chapter 5 normal probability distributions. chapter 5 normal probability distributions section 5-1...

Chapter 5

Normal Probability Distributions

Chapter 5Normal Probability DistributionsSection 5-1 – Introduction to Normal Distributions and the Standard Normal DistributionA. The normal distribution is the most important of the continuous

probability distributions.1. Definition: A normal distribution is a continuous

probability distribution for a random variable x.

a. The graph of a normal distribution is called a normal curve.

2. A normal distribution has the following properties:

a. The mean, median, and mode are equal (or VERY close to

equal).b. The normal curve is bell-

shaped and is symmetric about the mean.

c. The total area under the normal curve is equal to one.

d. The normal curve approaches, but never touches, the x-axis

as it gets further away from the mean.

Chapter 5Normal Probability DistributionsSection 5-1 – Introduction to Normal Distributions and the Standard Normal Distribution

e. The graph curves downward within one standard deviation of

the mean, and it curves upward outside of one standard

deviation from the mean.1) The

points where the curve changes from curving

upward to curving downward are called inflection

points.B. We know that a discrete probability can be graphed with a

histogram (although we didn’t emphasize this in Chapter 4).

1. For a continuous probability distribution, you can use a

probability density function (pdf).a. A probability density function

has two requirements:1) The total area under

the curve has to equal one.2) The function

can never be negative.


b. We can graph a normal curve with the mean of μ and

standard deviation of σ by using the normal probability

density function:

1) Notice that this entire equation depends completely on

what μ and σ are, since e and π are constants.

2. A normal distribution can have ANY mean and ANY POSITIVE standard deviation.

a. These two parameters completely determine the shape of the

normal curve.1) The mean

gives the axis of symmetry.2) The standard

deviation describes how spread out (or

bunched up) the data is.

Chapter 5Normal Probability DistributionsSection 5-1 – Introduction to Normal Distributions and the Standard Normal DistributionC. The Standard Normal Curve

1. There are infinitely many normal distributions, because there are infinitely many possible combinations of means and standard

deviations.a. The standard normal

distribution has a mean of zero and a standard deviation of 1.

1) The horizontal scale of the graph of the standard normal

distribution corresponds to z-scores.

a)Remember that z-scores are measures of position that

indicate the number of standard deviations values lie

away from the mean.

1. z = x minus the mean over the standard deviation, or


2. The standard normal distribution has the following properties:

a. The cumulative area under the curve is close to 0 for z-scores

close to -3.49.b. The cumulative area increases

as the z- scores increase.c. The cumulative area for z = 0 is

0.5000.d. The cumulative area is close to

1 for z-scores close to 3.49.3. To find the corresponding area under the curve

for any given (or calculated) z-score, there are two main methods.

a. The easiest, and the one I suggest, is to use the TI-84

calculator.1) 2nd VARS

normalcdf (lower boundary, upper boundary)a)

If you want the area to the left of a z-score, use -10,000

as your lower boundary and the z-score you are

interested in as your upper boundary.


b)If you want the area to the right of a z-score, use z-

score you are interested in as your lower boundary and

use 10,000 as your upper boundary.c)

If you want the area between two z-scores, use them

both (smaller one as lower, larger one as upper).b. The other way, which will be

demonstrated in class, is to use the z-score chart.

1) You should have done this in Algebra 2, so it should be a

quick review for us.


4. Remember that in Section 2.4 we learned from the Empirical Rule that values lying more than 2 standard deviations from the

mean are considered to be unusual. a. We also learned that values

lying more than 3 standard deviations from mean are very

unusual, or outliers. b. In terms of z-scores, this

means that a z-score of less than -2 or greater than 2 means an unusual event.

1) A z-score of less than -3 or greater than 3 means a very

unusual event. (Outlier)

Section 5-1Normal Probability Distributions

Examples

Selected Even Problems, Using the TI-83#22

Find the area to the left of z = 0.08.2nd VARS normalcdf(-10000,.08) = .5319

#32Find the area to the right of z = 2.51

2nd VARS normalcdf(2.51,10000) = .0060#36

Find the area between z = -0.51 and z = 02nd VARS normalcdf(-0.51,0) = .1950

#40Find the area to the left of z = -1.96 or to the right of z = 1.96

Remember that we ADD probabilities for OR questions.2nd VARS normalcdf(-10000,-1.96) = .0252nd VARS normalcdf(1.96,10,000) = .025.025 + .025 = .0500

You are performing a study about the height of 20-29 year old men. A previous study found the height to be normally distributed, with a mean of 69.6 inches and a standard deviation of 3.0 inches. You randomly sample 30 men and find their heights (in inches) to be as follows:

72.1 71.2 67.9 67.3 69.5 68.6 68.8 69.4 73.5 67.169.2 75.7 71.1 69.6 70.7 66.9 71.4 62.9 69.2 64.968.2 65.2 69.7 72.2 67.5 66.6 66.5 64.2 65.4 70.0

A) Draw a frequency histogram to display these data points using seven classes. Is it reasonable to assume that the heights are normally distributed? Why?

B) Find the mean and standard deviation of your sample.C) Compare the mean and standard deviation of your sample with those in the

previous study. Discuss the differences.

Page 250, # 42

You are performing a study about the height of 20-29 year old men. A previous study found the height to be normally distributed, with a mean of 69.6 inches and a standard deviation of 3.0 inches. You randomly sample 30 men and find their heights (in inches) to be as follows:

72.1 71.2 67.9 67.3 69.5 68.6 68.8 69.4 73.5 67.169.2 75.7 71.1 69.6 70.7 66.9 71.4 62.9 69.2 64.968.2 65.2 69.7 72.2 67.5 66.6 66.5 64.2 65.4 70.0

Page 250, # 42

Entering the 30 data points into the TI-83, using STAT and Edit, we can calculate the mean, standard deviation, and median.STAT, Calc, 1-Var Stats gives us what we need.

The mean is 68.75, the standard deviation is 2.85, and the median is 69.00.

LL UL LB UB MdPt Freq. Rel. Freq.

Cum. Freq.

Max Value:

75.7Min Value:

62.9Range: 75.7 – 62.9 =

12.8Class Width: 12.8/7 =

2Remember to ROUND UP!!

First Lower Limit is the Minimum Value!!!Add Class Width Down

64.9

66.9

68.9

70.9

72.9

62.9 64.8 First Upper Limit is one unit less than the 2nd Lower Limit (Remember, our units are tenths, not whole numbers).

Add Class Width Down

66.8

68.8

70.8

72.8

74.8

72.1 71.2 67.9 67.3 69.5 68.6 68.8 69.4 73.5 67.169.2 75.7 71.1 69.6 70.7 66.9 71.4 62.9 69.2 64.968.2 65.2 69.7 72.2 67.5 66.6 66.5 64.2 65.4 70.0

74.9 76.8

LL UL LB UB MdPt Freq. Rel. Freq.

Cum. Freq.

64.9

66.9

68.9

70.9

72.9

62.9 64.8

66.8

68.8

70.8

72.8

74.8

72.1 71.2 67.9 67.3 69.5 68.6 68.8 69.4 73.5 67.169.2 75.7 71.1 69.6 70.7 66.9 71.4 62.9 69.2 64.968.2 65.2 69.7 72.2 67.5 66.6 66.5 64.2 65.4 70.0

74.9 76.8

Subtract one-half unit from lower limits to get lower boundaries.

REMEMBER that our units are tenths!! One-half of a tenth is 5 hundredths (.05) Add one-half unit to upper limits to get upper boundaries

62.85

64.85

66.85

68.85

70.85

72.8574.85

64.85

66.85

68.85

70.85

72.85

74.8576.85

Find the mean of the limits (or boundaries) to find the midpoint of each class.

63.85

65.85

67.85

69.85

71.85

73.8575.85

Count how many data points fit in each class and enter that into the Frequency column

LL UL LB UB MdPt Freq.

64.9

66.9

68.9

70.9

72.9

62.9 64.8

66.8

68.8

70.8

72.8

74.874.9 76.8

62.85

64.85

66.85

68.85

70.85

72.8574.85

64.85

66.85

68.85

70.85

72.85

74.8576.85

63.85

65.85

67.85

69.85

71.85

73.8575.85

72.1 71.2 67.9 67.3 69.5 68.6 68.8 69.4 73.5 67.169.2 75.7 71.1 69.6 70.7 66.9 71.4 62.9 69.2 64.968.2 65.2 69.7 72.2 67.5 66.6 66.5 64.2 65.4 70.0

2

5

8

8

5

11

8 8

5 5

21 1

Looking at the histogram drawn from the frequency table, it is easy to see that the data is almost perfectly bell-shaped, symmetrical and centered about the mean.

Draw the histogram using the frequencies obtained from the table we just did.

The mean, median, and mode are also very closely bunched together.Mean is 68.75, median is 69.00 and the mode is 69.2

For these reasons, it is reasonable to assume that the heights are normally distributed.

62.85 64.85 66.85 68.85 70.85 72.85 74.85 76.85

The last part of the question was to compare the mean and standard deviation of your sample with those in the previous study. Discuss the differences.

Our mean and standard deviation were 68.75 and 2.85.The previous study had a mean of 69.6 and a standard deviation of 3.0.This means that our sample of men was shorter than the previous study, but that they were also more closely bunched together in height.

Your assignments are:

Classwork:Pages 248-250, #1-8 All, and #9-41 Odd

Homework:Pages 250-252, #43-61 Odd

Chapter 5Normal Probability DistributionsSection 5-2 – Normal Distributions: Finding Probabilities A. To find the probability of any particular x value in a normal

distribution, we need to convert that x-value to a corresponding z- score.

1. Remember that the z-score is simply a measure of position

indicating how many standard deviations away from the

mean an x-value is. , or z is the difference of x and the

mean, divided by the standard deviation.B. Once we have the z-score, we can follow the procedures in Section 1 for finding the probability.C. Another option is that the TI-84 calculator can also figure

probabilities using x-values, without having to first calculate the z- scores.

Chapter 5Normal Probability DistributionsSection 5-2 – Normal Distributions: Finding Probabilities

1. 2nd VARS normalcdf (lower, upper, mean, standard deviation), in that order will give us the area under the curve.

a. For finding the area to the left of an x-value, use the smallest

possible x-value, or some x-value that is at least 5 standard

deviations below the mean, as the lower value, then use the given x as the upper

value, enter the mean, and the standard deviation and let the calculator do the rest!!

b. For finding the area to the right of an x-value, use the given x

as the lower value, and some artificially high number (at least 5 standard deviations

above the mean) as the upper value, then enter the mean and standard

deviation.c. For finding the area between

two given x values, use the smaller value as the lower value, the

bigger one as the upper value, then enter the mean and standard deviation.

Chapter 5Normal Probability DistributionsSection 5-2 – Normal Distributions: Finding ProbabilitiesD. Once we have calculated the probability of a particular x-value

occurring once, we can use that probability to figure how many

times that x-value might occur out of n number of trials.1. For example, if we know that there is a .7333

probability that a shopper will spend between 24 and 54 minutes in a grocery

store, then we can predict that if we have 200 shoppers, about 147 of them will be in the store between 24 and 54 minutes

(200*.7333 = 146.67, or about 147).

Chapter 5Normal Probability DistributionsSection 5-2 – Normal Distributions: Finding ProbabilitiesExamples: Page 256 #1-6

1)5)

2)6)

3)

4)


1)

86 91 96 101817671


1)2nd VARS normalcdf

(lower limit, upper limit, mean, standard

deviation)Use a number at

least 5 standard deviations below the mean for the

lower limit.86 -

(5*5) = 86 – 25 = 61.Use the given x

value as the upper limit.2nd VARS normalcdf (61,

80, 86, 5) = .1151

Chapter 5Normal Probability DistributionsSection 5-2 – Normal Distributions: Finding ProbabilitiesExamples: 6 #1-6

2)

86 91 96 101817671





least 5 standard deviations below the mean for the

lower limit.86 -

(5*5) = 86 – 25 = 61.Use the given x

value as the upper limit.2nd VARS normalcdf (61,

100, 86, 5) = .9974


2)

86 91 96 101817671





least 5 standard deviations ABOVE the mean for the

UPPER limit.86 +

(5*5) = 86 + 25 = 111.Use the given x

value as the Lower limit.2nd VARS normalcdf (92,

111, 86, 5) = .1151


2)

86 91 96 101817671





least 5 standard deviations ABOVE the mean for the

UPPER limit.86 +

(5*5) = 86 + 25 = 111.Use the given x

value as the Lower limit.2nd VARS normalcdf (75,

111, 86, 5) = .9861


2)

86 91 96 101817671




deviation) Use the given x values as the Lower and Upper

Limits.2nd VARS normalcdf (70,

80, 86, 5) = .1144


2)

86 91 96 101817671




deviation) Use the given x values as the Lower and Upper

Limits.2nd VARS normalcdf (85,

95, 86, 5) = .5433

Your assignments are:

Classwork:Pages 256-258, #7-20 All

Homework:Pages 258-259, #21-30 All

chapter 5 normal probability distributions. chapter 5 normal probability distributions section 5-1...

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