7- 1 Chapter Seven The Normal Probability The Normal Probability Distribution 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
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Chapter SevenThe Normal Probability The Normal Probability DistributionDistributionGOALS
When you have completed this chapter, you will be able to:ONEList the characteristics of the normal probability distribution.
TWO Define and calculate z values.
THREEDetermine the probability an observation will lie between two points using the standard normal distribution.
FOURDetermine the probability an observation will be above or below a given value using the standard normal distribution.
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Chapter Seven continued
GOALSWhen you have completed this chapter, you will be able to:
FIVE Compare two or more observations that are on different probability distributions.
SIXUse the normal distribution to approximate the binomial probability distribution.
The Normal Probability The Normal Probability DistributionDistribution
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Characteristics of a Normal Probability Distribution
• The normal curve is bell-shaped and has a single peak at the exact center of the distribution.
The arithmetic mean, median, and mode of the distribution are equal and located at the peak. Thus half the area under the curve is above the mean and half is below it.
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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.
• The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.
• It is also called the z distribution.
X
z
A z-value is the distance between a selected value, designated X, and the population mean , divided by the population standard deviation, . The formula is:
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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 $2,200?
00.2200$
000,2$200,2$
X
z
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EXAMPLE 1 continued
• A z-value of 1 indicates that the value of $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 $2000.
50.1200$
200,2$700,1$
X
z
What is the z-value of $1,700.
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Areas Under the Normal Curve
• About 68 percent of the area under the normal curve is within one standard deviation of the mean.
• About 95 percent is within two standard deviations
of the mean. 2
• Practically all is within three standard deviations of the mean.
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0 . 4
0 . 3
0 . 2
0 . 1
. 0
x
f(
x
r a l i t r b u i o n : = 0 , = 1Areas Under the Normal Curve
EXAMPLE 2The daily water usage per person in New Providence, New Jersey is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons. About 68 percent of those living in New Providence will use how many gallons of water?
• About 68% of the daily water usage will lie between 15 and 25 gallons.
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EXAMPLE 3• What is the probability that a person from New
Providence selected at random will use between 20 and 24 gallons per day?
00.05
2020
X
z
80.05
2024
X
z
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Example 3 continued
• The area under a normal curve between a z-value of 0 and a z-value of 0.80 is 0.2881.
• We conclude that 28.81 percent of the residents use between 20 and 24 gallons of water per day.
• What percent of the population use between 18 and 26 gallons per day?
40.05
2018
X
z
20.15
2026
X
z
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Example 3 continued
• The area associated with a z-value of –0.40 is .1554.
• The area associated with a z-value of 1.20 is .3849.
• Adding these areas, the result is .5403.
• We conclude that 54.03 percent of the residents use between 18 and 26 gallons of water per day.
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EXAMPLE 4• Professor Mann has determined that the scores
in his statistics course are approximately normally distributed 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?
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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 1.04.
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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.
2.772.572)5(04.1725
7204.1
X
X
Those with a score of 77.2 or more earn an A.
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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.
The normal probability distribution is generally a good approximation to the binomial probability distribution when n and n(1- ) are both greater than 5.
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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.
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Continuity Correction Factor• The value .5 subtracted or added, depending
on the problem, to a selected value when a binomial probability distribution (a discrete probability distribution) is being approximated by a continuous probability distribution (the normal distribution).
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EXAMPLE 5A recent study by a marketing research firm showed that 15% of American households owned a video camera. For a sample of 200 homes, how many of the homes would you expect to have video cameras?
This is the mean of a binomial distribution.
n (. )( )15 200 30
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EXAMPLE 5 continued
• What is the variance?
2 1 30 1 15 255 n ( ) ( )( . ) .
0498.55.25
What is the standard deviation?
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Example 5
What is the probability that less than 40 homes in the sample have video cameras?
• We use the correction factor, so X is 39.5. • The value of z is 1.88.
88.10498.5
0.305.39
X
z
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Example 5 continued
• From Appendix D the area between 0 and 1.88 on the z scale is .4699.
• So the area to the left of 1.88 is .5000 + .4699 = .9699.
• The likelihood that less than 40 of the 200 homes have a video camera is about 97%.
