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Normal Probability Distributions

Chapter Outline

5.1 Introduction to Normal Distributions and the
Standard Normal Distribution
5.2 Normal Distributions: Finding Probabilities
5.3 Normal Distributions: Finding Values
5.4 Sampling Distributions and the Central Limit
Theorem
5.5 Normal Approximations to Binomial
Distributions

Larson/Farber 4th ed

Section 5.1

Introduction to Normal Distributions

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Section 5.1 Objectives

Interpret graphs of normal probability distributions
Find areas under the standard normal curve

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Properties of a Normal Distribution

Continuous random variable

Has an infinite number of possible values that can be represented by an interval on the number line.

Continuous probability distribution

The probability distribution of a continuous random variable.

The time spent studying can be any number between 0 and 24.

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Hours spent studying in a day

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Properties of Normal Distributions

Normal distribution

A continuous probability distribution for a random variable, x.
The most important continuous probability distribution in statistics.
The graph of a normal distribution is called the normal curve.

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Properties of Normal Distributions

The mean, median, and mode are equal.

The normal curve is bell-shaped and symmetric about the mean.

The total area under the curve is equal to one.

The normal curve approaches, but never touches the x-axis as it extends farther and farther away from the mean.

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Total area = 1

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Properties of Normal Distributions

Between μ – σ and μ + σ (in the center of the curve), the graph curves downward. The graph curves upward to the left of μ – σ and to the right of μ + σ. The points at which the curve changes from curving upward to curving downward are called the inflection points.

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μ  3σ

μ + σ

μ  2σ

μ  σ

μ + 2σ

μ + 3σ

Inflection points

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Means and Standard Deviations

A normal distribution can have any mean and any positive standard deviation.
The mean gives the location of the line of symmetry.
The standard deviation describes the spread of the data.

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μ = 3.5

σ = 1.5

μ = 3.5

σ = 0.7

μ = 1.5

σ = 0.7

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Example: Understanding Mean and Standard Deviation

Which curve has the greater mean?

Solution:

Curve A has the greater mean (The line of symmetry of curve A occurs at x = 15. The line of symmetry of curve B occurs at x = 12.)

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Example: Understanding Mean and Standard Deviation

Which curve has the greater standard deviation?

Solution:

Curve B has the greater standard deviation (Curve B is more spread out than curve A.)

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Example: Interpreting Graphs

The heights of fully grown white oak trees are normally distributed. The curve represents the distribution. What is the mean height of a fully grown white oak tree? Estimate the standard deviation.

Solution:

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μ = 90 (A normal curve is symmetric about the mean)

σ = 3.5 (The inflection points are one standard deviation away from the mean)

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The Standard Normal Distribution

Standard normal distribution

A normal distribution with a mean of 0 and a standard deviation of 1.

Any x-value can be transformed into a z-score by using the formula

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3

2

1

Area = 1

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The Standard Normal Distribution

If each data value of a normally distributed random variable x is transformed into a z-score, the result will be the standard normal distribution.

Use the Standard Normal Table to find the cumulative area under the standard normal curve.

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Normal Distribution

m=0

s=1

Standard Normal Distribution

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Properties of the Standard Normal Distribution

The cumulative area is close to 0 for z-scores close to z = 3.49.

The cumulative area increases as the z-scores increase.

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z = 3.49

Area is close to 0

3

2

1

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Properties of the Standard Normal Distribution

The cumulative area for z = 0 is 0.5000.

The cumulative area is close to 1 for z-scores close to z = 3.49.

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z = 3.49

Area is close to 1

Area is 0.5000

z = 0

3

2

1

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Example: Using The Standard Normal Table

Find the cumulative area that corresponds to a z-score of 1.15.

The area to the left of z = 1.15 is 0.8749.

Move across the row to the column under 0.05

Solution:

Find 1.1 in the left hand column.

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Example: Using The Standard Normal Table

Find the cumulative area that corresponds to a z-score of -0.24.

Solution:

Find -0.2 in the left hand column.

The area to the left of z = -0.24 is 0.4052.

Move across the row to the column under 0.04

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Finding Areas Under the Standard Normal Curve

Sketch the standard normal curve and shade the appropriate area under the curve.

Find the area by following the directions for each case shown.

To find the area to the left of z, find the area that corresponds to z in the Standard Normal Table.

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Use the table to find the area for the z-score

The area to the left of z = 1.23 is 0.8907

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Finding Areas Under the Standard Normal Curve

To find the area to the right of z, use the Standard Normal Table to find the area that corresponds to z. Then subtract the area from 1.

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Subtract to find the area to the right of z = 1.23: 1  0.8907 = 0.1093.

Use the table to find the area for the z-score.

The area to the left of z = 1.23 is 0.8907.

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Finding Areas Under the Standard Normal Curve

To find the area between two z-scores, find the area corresponding to each z-score in the Standard Normal Table. Then subtract the smaller area from the larger area.

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Subtract to find the area of the region between the two z-scores: 0.8907  0.2266 = 0.6641.

The area to the left of z = 0.75 is 0.2266.

The area to the left of z = 1.23 is 0.8907.

Use the table to find the area for the z-scores.

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Example: Finding Area Under the Standard Normal Curve

Find the area under the standard normal curve to the left of z = -0.99.

From the Standard Normal Table, the area is equal to 0.1611.

Solution:

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0.99

0.1611

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Example: Finding Area Under the Standard Normal Curve

Find the area under the standard normal curve to the right of z = 1.06.

From the Standard Normal Table, the area is equal to 0.1446.

Solution:

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1  0.8554 = 0.1446

1.06

0.8554

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Find the area under the standard normal curve between z = 1.5 and z = 1.25.

Example: Finding Area Under the Standard Normal Curve

From the Standard Normal Table, the area is equal to 0.8276.

Solution:

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1.25

1.50

0.8944

0.0668

0.8944 0.0668 = 0.8276

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Section 5.1 Summary

Interpreted graphs of normal probability distributions
Found areas under the standard normal curve

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Section 5.2

Normal Distributions: Finding Probabilities

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Section 5.2 Objectives

Find probabilities for normally distributed variables

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Probability and Normal Distributions

If a random variable x is normally distributed, you can find the probability that x will fall in a given interval by calculating the area under the normal curve for that interval.

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P(x < 600) = Area

600

μ =500

μ = 500

σ = 100

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Probability and Normal Distributions

P(x < 500) = P(z < 1)

Normal Distribution

Standard Normal Distribution

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Same Area

600

μ =500

P(x < 600)

μ = 500 σ = 100

μ = 0

μ = 0 σ = 1

P(z < 1)

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Example: Finding Probabilities for Normal Distributions

A survey indicates that people use their computers an average of 2.4 years before upgrading to a new machine. The standard deviation is 0.5 year. A computer owner is selected at random. Find the probability that he or she will use it for fewer than 2 years before upgrading. Assume that the variable x is normally distributed.

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Solution: Finding Probabilities for Normal Distributions

P(x < 2) = P(z < -0.80) = 0.2119

Normal Distribution

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2.4

P(x < 2)

μ = 2.4 σ = 0.5

-0.80

μ = 0 σ = 1

P(z < -0.80)

Standard Normal Distribution

0.2119

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Example: Finding Probabilities for Normal Distributions

A survey indicates that for each trip to the supermarket, a shopper spends an average of 45 minutes with a standard deviation of 12 minutes in the store. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store. Find the probability that the shopper will be in the store for between 24 and 54 minutes.

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Solution: Finding Probabilities for Normal Distributions

P(24 < x < 54) = P(-1.75 < z < 0.75)

= 0.7734 – 0.0401 = 0.7333

Normal Distribution
μ = 45 σ = 12

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P(24 < x < 54)

0.0401

-1.75

P(-1.75 < z < 0.75)

Standard Normal Distribution
μ = 0 σ = 1

0.75

0.7734

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Example: Finding Probabilities for Normal Distributions

Find the probability that the shopper will be in the store more than 39 minutes. (Recall μ = 45 minutes and
σ = 12 minutes)

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Solution: Finding Probabilities for Normal Distributions

P(x > 39) = P(z > -0.50) = 1– 0.3085 = 0.6915

Normal Distribution
μ = 45 σ = 12

Standard Normal Distribution
μ = 0 σ = 1

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P(x > 39)

0.3085

P(z > -0.50)

-0.50

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Example: Finding Probabilities for Normal Distributions

If 200 shoppers enter the store, how many shoppers would you expect to be in the store more than 39 minutes?

Solution:

Recall P(x > 39) = 0.6915

200(0.6915) =138.3 (or about 138) shoppers

Larson/Farber 4th ed

Example: Using Technology to find Normal Probabilities

Assume that cholesterol levels of men in the United States are normally distributed, with a mean of 215 milligrams per deciliter and a standard deviation of 25 milligrams per deciliter. You randomly select a man from the United States. What is the probability that his cholesterol level is less than 175? Use a technology tool to find the probability.

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Solution: Using Technology to find Normal Probabilities

Must specify the mean, standard deviation, and the x-value(s) that determine the interval.

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Section 5.2 Summary

Found probabilities for normally distributed variables

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Section 5.3

Normal Distributions: Finding Values

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Section 5.3 Objectives

Find a z-score given the area under the normal curve
Transform a z-score to an x-value
Find a specific data value of a normal distribution given the probability

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Finding values Given a Probability

In section 5.2 we were given a normally distributed random variable x and we were asked to find a probability.
In this section, we will be given a probability and we will be asked to find the value of the random variable x.

probability

5.2

5.3

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Example: Finding a z-Score Given an Area

Find the z-score that corresponds to a cumulative area of 0.3632.

Solution:

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0.3632

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Solution: Finding a z-Score Given an Area

Locate 0.3632 in the body of the Standard Normal Table.

The values at the beginning of the corresponding row and at the top of the column give the z-score.

The z-score is -0.35.

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Example: Finding a z-Score Given an Area

Find the z-score that has 10.75% of the distribution’s area to its right.

Solution:

Because the area to the right is 0.1075, the cumulative area is 0.8925.

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0.1075

1 – 0.1075

= 0.8925

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Solution: Finding a z-Score Given an Area

Locate 0.8925 in the body of the Standard Normal Table.

The values at the beginning of the corresponding row and at the top of the column give the z-score.

The z-score is 1.24.

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Example: Finding a z-Score Given a Percentile

Find the z-score that corresponds to P5.

Solution:

The z-score that corresponds to P5 is the same z-score that corresponds to an area of 0.05.

The areas closest to 0.05 in the table are 0.0495 (z = -1.65) and 0.0505 (z = -1.64). Because 0.05 is halfway between the two areas in the table, use the z-score that is halfway between -1.64 and -1.65. The z-score is -1.645.

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0.05

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Transforming a z-Score to an x-Score

To transform a standard z-score to a data value x in a given population, use the formula

x = μ + zσ

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Example: Finding an x-Value

The speeds of vehicles along a stretch of highway are normally distributed, with a mean of 67 miles per hour and a standard deviation of 4 miles per hour. Find the speeds x corresponding to z-sores of 1.96, -2.33, and 0.

Solution: Use the formula x = μ + zσ

z = 1.96: x = 67 + 1.96(4) = 74.84 miles per hour
z = -2.33: x = 67 + (-2.33)(4) = 57.68 miles per hour
z = 0: x = 67 + 0(4) = 67 miles per hour

Notice 74.84 mph is above the mean, 57.68 mph is below the mean, and 67 mph is equal to the mean.

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Example: Finding a Specific Data Value

Scores for a civil service exam are normally distributed, with a mean of 75 and a standard deviation of 6.5. To be eligible for civil service employment, you must score in the top 5%. What is the lowest score you can earn and still be eligible for employment?

Solution:

An exam score in the top 5% is any score above the 95th percentile. Find the z-score that corresponds to a cumulative area of 0.95.

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1 – 0.05

= 0.95

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Solution: Finding a Specific Data Value

From the Standard Normal Table, the areas closest to 0.95 are 0.9495 (z = 1.64) and 0.9505 (z = 1.65). Because 0.95 is halfway between the two areas in the table, use the z-score that is halfway between 1.64 and 1.65. That is, z = 1.645.

1.645

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Solution: Finding a Specific Data Value

Using the equation x = μ + zσ

x = 75 + 1.645(6.5) ≈ 85.69

1.645

The lowest score you can earn and still be eligible for employment is 86.

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85.69

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Section 5.3 Summary

Found a z-score given the area under the normal curve
Transformed a z-score to an x-value
Found a specific data value of a normal distribution given the probability

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Section 5.4

Sampling Distributions and the Central Limit Theorem

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Section 5.4 Objectives

Find sampling distributions and verify their properties
Interpret the Central Limit Theorem
Apply the Central Limit Theorem to find the probability of a sample mean

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Sampling Distributions

Sampling distribution

The probability distribution of a sample statistic.
Formed when samples of size n are repeatedly taken from a population.
e.g. Sampling distribution of sample means

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Sampling Distribution of Sample Means

Population with μ, σ

The sampling distribution consists of the values of the sample means,

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Sample 1

Sample 5

Sample 2

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The standard deviation of the sample means, , is equal to the population standard deviation, σ divided by the square root of the sample size, n.

The mean of the sample means, , is equal to the population mean μ.

Properties of Sampling Distributions of Sample Means

Called the standard error of the mean.

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Example: Sampling Distribution of Sample Means

The population values {1, 3, 5, 7} are written on slips of paper and put in a box. Two slips of paper are randomly selected, with replacement.

Find the mean, variance, and standard deviation of the population.

Solution:

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Example: Sampling Distribution of Sample Means

Graph the probability histogram for the population values.

All values have the same probability of being selected (uniform distribution)

Solution:

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Population values

Probability

0.25

P(x)

Probability Histogram of Population of x

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Example: Sampling Distribution of Sample Means

List all the possible samples of size n = 2 and calculate the mean of each sample.

3, 7

3, 5

3, 3

3, 1

1, 7

1, 5

1, 3

1, 1

7, 7

7, 5

7, 3

7, 1

5, 7

5, 5

5, 3

5, 1

These means form the sampling distribution of sample means.

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Sample

Solution:

Sample

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Example: Sampling Distribution of Sample Means

Construct the probability distribution of the sample means.

Probability

Solution:

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f	Probability
1	1	0.0625
2	2	0.1250
3	3	0.1875
4	4	0.2500
5	3	0.1875
6	2	0.1250
7	1	0.0625

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Example: Sampling Distribution of Sample Means

Find the mean, variance, and standard deviation of the sampling distribution of the sample means.

Solution:

The mean, variance, and standard deviation of the 16 sample means are:

These results satisfy the properties of sampling distributions of sample means.

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Example: Sampling Distribution of Sample Means

Graph the probability histogram for the sampling distribution of the sample means.

The shape of the graph is symmetric and bell shaped. It approximates a normal distribution.

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Sample mean

Probability

0.25

P(x)

Probability Histogram of Sampling Distribution of

0.20

0.15

0.10

0.05

Solution:

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The Central Limit Theorem

If samples of size n  30, are drawn from any population with mean =  and standard deviation = ,

then the sampling distribution of the sample means approximates a normal distribution. The greater the sample size, the better the approximation.

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The Central Limit Theorem

If the population itself is normally distributed,

the sampling distribution of the sample means is normally distribution for any sample size n.

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The Central Limit Theorem

In either case, the sampling distribution of sample means has a mean equal to the population mean.

The sampling distribution of sample means has a variance equal to 1/n times the variance of the population and a standard deviation equal to the population standard deviation divided by the square root of n.

Variance

Standard deviation (standard error of the mean)

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The Central Limit Theorem

Any Population Distribution

Normal Population Distribution

Distribution of Sample Means, n ≥ 30

Distribution of Sample Means, (any n)

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Example: Interpreting the Central Limit Theorem

Phone bills for residents of a city have a mean of $64 and a standard deviation of $9. Random samples of 36 phone bills are drawn from this population and the mean of each sample is determined. Find the mean and standard error of the mean of the sampling distribution. Then sketch a graph of the sampling distribution of sample means.

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Solution: Interpreting the Central Limit Theorem

The mean of the sampling distribution is equal to the population mean

The standard error of the mean is equal to the population standard deviation divided by the square root of n.

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Solution: Interpreting the Central Limit Theorem

Since the sample size is greater than 30, the sampling distribution can be approximated by a normal distribution with

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Example: Interpreting the Central Limit Theorem

The heights of fully grown white oak trees are normally distributed, with a mean of 90 feet and standard deviation of 3.5 feet. Random samples of size 4 are drawn from this population, and the mean of each sample is determined. Find the mean and standard error of the mean of the sampling distribution. Then sketch a graph of the sampling distribution of sample means.

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Solution: Interpreting the Central Limit Theorem

The mean of the sampling distribution is equal to the population mean

The standard error of the mean is equal to the population standard deviation divided by the square root of n.

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Solution: Interpreting the Central Limit Theorem

Since the population is normally distributed, the sampling distribution of the sample means is also normally distributed.

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Probability and the Central Limit Theorem

To transform x to a z-score

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Example: Probabilities for Sampling Distributions

The graph shows the length of time people spend driving each day. You randomly select 50 drivers age 15 to 19. What is the probability that the mean time they spend driving each day is between 24.7 and 25.5 minutes? Assume that σ = 1.5 minutes.

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Solution: Probabilities for Sampling Distributions

From the Central Limit Theorem (sample size is greater than 30), the sampling distribution of sample means is approximately normal with

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Solution: Probabilities for Sampling Distributions

Normal Distribution
μ = 25 σ = 0.21213

25.5

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P(24.7 < x < 25.5)

24.7

-1.41

P(-1.41 < z < 2.36)

Standard Normal Distribution
μ = 0 σ = 1

2.36

0.9909

0.0793

P(24 < x < 54) = P(-1.41 < z < 2.36)

= 0.9909 – 0.0793 = 0.9116

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Example: Probabilities for x and x

A bank auditor claims that credit card balances are normally distributed, with a mean of $2870 and a standard deviation of $900.

Solution:

You are asked to find the probability associated with a certain value of the random variable x.

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What is the probability that a randomly selected credit card holder has a credit card balance less than $2500?

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Solution: Probabilities for x and x

P( x < 2500) = P(z < -0.41) = 0.3409

Normal Distribution
μ = 2870 σ = 900

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2870

P(x < 2500)

2500

-0.41

P(z < -0.41)

Standard Normal Distribution
μ = 0 σ = 1

0.3409

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Example: Probabilities for x and x

You randomly select 25 credit card holders. What is the probability that their mean credit card balance is less than $2500?

Solution:

You are asked to find the probability associated with a sample mean .

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Solution: Probabilities for x and x

Normal Distribution
μ = 2870 σ = 180

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-2.06

P(z < -2.06)

Standard Normal Distribution
μ = 0 σ = 1

0.0197

2870

P(x < 2500)

2500

P( x < 2500) = P(z < -2.06) = 0.0197

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Solution: Probabilities for x and x

There is a 34% chance that an individual will have a balance less than $2500.
There is only a 2% chance that the mean of a sample of 25 will have a balance less than $2500 (unusual event).
It is possible that the sample is unusual or it is possible that the auditor’s claim that the mean is $2870 is incorrect.

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Section 5.4 Summary

Found sampling distributions and verify their properties
Interpreted the Central Limit Theorem
Applied the Central Limit Theorem to find the probability of a sample mean

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Section 5.5

Normal Approximations to Binomial Distributions

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Section 5.5 Objectives

Determine when the normal distribution can approximate the binomial distribution
Find the correction for continuity
Use the normal distribution to approximate binomial probabilities

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Normal Approximation to a Binomial

Normal Approximation to a Binomial Distribution

If np  5 and nq  5, then the binomial random variable x is approximately normally distributed with
mean μ = np
standard deviation

The normal distribution is used to approximate the binomial distribution when it would be impractical to use the binomial distribution to find a probability.

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Normal Approximation to a Binomial

Binomial distribution: p = 0.25

As n increases the histogram approaches a normal curve.

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Fifty-one percent of adults in the U.S. whose New Year’s resolution was to exercise more achieved their resolution. You randomly select 65 adults in the U.S. whose resolution was to exercise more and ask each if he or she achieved that resolution.

Example: Approximating the Binomial

Decide whether you can use the normal distribution to approximate x, the number of people who reply yes. If you can, find the mean and standard deviation.

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Solution: Approximating the Binomial

You can use the normal approximation

n = 65, p = 0.51, q = 0.49

np = (65)(0.51) = 33.15 ≥ 5

nq = (65)(0.49) = 31.85 ≥ 5

Mean: μ = np = 33.15
Standard Deviation:

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Fifteen percent of adults in the U.S. do not make New Year’s resolutions. You randomly select 15 adults in the U.S. and ask each if he or she made a New Year’s resolution.

Example: Approximating the Binomial

Decide whether you can use the normal distribution to approximate x, the number of people who reply yes. If you can find, find the mean and standard deviation.

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Solution: Approximating the Binomial

You cannot use the normal approximation

n = 15, p = 0.15, q = 0.85

np = (15)(0.15) = 2.25 < 5

nq = (15)(0.85) = 12.75 ≥ 5

Because np < 5, you cannot use the normal distribution to approximate the distribution of x.

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Correction for Continuity

The binomial distribution is discrete and can be represented by a probability histogram.
To calculate exact binomial probabilities, the binomial formula is used for each value of x and the results are added.
Geometrically this corresponds to adding the areas of bars in the probability histogram.

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Correction for Continuity

When you use a continuous normal distribution to approximate a binomial probability, you need to move 0.5 unit to the left and right of the midpoint to include all possible x-values in the interval (correction for continuity).

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Exact binomial probability

P(x = c)

Normal approximation

P(c – 0.5 < x < c + 0.5)

c+ 0.5

c– 0.5

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Example: Using a Correction for Continuity

Use a correction for continuity to convert the binomial intervals to a normal distribution interval.

The probability of getting between 270 and 310 successes, inclusive.

Solution:

The discrete midpoint values are 270, 271, …, 310.
The corresponding interval for the continuous normal distribution is

269.5 < x < 310.5

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Example: Using a Correction for Continuity

Use a correction for continuity to convert the binomial intervals to a normal distribution interval.

The probability of getting at least 158 successes.

Solution:

The discrete midpoint values are 158, 159, 160, ….
The corresponding interval for the continuous normal distribution is

x > 157.5

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Example: Using a Correction for Continuity

Use a correction for continuity to convert the binomial intervals to a normal distribution interval.

The probability of getting less than 63 successes.

Solution:

The discrete midpoint values are …,60, 61, 62.
The corresponding interval for the continuous normal distribution is

x < 62.5

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Using the Normal Distribution to Approximate Binomial Probabilities

Verify that the binomial distribution applies.

Determine if you can use the normal distribution to approximate x, the binomial variable.

Find the mean  and standard deviation for the distribution.

Is np  5?
Is nq  5?

Specify n, p, and q.

Larson/Farber 4th ed

In Words In Symbols

Larson/Farber 4th ed

Using the Normal Distribution to Approximate Binomial Probabilities

Apply the appropriate continuity correction. Shade the corresponding area under the normal curve.

Find the corresponding z-score(s).

Find the probability.

Add or subtract 0.5 from endpoints.

Use the Standard Normal Table.

Larson/Farber 4th ed

In Words In Symbols

Larson/Farber 4th ed

Example: Approximating a Binomial Probability

Fifty-one percent of adults in the U. S. whose New Year’s resolution was to exercise more achieved their resolution. You randomly select 65 adults in the U. S. whose resolution was to exercise more and ask each if he or she achieved that resolution. What is the probability that fewer than forty of them respond yes? (Source: Opinion Research Corporation)

Solution:

Can use the normal approximation (see slide 89)

μ = 65∙0.51 = 33.15

Larson/Farber 4th ed

Solution: Approximating a Binomial Probability

Apply the continuity correction:

Fewer than 40 (…37, 38, 39) corresponds to the continuous normal distribution interval x < 39.5

P(z < 1.58) = 0.9429

Larson/Farber 4th ed

39.5

μ =33.15

P(x < 39.5)

Normal Distribution

μ = 33.15 σ = 4.03

1.58

μ =0

P(z < 1.58)

Standard Normal

μ = 0 σ = 1

0.9429

Larson/Farber 4th ed

Example: Approximating a Binomial Probability

A survey reports that 86% of Internet users use Windows® Internet Explorer ® as their browser. You randomly select 200 Internet users and ask each whether he or she uses Internet Explorer as his or her browser. What is the probability that exactly 176 will say yes? (Source: 0neStat.com)

Solution:

Can use the normal approximation

np = (200)(0.86) = 172 ≥ 5 nq = (200)(0.14) = 28 ≥ 5

μ = 200∙0.86 = 172

Larson/Farber 4th ed

Apply the continuity correction:

Exactly 176 corresponds to the continuous normal distribution interval 175.5 < x < 176.5

Solution: Approximating a Binomial Probability

P(0.71 < z < 0.92) = 0.8212 – 0.7611 = 0.0601

Larson/Farber 4th ed

0.7611

0.92

μ =0

P(0.71 < z < 0.92)

Standard Normal

μ = 0 σ = 1

0.71

0.8212

176.5

μ =172

P(175.5 < x < 176.5)

Normal Distribution

μ = 172 σ = 4.91

175.5

Larson/Farber 4th ed

Section 5.5 Summary

Determined when the normal distribution can approximate the binomial distribution
Found the correction for continuity
Used the normal distribution to approximate binomial probabilities

Larson/Farber 4th ed

Value-Mean

Standard deviation

600500

100

===

22.4

0.80

0.5

===-

2445

175

===

5445

075

===

3945

050

===

12345

,,,,,...

xxxxx

Mean:

Varianc:

()

S-

Standard Deviat

ion

236

=»

251581

=»

52236

1581

==»»

1.5

===

1.5

3.5

1.75

===

1.75

Value-Mean

StandardError

===

1.5

0.21213

==»

24725

141

===

25525

236

===

25002870

041

900

==»

2870

900

180

===

25002870

206

900

==»

σnpq

650.510.494.03

==××»

σnpq

npq

650.510.494.03

=××»

3953315

158

403

==»

.-.

2000.860.144.91

=××»

1755172

071

491

==»

1765172

092

491

==»