Transcript Using the Normal Curve and z

6
Normal Probability
Distributions
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6.3
Applications of Normal
Distributions
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Probabilities and Normal Curves
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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)?
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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).
(6.3)
The z values are shown on the figure.
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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?
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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.
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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).
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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.
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