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14
Descriptive Statistics
What a Data Set Tells Us
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section
Section14.4,
1.1, Slide
Slide11
14.4 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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 2
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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 3
The Normal Distribution
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 4
The Normal Distribution
We represent the mean by μ and the standard
deviation by σ.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 5
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)
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, 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.
(continued on next slide)
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 7
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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 8
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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 9
z-Scores
Below is a portion of a table that gives the area
under the standard normal curve between the mean
and a z-score.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 10
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)
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 11
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)
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 12
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)
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 13
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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 14
Converting Raw Scores to z-Scores
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 15
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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
(continued on next slide)
Section 14.4, Slide 16
Converting Raw Scores to z-Scores
• Solution (a): We have
We compute
(continued on next slide)
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 17
Converting Raw Scores to z-Scores
• Solution (b): We have
We compute
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 18
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)
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 19
Applications
Using a table, we have
that A = 0.460 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)
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 20
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)
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 21
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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 22
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)
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 23
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)
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 24
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 z-score 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)
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 25
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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Section 14.4, Slide 26