Transcript Section 5.2 ~ Properties of the Normal Distribution

Section 5.2 ~
Properties of the Normal Distribution
Introduction to Probability and Statistics
Ms. Young
Sec. 5.2
Objective

In this section you will learn how to interpret the
normal distribution in terms of the 68-95-99.7 rule,
standard scores, and percentiles.
Sec. 5.2
Properties of the Normal Distribution
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
Recall the following notations:

The population mean is represented using the Greek letter mu

The population standard deviation is represented using the Greek letter

sigma

The following diagram is an example of a distribution that represents
the replacement time (in years) for a TV
Sec. 5.2
Properties of the Normal Distribution
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Knowing the mean and standard deviation of a distribution allows
us to say a lot about where the data values lie

Example:

The mean TV replacement time is 8.2 years with a standard deviation of
1.1 years
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About two-thirds of all values fall within one standard deviation, so the
replacement time for about two-thirds of the TV’s is between 7.1 years and
9.3 years (8.2 - 1.1 = 7.1 and 8.2 + 1.1 = 9.3)
About 95% of the data fall within two standard deviations of the mean, so
the replacement time for 95% of the TV’s in this sample is between 6.0 years
and 10.4 years (8.2 – 2.2 = 6.0 and 8.2 + 2.2 10.4)
Sec. 5.2
The 68-95-99.7 Rule

The 68-95-99.7 Rule gives precise guidelines for the
percentage of data values that lie within 1, 2, and 3
standard deviations of the mean for any normal distribution
About 68% (more precisely, 68.3%), or just over two-thirds, of
the data points fall within 1 standard deviation of the mean
 About 95% (more precisely, 95.4%) of the data points fall
within 2 standard deviations of the mean
 About 99.7% of the data points fall within 3 standard
deviations of the mean

Sec. 5.2
Example 1

If there are 600 total values. Approximately,
how many values occur within 1 standard of
deviation from the mean? 2 standards of
deviation? 3 standards of deviation?
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Since 68% of the data values lie within 1 standard
deviation, 408 values fall within 1 standard deviation
(.68 * 600 = 408)
Since 95% of the data values lie within 2 standard
deviations, 570 values fall within 2 standard deviations
(.95 * 600 = 570)
Since 99.7% of the data values lie within 3 standard
deviations, 598.2 values fall within 3 standard deviations
(.997 * 600 = 598.2)
Sec. 5.2
Example 2

Vending machines can be adjusted to reject coins above and
below certain weights. The weights of legal U.S. quarters
have a normal distribution with a mean of 5.67 grams and a
standard deviation of 0.0700 gram. If a vending machine is
adjusted to reject quarters that weigh more than 5.81
grams and less than 5.53 grams, what percentage of legal
quarters will be rejected by the machine?
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A weight of 5.81 is 0.14 gram (5.81 – 5.67 = 0.14), or 2 standard
deviations above the mean and a weight of 5.53 is 0.14 gram
(5.53 - 5.67 = -0.14), or 2 standard deviations below the mean
Since 95% of the data values fall within 2 standard deviation of
the mean, the machine will accept 95% of the quarters inserted
and will reject 5%
Sec. 5.2
Unusual Values
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Unusual values are values that are more than 2 standard
deviations away from the mean
Example 3 ~ Traveling and Pregnancy

Suppose your friend is pregnant and is trying to determine if she
should schedule an important business meeting two weeks before her
due date. Data suggests that the number of days between the birth
date and due date is normally distributed with a mean of μ = 0 days and
a standard deviation of σ = 15 days. How would you help your friend
make the decision? Would a birth two weeks before the due date be
considered “unusual”?
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A birth 2 weeks before the due date is approximately 1 standard deviation
below the mean
Since 68% of births will occur within 15 days of the due date (in either
direction), that means that a total of 32% of births will occur more than 15
days from the due date (in either direction)

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Since the distribution is symmetric, that means that 16% of births will occur more
than 15 days early
To help your friend decide if it would be safe to schedule a meeting 14 days
before her due date, you could tell her that 16% of births will occur more
than 15 days early, which equates to about 1 in 6
Since a birth date that is two weeks early falls well within 2 standard
deviations, this would not be considered unusual
Sec. 5.2
Example 4
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Suppose you measure your heart rate at noon every day for
a year and record the data. You discover that the data has
a normal distribution with a mean of 66 and a standard of
deviation of 4. On how many days was your heart rate below
58 beats per minute?
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58 falls 2 standard deviations below the mean (58 – 66 = -8), so
95% of time your heart rate will be between 58 bpm and 74 bpm,
therefore 5% of the time your heart rate will either be above 74
bpm or below 58 bpm
Since the distribution is symmetric, the area below 2 standard
deviations would be 2.5% (5/2 = 2.5), so 2.5% of the time your
heart rate will fall below 58 bpm
Since there are 365 days in a year (with the exception of leap
year), approximately 9 days (.025 * 365 = 9.125) out of the year
your heart rate will be below 58 bpm
Sec. 5.2
Example 5
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Suppose that your fourth-grade daughter is told that her
height is 1 standard deviation above the mean for her age
and sex. What is her percentile for height? Assume that
heights of fourth-grade girls are normally distributed
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Since 68% of fourth-grade girls lie within 1 standard deviation
(both directions), then 34% of fourth-grade girls lie within 1
standard deviation above the mean
Since the data is symmetric, 50% of fourth-grade girls will fall
below the mean
If your daughter’s height is 1 standard deviation above the mean,
then she is taller than 84% (50% + 34% = 84%) of fourth-grade
girls, which means that she is in the 84th percentile
Sec. 5.2
Standard Scores
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A standard score (or z-score) is the number of standard
deviations a data value lies above or below the mean

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The standard score is represented using z
The standard score is positive for values above the mean
and negative for values below the mean

Examples ~
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A value that is 2 standard deviations above the mean has a
standard score of 2, or z = 2
A value that is 2 standard deviations below the mean has a
standard score of -2, or z = -2
The standard score of the mean will be z = 0, since it’s 0 standard
deviations from the mean
The standard score is found using the following formula:
data value - mean
z = standard score =
standard deviation
z
x

Sec. 5.2
Example 6
z = standard score =

data value - mean
standard deviation
The Stanford-Binet IQ test is scaled so that scores have a
mean of 100 and a standard deviation of 16. Find the standard
scores (z-scores) for IQs of 85, 100, and 125.
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Standard score for 85:
z
85  100
 0.94
16
z
100  100
 0.00
16
z
125  100
 1.56
16
Standard score for 100:
Standard score for 125:
Sec. 5.2
Standard Scores and Percentiles
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Once you know the standard score of a data value, the
properties of the normal distribution allow you to find its
percentile
This is typically done with a standard score table (refer
to Appendix A on p.446)

The table gives the percentage of values less than that
value
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Ex. ~ a value that has a standard score of 0.15, has 55.96% of
the total values falling below it, and it therefore lies in the 55th
percentile
Sec. 5.2
Example 7
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Cholesterol levels in men 18 to 24 years of age are normally
distributed with a mean of 178 and a standard deviation of 41.

What is the percentile for a 20-year old man with a cholesterol
level of 190?
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The standard score for a cholesterol level of 190 is:
z
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
190  178
 0.29
41
If you look at the chart, you see that a standard score of .29 has
61.41% of values falling below, which means that it falls in about the
61st percentile
What cholesterol level corresponds to the 90th percentile, the level
at which treatment may be necessary?
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According to the chart, a standard score of 1.29 has 90.15% of the
values falling below it making it fall in the 90th percentile
If the mean cholesterol level is 178 and the standard deviation is 41,
then 1.29 standard deviations would equate to a level of 52.89 (41 *
1.29 = 52.89)
A level of about 231 (178 + 52.89 = 230.89) would be in the 90th
percentile
Sec. 5.2
Example 8
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The heights of American women aged 18 to 24 are normally distributed
with a mean of 65 inches and a standard deviation of 2.5 inches. In
order to serve in the U.S. Army, women must be between 58 inches and
80 inches tall. What percentage of women are ineligible to serve based
on their height?
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If a woman is shorter than 58 inches or taller than 80 inches she would be
ineligible to serve in the Army
The standard score for a height of 58 inches is:
z
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After looking at the table, you can see that .26% of women are too short to serve
in the Army
The standard score for a height of 80 inches is:
z
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58  65
 2.8
2.5
80  65
 6.0
2.5
Since a standard score of 6.0 does not appear on the table, it means that it is
above the 99.98th percentile in which case fewer than 0.02% (100% - 99.98% =
0.02%) of women are too tall to serve in the Army
All together, fewer than .28% (.26% + .02% = .28%) of women are ineligible to
serve in the Army