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Chapter 13
Chapter 13: Statistics
13.1
13.2
13.3
13.4
13.5
13.6
Statistics
Visual Displays of Data
Measures of Central Tendency
Measures of Dispersion
Measures of Position
The Normal Distribution
Regression and Correlation
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Chapter 1
The Normal Distribution
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Section 13-5
Discrete and Continuous Random Variables
Definition and Properties of a Normal Curve
A Table of Standard Normal Curve Areas
Interpreting Normal Curve Areas
The Normal Distribution
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Random Variables
A variable is a characteristic or attribute that can
assume different values.*
A random variable is a variable with value
assigned by the result of an experiment.
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Discrete and Continuous Random
Variables
A random variable that can take on only
certain fixed values is called a discrete
random variable. A variable whose values
are not restricted in this way is a continuous
random variable.
*Allan G. Bluman, Elementary Statistics, 4th ed,. ©2008
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Definition and Properties of a Normal
Curve
Normal Curves
B
A normal curve is a symmetric, bell-shaped curve.
Any random variable whose distribution curve has
this characteristic shape is said to have a normal
distribution.
A
S
C
On a normal curve, if the quantity shown on the
horizontal axis is the number of standard deviations
from the mean, rather than values of the random
variable itself, then we call the curve the standard
normal curve.
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0
S is standard, with mean = 0, standard deviation = 1
A has mean < 0, standard deviation = 1
B has mean = 0, standard deviation < 1
C has mean > 0, standard deviation > 1
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Properties of Normal Curves
Empirical Rule
The graph of a normal curve is bell-shaped and
symmetric about a vertical line through its center.
The mean, median, and mode of a normal curve are all
equal and occur at the center of the distribution.
Empirical Rule About 68% of all data values of a
normal curve lie within 1 standard deviation of the
mean (in both directions), and about 95% within 2
standard deviations, and about 99.7% within 3
standard deviations.
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Example: Applying the Empirical Rule
A Table of Standard Normal Curve Areas
Suppose that 280 sociology students take an exam and
that the distribution of their scores can be treated as
normal. Find the number of scores falling within 2
standard deviations of the mean.
To answer questions that involve regions other than
1, 2, or 3 standard deviations of the mean we can
refer to the table on page 829 or other tools such as a
computer or calculator.
Solution
A total of 95% of all scores lie within 2 standard
deviations of the mean.
(.95)(280)= 266 scores
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A Table of Standard Normal Curve Areas
Recall that the z-score of a data item is the
number of standard deviations away from the
mean that data item is.
Example: Applying the Normal Curve
Table
Use the table to find the percent of all scores that
lie between the mean and the following values.
The table on p. 829 gives the fraction of all scores
in a normal distribution that lie between the mean
and z standard deviations from the mean.
Because of symmetry, the table can be used for
values above the mean or below the mean.
a) 1.5 standard deviation above the mean
b) 2.62 standard deviations below the mean
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Example: Applying the Normal Curve
Table
Example: Applying the Normal Curve
Table
Solution (continued)
Solution
a) Here z = 1.50. Find 1.50 in the z column. The
table entry is .433, so 43.3% of all values lie between
the mean and 1.5 standard deviations above the mean.
b) Even though it is below the mean, the table still
works because of the symmetry. Find 2.62 in the z
column. The table entry is .496, so 49.6% of all
values lie between the mean and 2.62 standard
deviations below the mean.
z = 1.5
x
z = –2.62
x
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© 2008 Pearson Addison-Wesley. All rights reserved
Example: Applying the Normal Curve
Table
Example: Applying the Normal Curve
Table
Find the total area indicated in the region in
color below.
Solution
z = –1.7
x
Find the total area indicated in the region in
color below.
z = 2.55
Solution
z = 2.55 leads to an area of .495 and z = 1.7 leads
to an area of .455. Add these areas to get
.495 + .455 = .950
z = .61
z = 2.63 leads to an area of .496 and z = .61 leads
to an area of .229. Subtract these areas to get
.496 – .229 = .267.
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z = 2.63
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Example: Applying the Normal Curve
Table
Interpreting Normal Curve Areas
Find the total area indicated in the region in
color below.
In a standard normal curve, the following three
quantities are equivalent.
1. Percentage (of total items that lie in an interval)
2. Probability (of a randomly chosen item lying in
an interval)
z = 2.14
Solution
3. Area (under the normal curve along an interval)
z = 2.14 leads to an area of .484. The total area
to the right of the mean is .500. Subtract these
values to get .500 – .484 = .016.
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Example: Finding z-Values for Given
Areas
Example: Probability with Volume
The volumes of soda in bottles from a small company
are distributed normally with mean 12 ounces and
standard deviation .15 ounce. If 1 bottle is randomly
selected, what is the probability that it will have more
than 12.33 ounces?
Assuming a normal distribution, find the z-value
meeting the condition that 39% of the area is to
the right of z.
11%
Solution
39%
Solution
12.33 corresponds to a z-value of 2.2. The probability
is equal to the area above z = 2.2, which is
.500 – .486 = .014 (or 1.4%).
Because 50% of the area lies to the right of the mean,
there must be 11% of the area between the mean and z.
From the table, A = .110 corresponds to z = .28.
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Example: Finding z-Values for Given
Areas
Assuming a normal distribution, find the z-value
meeting the condition that 76% of the area is to
the left of z.
26%
Solution
50%
24%
There must be 26% of the area between the mean
and z. From the table, A = .260 corresponds closely
to z = .71.
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