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6 Normal Probability Distributions

Copyright © Cengage Learning. All rights reserved.

Applications of NormalDistributions

6.3

3

Probabilities and Normal Curves

4

Probabilities and Normal Curves

To demonstrate the process of converting to a standard normal curve to find probabilities, let’s consider IQ scores.

IQ scores are normally distributed with a mean of 100 and a standard deviation of 16.

If a person is picked at random, what is the probability that his or her IQ is between 100 and 115? That is, what is

P (100 < x < 115)?

5

Probabilities and Normal Curves

P(100 < x < 115) is represented by the shaded area

in the figure below. The variable x must be standardized using formula (6.3).

The z values are shown on the figure.

(6.3)

6

Probabilities and Normal Curves

Therefore,

P (100 < x < 115) = P (0.00 < z < 0.94) = 0.3264.

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Probabilities and Normal Curves

Thus, the probability is 0.3264 (found by using Table 3 in Appendix B) that a person picked at random has an IQ between 100 and 115.

What if we need to determine a probability for “any” normal curve?

How do we calculate probability under “any” normal curve?

8

Probabilities and Normal Curves

Let’s continue to use the example of IQ scores and try to find the probability that a person selected at random will have an IQ greater than 90.

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Probabilities and Normal Curves

P (x > 90) = P (z > – 0.63)

= 1.0000 – 0.2643 = 0.7357

Thus, the probability is 0.7357 that a person selected at random will have an IQ greater than 90.

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Using the Normal Curve and z

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Using the Normal Curve and z

The normal table, Table 3, can be used to answer many kinds of questions that involve a normal distribution.

Many problems call for the location of a “cut-off point,” a particular value of x such that there is exactly a certain percentage in a specified area.

Determine Data Values

In a large class, suppose your instructor tells you that you need to obtain a grade in the top 10% of your class to get an A on a particular exam.

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Using the Normal Curve and z

From past experience, she is able to estimate that the mean and standard deviation on this exam will be 72 and 13, respectively.

What will be the minimum grade needed to obtain an A? (Assume that the grades will have an approximately normal distribution.)

We can use the normal curve and z to determine that data value.

13

Using the Normal Curve and z

Start by converting the 10% to information that is compatible with Table 3 by subtracting:

10% = 0.1000;

1.0000 – 0.1000 = 0.9000

Look in Table 3 to find the value of z associated with the area entry closest to 0.9000; it is z = 1.28. Thus,

P (z > 1.28) = 0.10

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Using the Normal Curve and z

Now find the x value that corresponds to z = 1.28 by

using formula (6.3):

x – 72 = (13)(1.28)

x = 72 + (13)(1.28) = 72 + 16.64 = 88.64, or 89

Thus, if you receive an 89 or higher (the data value), you

can expect to be in the top 10% (which means you can expect to receive an A).

15

Using the Normal Curve and z

Determine Percentiles

Just as you can use the normal curve and z to find data values, you can also use them to find percentiles.

Let’s return to the example of IQ scores and find the 33rd percentile for IQ scores when = 100 and = 16.

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Using the Normal Curve and z

P (z < P33) = 0.3300

33rd percentile is at z = – 0.44

Now we convert the 33rd percentile of the z-scores, – 0.44, to an x-score using formula (6.3):

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Using the Normal Curve and z

x – 100 = (16)(– 0.44)

x = 100 – 7.04 = 92.96

Thus, 92.96 is the 33rd percentile for IQ scores.

Determine Population Parameters

The normal curve and z can also be used to determine population parameters.

That is, when given related information, you can find the standard deviation, .

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Using the Normal Curve and z

To illustrate how this can be done, let’s look at the incomes of junior executives in a large corporation, which are approximately normally distributed.

A pending cutback will not discharge those junior executives with earnings within $4,900 of the mean. If this represents the middle 80% of the incomes, what is the standard deviation for the salaries of this group of junior executives?

Table 3 indicates that the middle 80%, or 0.8000, of a normal distribution is bounded by –1.28 and 1.28.

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Using the Normal Curve and z

Consider point B shown in the figure below: 4,900 is the difference between the x-value at B and the value of the mean, the numerator of formula (6.3) [x – = 4,900].

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Using the Normal Curve and z

Using formula (6.3), we can find the value of :

= 3,828.125 = $3,828

That is, the current standard deviation for the salaries of junior executives is $3,828.