Download Chapter 9 Statistics

Chapter 9
Statistics
Section 9.2
Measures of Variation
Who Has Better Scores?

Adam and Bonnie are comparing their quiz
scores in an effort to determine who is the
“best”.
Help them decide by calculating the mean,
median, and mode for each.
Adam’s Scores
85
60
105
85
72
100
Bonnie’s Scores
81
85
86
85
90
80
The Winner?
Adam
Bonnie
Mean
84.5
84.5
Median
85
85
Mode
85
85
So, who has the better quiz scores?
Another Way to Compare

Sometimes the measures of central
tendency (mean, median, mode) aren’t
enough to adequately describe the data.

We also need to take into account the
consistency, or spread, of the data.
Range

The range of the data is the difference between
the largest and smallest number in a sample.

Find the range of Adam and Bonnie’s scores.
Adam: 105 – 60 = 45
 Bonnie: 90 – 80 = 10


Based on the range, Bonnie’s scores are more
consisent, and some might argue therefore,
better than Adam’s.
Another Measure of Dispersion

The most useful measure of variation
(spread) is the standard deviation.

First, we will look at the deviations from
the mean.
Deviations from the Mean

The deviation from the mean is the difference
between a single data point and the calculated
mean of the data.
X X
Data point close to mean: small deviation
 Data point far from mean: large deviation
 Sum of deviations from mean is always zero.
 Mean of the deviations is always zero.

Deviations from Mean
for Adam and Bonnie
Adam
Bonnie
Data
Point
Deviation from Mean
Data
Point
Deviation from Mean
85
85 - 84.5 = 0.5
81
81 - 84.5 = -3.5
60
60 – 84.5 = -24.5
85
85 – 84.5 = 0.5
105 105 – 84.5 = 20.5
86
86 – 84.5 = 1.5
85
85 – 84.5 = 0.5
85
85 – 84.5 = 0.5
72
72 – 84.5 = -12.5
90
90 – 84.5 = 5.5
100 100 – 84.5 = 15.5
80
80 – 84.5 = -4.5
Sum = 0
Sum = 0
Variance
Because the average of the mean deviation is
always zero, we must modify our approach using
the variance.
 The variance is the mean of the squares of the
deviation.

Variance for
Adam and Bonnie’s Scores


Using the deviations from the mean we have already
calculated for Adam and Bonnie, we will find the variance
for each.
(.5)2  (24.5)2  (20.5)2  (.5) 2  (12.5) 2  (15.5) 2
Adam : s² =
6 1
1417.5
s² =
5

= 283.5
(3.5)2  (.5)2  (1.5) 2  (.5) 2  (5.5) 2  (4.5) 2
Bonnie: s² =
6 1
s² =
65.5
5
= 13.1
Standard Deviation
To find the variance, we squared the deviations
from the mean, so the variance is in squared
units.
 To return to the same units as the data, we use
the square root of the variance, the standard
deviation.

s s
Standard Deviation
2
Variance
Adam and Bonnie’s Standard Deviation

Adam:
s  s  283.5  16.84

Bonnie:
s  s  13.1  3.62

Based on the standard deviation, Bonnie’s scores
are better because there is less dispersion. In
other words, she is more consistent than Adam.
2
2
Formulas for Variance and Standard
Deviation
Example 1

The number of
homicide victims in
Vermont from 1992
through 2001 is given
in the table at right.
(Source: http://170.222.24.9/cjs/crime_01/homicide_01.html)

Find the mean,
median, mode, and
standard deviation of
the data.
Year
Homicide
Victims
1992
1993
1994
21
15
5
1995
1996
1997
1998
1999
2000
2001
13
11
9
12
17
12
11
Sample vs. Population

The mean, variance, and standard deviation of a random
sample is referred to as the sample mean ( X ), sample
variance (s²), and sample standard deviation (s).

The sample mean, variance, standard deviation, etc. can
only give us an approximation to the population mean (µ),
the population variance (σ²), and the population standard
deviation (σ).

The main difference lies in the denominator of the
formulas for standard deviation. When the value of the
sample, n, is large, the sample standard deviation gives a
good estimate of the population standard deviation.
Grouped Distributions
Example 2

Mr. Smith recently
gave a math test and
organized his scores
into the table at right.
Help Mr. Smith
determine the class
average, the median
score, and standard
deviation.
Score
Frequency
40 - 49
2
50 - 59
3
60 - 69
6
70 - 79
12
80 - 89
7
90 - 100
5
Chebyshev’s Theorem

Chebyshev’s Theorem states that for any set
of numbers, the fraction (or probability) that
will lie within k standard deviations of the
mean (for k > 1) is at least
__1__
1 k²
Example 3

Use Chebyshev’s Theorem to find the
fraction of all the numbers of a data set
that must lie within 4 standard deviations
from the mean.
Example 4

In a certain distribution of numbers, the
mean is 50 with a standard deviation of 6.
Use Chebyshev’s Theorem to tell the
probability that a number lies in each
interval.
a.) between 38 and 62
b.) less than 38 or more than 62