# Unit 4 Measures of Dispersion 1 question data set and calculate the standard deviation

Take the numbers of second ten cases (64, 56, 17, 38, 94, 78, 101, 71, 63, 65) from Exercise 2 (p. 92) as your data set and calculate the standard deviation. Show your work step by step as that of in-class exercise 4.3.  Unit 4 Measures of Dispersion
A. Introduction
1. Unit 3 has discussed the central tendency. We are going to examined measures of
dispersion in Unit 4.
2. Measures of dispersion (variation or variability) answer the question—how typical is the
average value?
3. Range, variance, and standard deviation are the three commonly used measures of
dispersion. Following table presents the statistics of students’ performance in an exam.
You may find the count is 30 (N = 30). It means that 30 students had participated this
exam. You may also find the value of range (44.00), standard deviation (9.78002), and
variance (95.64888). How can we calculate those values? How can we interpret those
values? Following sections will provide useful guidelines.
B. Range
1. The Range is the difference between the minimum value (L) and maximum value (H) in
the data set. The Range = H – L
2. In the above table, the minimum value (L) is 56 and the maximum value (H) is 100. As
such, the range is 44 (100 – 56). It is that simple!
3. The range can sometimes be misleading when there are extremely high or low values. So
we may be better off using variance or standard deviation.
C. Variance
1. Variance is the expectation of the squared deviation of a random variable from its mean.
Informally, it measures how far a set of (random) numbers are spread out from their
average value.
2. Because the computation of variance is based upon the mean, it requires interval or ratio
level data. Following presents the formula of variance for population and sample:
(1) Population: ² = f(x – )²/N
(2) Sample: s² = f(x – x-bar)²/N – 1
1
3. The main difference between the formula of population and sample is the denominator.
The denominator in the formula of population is “N” while and the denominator in the
formula of sample is “N – 1,” a concept named degree of freedom that will be introduced
in Unit 6. Hence, the text uses a “compromised” formula: s² = f(x – x-bar)²/N
In this unit and its homework, we will use this compromised formula to calculate variance.
D. Standard Deviation
1. The standard deviation is a measure of the amount of variation or dispersion of a set of
values. A low standard deviation indicates that the values tend to be close to the mean of
the set, while a high standard deviation indicates that the values are spread out over a
wider range.
2. In short, the standard deviation is the positive square root of
the variance. Following presents the formula of standard
deviation for population and sample, respectively:
(1) Population:  = ²
(2) Sample: s = s²
(3) Let’s take another look of the statistics of students’
performance in an exam, in which standard deviation is
9.78002 and variance is 95.64888. Use your calculator to
take a positive square root of the variance (95.64888),
you will see the standard deviation as 9.78002 (rounding).
Is it a coincident? No, this will always be the case as it is a
property of the standard deviation.
In-Class Exercise 4.1: Following table presents the statistical approach that may fit to the
program of SPSS (Statistical Package for Social Science). This table consists of five columns,
and you’ve learned the first three columns in Unit 3. The fourth column presents the dispersion
from the mean (x-bar) of each measurement (X). The fifth column presents the frequency
multiplying the dispersion from the mean of each measurement (f(x – x-bar)²). How can we find
out the variance and standard deviation?
How to calculate variance and standard deviation of table 5.1 (when all f is one)
x
f
fx
(x – x-bar)
f(x – x-bar)²
6
1
6
(6 – 4) = 2
4
5
1
5
(5 – 4) = 1
1
4
1
4
(4 – 4) = 0
0
3
1
3
(3 – 4) = –1
1
2
1
2
(2 – 4) = –2
4
N=5
∑fX = 20
∑f(X – X-bar) = 0 ∑f(X – X-bar)²=10
First, we need to find out the mean (you’ve learned that from prior chapter) for the
calculation of the fourth column. Mean: X-bar = ∑fX / N = 20 / 5 = 4. Then we find out
that the sum of dispersions of individual measurements from the mean is zero. To
continue our calculation, we have to square the value of each dispersion in the fifth
2
column. Then you will obtain 10 by adding each squared dispersions (4+1+0+1+4 = 10).
Now you may calculate the variance as below (the same color is sued helping you to
trace):
Variance: s² = f(x – x-bar)²/N = 10 / 5 = 2
After you obtain the variance, you just take a positive square root of the variance, you
will obtain the standard deviation as below:
Standard deviation: s = s² = 2 = 1.4142135624 = 1.4 (rounding to the nearest tenth)
In-Class Exercise 4.2: Following table presents the statistical approach that may fit to the
program of SPSS (Statistical Package for Social Science) when not all frequency is one. This
table consists of six columns, and you’ve learned the first three columns in Unit 3. The fourth
column presents the dispersion from the mean (x-bar) of each measurement (X). The fifth
column presents the squared dispersions ((x – x-bar)²). The sixth column presents the frequency
multiplying the dispersion from the mean of each measurement (f(x – x-bar)²). How can we find
out the variance and standard deviation?
How to calculate variance and standard deviation of table 5.2 (when not all f is one)
x
f
fx
(x – x-bar)
(x – x-bar)²
f(x – x-bar)²
10
1
10
(10 – 4.9) = 5.1
26.01
26.01
9
2
18
(9 – 4.9) = 4.1
16.81
33.62
8
3
24
(8 – 4.9) = 3.1
9.61
28.83
7
1
7
(7 – 4.9) = 2.1
4.41
4.41
6
2
12
(6 – 4.9) = 1.1
1.21
2.42
5
12
60
(5 – 4.9) = 0.1
0.01
0.12
4
1
4
(4 – 4.9) = –0.9
0.81
0.81
3
1
3
(3– 4.9) = –1.9
3.61
3.61
2
3
6
(2 – 4.9) = –2.9
8.41
25.23
1
2
2
(1 – 4.9) = –3.9
15.21
30.42
0
2
0
(0 – 4.9) = –4.9
24.01
48.02
N = 30 ∑fX = 146 ∑f(X – X-bar) = 1.1
∑f(X – XShould be 0, because
bar)² = 203.5
of rounding to the
nearest tenth of the
mean
Mean: X-bar = ∑fX / N = 146 / 30 = 4.8666667 = 4.9 (rounding to the nearest tenth)
Variance: s² = f(x – x-bar)²/N = 203.5 / 30 = 6.7833333333
Standard deviation: s = s² = 6.7833333333 = 2.6044833141 = 2.6 (rounding to the
nearest tenth)
3
In-Class Exercise 4.3: Extract the first ten cases from Exercise 2 (p. 92) as your data set and
calculate the standard deviation. Show your work step by step.
The value of first ten cases: 70, 35, 86, 81, 63, 71, 58, 53, 99, 85
Step1: Rank the value of first ten cases (x) as the first column as below
Step 2: Make the table as below:
x
f
fx
(x – x-bar)
35
1
35
35 – 70 = –35
53
1
53
53 – 70 = –17
58
1
58
58 – 70 = –12
63
1
63
63 – 70 = –7
70
1
70
70 – 70 = 0
71
1
71
71 – 70 = 1
81
1
81
81 – 70 = 11
85
1
85
85 – 70 = 15
86
1
86
86 – 70 = 16
99
1
99
99 – 70 = 29
N = 10 ∑fX = 701 ∑f(X – X-bar) = 1
Should be 0, because of
rounding to the nearest
tenth of the mean
f(x – x-bar)²
1225
289
144
49
0
1
121
225
256
841
∑f(X – X-bar)² = 3151
Step 3: Calculate the mean:
X-bar = ∑fX / N = 701 / 10 = 70.1 = 70 (rounding to the nearest integer)
Step 4: Complete the column (x – x-bar) = 1 (Note: the sum of dispersion should be zero because
of rounding of mean in step 3)
Step 5: Complete the column (x – x-bar)² = 3151
Step 6: Calculate the variance
s² = f(x – x-bar)²/N = 3151/10 = 315.1
Step 7: Calculate the standard deviation
s = s² = 315.1 = 17.75105631 = 17.8 (rounding to the nearest tenth)
4

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