1. Chapter 6
Created by Bethany Stubbe and Stephan Kogitz

2. The Normal Probability Distribution
Chapter Six
The Normal Probability Distribution
GOALS
When you have completed this chapter, you will be able to:
ONE List the characteristics of the normal probability distribution.
TWO Define and calculate z values.
THREE Determine the probability an observation will lie between two points using the standard normal distribution.
FOUR Determine the probability an observation will be above or below a given value using the standard normal distribution.

3. The Normal Probability Distribution
Chapter Six continued
The Normal Probability Distribution
GOALS
When you have completed this chapter, you will be able to:
FIVE Compare two or more observations that are on different probability distributions.
SIX Use the normal distribution to approximate the binomial probability distribution.

4. Characteristics of a Normal Probability Distribution
The characteristics are:
It is bell-shaped and has a single peak at the exact center of the distribution.
The arithmetic mean, median, and mode are equal and located at the peak.
Half the area under the curve is above the mean and half is below it.

5. Characteristics of a Normal Probability Distribution
The normal probability distribution is symmetrical about its mean.
The normal probability distribution is asymptotic. That is the curve gets closer and closer to the X-axis but never actually touches it.

6. Characteristics of a Normal Distribution- 5 . 4 3 2 1 x f ( r a l i t b u o n : m = , s2
Characteristics of a Normal Distribution
Normal
curve is
symmetrical
Theoretically,
curve
extends to
infinity a
Mean, median, and
mode are equal
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Copyright © 2003 by The McGraw-Hill Companies, Inc. All rights reserved

7. The Standard Normal Probability Distribution
The standard normal distribution has mean of 0 and a standard deviation of 1. It is also called the z distribution.

8. The Standard Normal Probability Distribution
A z-value or standard normal value is the distance between a selected value, designated X, and the population mean μ, divided by the population standard deviation, σ. The formula is:

9. EXAMPLE 1
The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $2,000 and a standard deviation of $200. What is the z-value for a salary of $2,200?

10. EXAMPLE 1 continued What is the z-value corresponding to $1,700?
A z-value of 1 indicates $2,200 is one standard deviation above the mean of $2,000. A z-value of –1.50 indicates that $1,700 is 1.5 standard deviation below the mean of $2000.

11. Areas Under the Normal Curve
About 68 percent of the area under the normal curve is within one standard deviation of the mean.
μ ± σ
About 95 percent is within two standard deviations of the mean.
μ± 2σ
Practically all is within three standard deviations of the mean.
μ ± 3σ

12. Areas Under the Normal Curve- 5 . 4 3 2 1 x f ( r a l i t b u o n : m = , s2
Areas Under the Normal Curve
Between:
±1σ %
±2σ %
±3σ %
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Copyright © 2003 by The McGraw-Hill Companies, Inc.,.All rights reserved

13. EXAMPLE 2
The monthly water usage per person in a New Brunswick town follows the a normal distribution with a mean of 20 cubic metres and a standard deviation of 5 cubic metres. About 68 percent of those living in the town will use how many cubic metres of water?
About 68% of the monthly water usage will lie between 15 and 25 cubic metres .

14. EXAMPLE 3
What is the probability that a person from the town selected at random will use between 20 and 24 cubic metres per month?

15. Example 3 continued
The area under a normal curve between a z-value of 0 and a z-value of 0.80 is
We conclude that percent of the residents use between 20 and 24 cubic metres of water per month.
See the following diagram.

16. EXAMPLE 3 P(0<z<.8) =.2881 0<x<.8 -4 -3 -2 -1 0 1 2 3 45 . 4 3 2 1 x f ( r a l i t b u o n : m = ,
EXAMPLE 3
P(0<z<.8)
=.2881
0<x<.8
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McGraw-Hill/Irwin

17. EXAMPLE 3 continued
What percent of the population use between 18 and 26 cubic metres per month?

18. Example 3 continued
The area associated with a z-value of –0.40 is
The area associated with a z-value of 1.20 is
Adding these areas, the result is
We conclude that percent of the residents use between 18 and 26 cubic metres of water per month.

19. EXAMPLE 4
Professor Mann reports the scores on Test 1 in his statistics course follow a normal distribution with a mean of 72 and a standard deviation of 5. He announces to the class that the top 15 percent of the scores will earn an A. What is the lowest score a student can earn and still receive an A?

20. Example 4 continued
To begin let X be the score that separates an A from a B.
If 15 percent of the students score more than X, then 35 percent must score between the mean of 72 and X.
The z-value associated corresponding to 35 percent is about See Appendix D.

21. Example 4 continued
We let z equal 1.04 and solve the standard normal equation for X. The result is the score that separates students that earned an A from those that earned a B.
Those with a score of 77.2 or more earn an A.

22. The Normal Approximation to the Binomial
The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n.

23. The Normal Approximation to the Binomial
The normal probability distribution is a good approximation to the binomial probability distribution when n π and n(1- π ) are both greater than 5.

24. The Normal Approximation continued
Recall for the binomial experiment:
There are only two mutually exclusive outcomes (success or failure) on each trial.
A binomial distribution results from counting the number of successes.
Each trial is independent.
The probability is fixed from trial to trial, and the number of trials n is also fixed.

25. Continuity Correction Factor
The value .5 subtracted or added, depending on the question, to a selected value when a discrete probability distribution is approximated by a continuous probability distribution.

26. EXAMPLE 5 This is the mean of a binomial distribution.
According to Statistics Canada, 15% of Canadian households used electronic banking in the year For a sample of 200 homes, how many of the households would you expect to have used electronic banking?
This is the mean of a binomial distribution.

27. EXAMPLE 5 continued Calculate the variance?
Calculate the standard deviation?

28. Example 5 continued
What is the probability that less than 40 households in the sample have used electronic banking?
We use the correction factor, so X is 39.5.
The value of z is 1.88.

29. Example 5 continued
From Appendix D the area between 0 and 1.88 is
So the area to the left of 1.88 is =
The likelihood that less than 40 of the 200 households have used electronic banking is about 97%.

30. EXAMPLE 5 P(z<1.88) =.5000+.4699 =.9699 z=1.88 0 1 2 3 4- 5 . 4 3 2 1 f ( x r a l i t b u o n : m = , s2
EXAMPLE 5
P(z<1.88) =
=.9699
z=1.88
Copyright ©2003 by The McGraw-Hill Companies, Inc., . All rights reserved
McGraw-Hill/Irwin