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Dr. Fowler  AFM  Unit 8-4
The Normal Distribution
• Understand the basic properties of the normal curve.
• Relate the area under a normal curve to z-scores.
• Make conversions between raw scores and z-scores.
• Use the normal distribution to solve applied problems.
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The Normal Distribution
The normal distribution describes many real-life data sets.
The histogram shown gives an idea of
the shape of a normal distribution.
The Normal Distribution
The Normal Distribution
We represent the mean by μ and the standard
deviation by σ.
The Normal Distribution
• Example: Suppose that the distribution of scores of
1,000 students who take a standardized intelligence
test is a normal distribution. If the distribution’s mean
is 450 and its standard deviation is 25,
a) how many scores do we expect to fall
between
425 and 475?
b) how many scores do we expect to fall
above
500?
(continued on next slide)
Section 15.4, Slide 6
The Normal Distribution
• Solution (a): 425 and 475 are each 1 standard deviation from the
mean. Approximately 68% of the scores lie within 1 standard
deviation of the mean. We expect about 0.68 × 1,000 = 680 scores
are in the range 425 to 475.
The Normal Distribution
Solution (b):
We know 5% of the scores lie more than 2 standard deviations above or
below the mean, so we expect to have
0.05 ÷ 2 = 0.025 of the scores to be above 500. Multiplying by 1,000, we can
expect that 0.025 * 1,000 = 25 scores to be above 500.
z-Scores
https://www.youtube.com/watch?v=mFYvUxOO2T4
z-Scores
The standard normal distribution has a mean of 0
and a standard deviation of 1.
There are tables (see next slide) that give the area
under this curve between the mean and a number
called a z-score. A z-score represents the number of
standard deviations a data value is from the mean.
For example, for a normal distribution with mean
450 and standard deviation 25, the value 500 is 2
standard deviations above the mean; that is, the
value 500 corresponds to a z-score of 2.
Converting Raw Scores to z-Scores
Section 15.4, Slide 11
Applications
• Example: Suppose you take a standardized test. Assume
that the distribution of scores is normal and you received a
score of 72 on the test, which had a mean of 65 and a
standard deviation of 4. What percentage of those who took
this test had a score below yours?
• Solution: We first find the z-score that corresponds to 72.
(continued on next slide)
z-Scores
Table
Of Normal
Distribution
Google
Applications
Using the z-score table, we have
that A = 0.4591 when z = 1.75.
The normal curve is symmetric,
so another 50% of the scores fall
below the mean. So, there are
50% + 46% = 96% of the scores
below 72.
(continued on next slide)
Applications
• Example: Consider the following information:
1911: Ty Cobb hit .420. Mean average was .266
with standard deviation .0371.
1941: Ted Williams hit .406. Mean average was .267
with standard deviation .0326.
1980: George Brett hit .390. Mean average was .261
with standard deviation .0317.
Assuming normal distributions, use z-scores to
determine which of the three batters was ranked the
highest in relationship to his contemporaries.
(continued on next slide)
Applications
• Solution:
Ty Cobb’s average of .420 corresponded to a
z-score of
Ted Williams’s average of .406 corresponded to
a z-score of
George Brett’s average of .390 corresponded to
a z-score of
Compared with his contemporaries, Ted Williams
ranks as the best hitter.
Excellent Job !!!
Well Done
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Converting Raw Scores to z-Scores
• Example: Suppose the
mean of a normal
distribution is 20 and its
standard deviation is 3.
a) Find the z-score
corresponding to the
raw score 25.
b) Find the z-score
corresponding to the
raw score 16.
(continued on next slide)
Converting Raw Scores to z-Scores
• Solution (a): We have
We compute
(continued on next slide)
Converting Raw Scores to z-Scores
• Solution (b): We have
We compute
Section 15.4, Slide 21
z-Scores
Command Summary
Calculates the inverse of the
cumulative normal distribution function.
Command Syntax
invNorm(probability[,μ, σ])
Menu Location
Press:
2ND DISTR to access the distribution
menu
3 to select invNorm(, or use arrows.
© 2010 Pearson Education, Inc. All rights reserved.
Section 15.4, Slide 22
z-Scores
• Example: Use a table to find the percentage of
the data (area under the curve) that lie in the
following regions for a standard normal
distribution:
a) between z = 0 and z = 1.3
b) between z = 1.5 and z = 2.1
c) between z = 0 and z = –1.83
(continued on next slide)
© 2010 Pearson Education, Inc. All rights reserved.
Section 15.4, Slide 23
z-Scores
• Solution (a): The area
under the curve between
z = 0 and z = 1.3 is shown.
Using a table we find this
area for the z-score 1.30.
We find that A is 0.403
when z = 1.30. We expect 40.3%, of the data to fall
between 0 and 1.3 standard deviations above the
mean.
(continued on next slide)
z-Scores
• Solution (b): The area under
the curve between z = 1.5
and z = 2.1 is shown. We first
find the area from z = 0 to z =
2.1 and then subtract the
area from z = 0 to z = 1.5
Using a table we get A = 0.482 when z = 2.1, and A = 0.433
when z = 1.5. The area is 0.482 – 0.433 = 0.049 or 4.9%
(continued on next slide)
Section 15.4, Slide 25
z-Scores
• Solution (c): Due to the
symmetry of the normal
distribution, the area
between z = 0 and z = –1.83
is the same as the area
between z = 0 and z = 1.83.
Using a table, we see that A = 0.466 when
z = 1.83. Therefore, 46.6% of the data values
lie between 0 and –1.83.
© 2010 Pearson Education, Inc. All rights reserved.
Section 15.4, Slide 26
Applications
• Example: A manufacturer plans to offer a warranty
on an electronic device. Quality control engineers
found that the device has a mean time to failure of
3,000 hours with a standard deviation of 500 hours.
Assume that the typical purchaser will use the
device for 4 hours per day. If the manufacturer does
not want more than 5% to be returned as defective
within the warranty period, how long should the
warranty period be to guarantee this?
(continued on next slide)
© 2010 Pearson Education, Inc. All rights reserved.
Section 15.4, Slide 27
Applications
• Solution: We need to
find a z-score such that
at least 95% of the area
is beyond this point. This
score is to the left of the
mean and is negative.
By symmetry we find the
z-score such that 95% of
the area is below this
score.
(continued on next slide)
Applications
50% of the entire area lies below the mean, so our
problem reduces to finding a z-score greater than 0
such that 45% of the area lies between the mean
and that z-score. If A = 0.450, the corresponding zscore is 1.64. 95% of the area underneath the
standard normal curve falls below z = 1.64. By
symmetry, 95% of the values lie above –1.64.
Since
, we obtain
(continued on next slide)
Applications
Solving the equation for x, we get
Owners use the device about 4 hours per day, so we
divide 2,180 by 4 to get 545 days. This is approximately
18 months if we use 31 days per month. The warranty
should be for roughly 18 months.