To find the probability z is between two given

values, find the cumulative areas for each and

subtract the smaller area from the larger.

Find P( -1.25 lt z lt 1.17)

1. P(z lt 1.17) 0.8790

2. P(z lt -1.25) 0.1056

3. P( -1.25 lt z lt 1.17) 0.8790 - 0.1056 0.7734

Summary

Probabilities and Normal Distributions

If a random variable, x is normally distributed,

the probability that x will fall within an

interval is equal to the area under the curve in

the interval.

IQ scores are normally distributed with a mean of

100 and standard deviation of 15. Find the

probability that a person selected at random will

have an IQ score less than 115.

To find the area in this interval, first find the

standard score equivalent to x 115.

Probabilities and Normal Distributions

Standard Normal Distribution

Find P(z lt 1)

P( z lt 1) 0.8413, so P( x lt115) 0.8413

Application

Monthly utility bills in a certain city are

normally distributed with a mean of 100 and a

standard deviation of 12. A utility bill is

randomly selected. Find the probability it is

between 80 and 115.

P(80 lt x lt 115)

P(-1.67 lt z lt 1.25)

0.8944 - 0.0475 0.8469

The probability a utility bill is between 80 and

115 is 0.8469.

Finding Percentiles

Monthly utility bills in a certain city are

normally distributed with a mean of 100 and a

standard deviation of 12. What is the smallest

utility bill that can be in the top 10 of the

bills?

z

Find the cumulative area in the table that is

closest to 0.9000 (the 90th percentile.) The area

0.8997 corresponds to a z-score of 1.28.

x 100 1.28(12) 115.36.

115.36 is the smallest value for the top 10.

Sampling Distributions

A sampling distribution is the probability

distribution of a sample statistic that is formed

when samples of size n are repeatedly taken from

a population. If the sample statistic is the

sample mean, then the distribution is the

sampling distribution of sample means.

Sample

Sample

Sample

Sample

Sample

Sample

The Central Limit Theorem

the sample means will have a normal distribution

The Central Limit Theorem

x

the distribution of means of sample size n , will

be normal with a mean standard deviation

Application

Distribution of means of sample size 60 , will

be normal.

Interpreting the Central Limit Theorem

The mean height of American men (ages 20-29) is ?

69.2. If a random sample of 60 men in this age

group is selected, what is the probability the

mean height for the sample is greater than 70?

Assume the standard deviation is 2.9.

mean

Find the z-score for a sample mean of 70

Interpreting the Central Limit Theorem

P (z gt 2.14) 1 - 0.9838 0.0162

z

There is a 0.0162 probability that a sample of 60

men will have a mean height greater than 70.

Application Central Limit Theorem

During a certain week the mean price of gasoline

in California was 1.164 per gallon. What is the

probability that the mean price for the sample of

38 gas stations in California is between 1.169

and 1.179? Assume the standard deviation

0.049.

standard deviation

Calculate the standard z-score for sample values

of 1.169 and 1.179.

Application Central Limit Theorem

P( 0.63 lt z lt 1.90) 0.9713 - 0.7357 0.2356

The probability is 0.2356 that the mean for the

sample is between 1.169 and 1.179.

Normal Approximations to the Binomial

Characteristics of a Binomial Experiment

- There are a fixed number of trials. (n)
- The n trials are independent and repeated under

identical conditions - Each trial has 2 outcomes, Success or Failure.
- The probability of success on a single trial is p

and the probability of failure is q. p q

1 - The central problem is to find the probability of

x successes out of n trials. Where x 0 or 1 or

2 n.

x is a count of the number of successes in n

trials.

Application

34 of Americans have type A blood. If 500

Americans are sampled at random, what is the

probability at least 300 have type A blood?

Using techniques of chapter 4 you could calculate

the probability that exactly 300, exactly

301exactly 500 Americans have A blood type

and add the probabilities.

Oryou could use the normal curve probabilities

to approximate the binomial probabilities.

Why do we require np ? 5 and nq ? 5?

n 5 p 0.25, q .75 np 1.25 nq 3.75

n 20 p 0.25 np 5 nq 15

4

n 50 p 0.25 np 12.5 nq 37.5

Binomial Probabilities

The binomial distribution is discrete with a

probability histogram graph. The probability that

a specific value of x will occur is equal to the

area of the rectangle with midpoint at x.

If n 50 and p 0.25 find P (14? x ? 16)

Add the areas of the rectangles with midpoints at

x 14, x 15, x 16.

0.111 0.089 0.065 0.265

P (14? x ? 16) 0.265

Correction for Continuity

Use the normal approximation to the binomial to

find P(14? x ? 16) if n 50 and p 0.25

Check that np 12.5 ? 5 and nq 37.5 ? 5.

The interval of values under the normal curve is

13.5 ? x ? 16.5.

To ensure the boundaries of each rectangle are

included in the interval, subtract 0.5 from a

left-hand boundary and add 0.5 to a right-hand

boundary.

Normal Approximation to the Binomial

Use the normal approximation to the binomial to

find P(14? x ? 16) if n 50 and p 0.25

Find the mean and standard deviation using

binomial distribution formulas.

Adjust the endpoints to correct for continuity

P(13.5 ? x ? 16.5)

Convert each endpoint to a standard score

P(0.33 ? z ? 1.31) 0.9049 - 0.6293 0.2756

Application

A survey of Internet users found that 75 favored

government regulations on junk e-mail. If 200

Internet users are randomly selected, find the

probability that fewer than 140 are in favor of

government regulation.

Since np150 ? 5 and nq 50 ? 5 use the normal

approximation to the binomial.

Use the correction for continuity P(x lt 139.5)

P(z lt -1.71) 0.0436

The probability that fewer than 140 are in favor

of government regulation is 0.0436