Measure of Dispersion or Variability
Lecture 4
Dr. Maher H.S. Alasady
The mean, mode and median do a nice job in telling where the center of
the data set is, but often we are interested in more. For example, a
pharmaceutical engineer develops a new drug that regulator iron in the blood.
Suppose he finds out that the average sugar content after taking the
medication is optimal level. This does not mean that the drug is effective.
There is a possibility that half of the patients have dangerously low sugar
content while the other half has dangerously high content. Instead of the drug
being an effective regulator, it is a deadly poison. What the pharmacist needs
is a measure of how far the data is spread apart. This is what the variance
and standard deviation do.
The most common measures of dispersion or variability are (Range,
Variance, Standard deviation and Coefficient of variation).
Range (R):
The range is the difference between the largest (XL) and the smallest
(XS) values in a set of observations.
R = X L - XS
Note: The range is poor measure of dispersion? Because it only takes into
account two of the values.
Variance:
The variance is the most commonly used to measure of spread in
biological statistics. For a population is defined as the sum of squares of the
deviation from the mean (SS), dividing by the total number of the deviations,
and by one less than the total number of the deviation (degree of freedom,
df) for a sample.
N
 
2
( X  )
i 1
N
2
or
2

 N
 
  Xi  
N
1
 2   Xi 2   i 1   ……..……….. (Population)

N  i 1
N




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n
S 
2
__
 ( Xi  X )
i 1
n 1
2
or
2

 n
 
  Xi  
 n
1
 Xi 2   i 1  
S2 

 ……..……..….. (Sample)
n  1  i 1
n




Note: The reason for dividing by (n-1), (df), in sample, because the sum of
deviations of the values from their mean is equal zero.
n
__
 ( Xi  X )  0
i 1
Why the separate formula for sample?
The formula for sample divides by n-1 to:
1. Correct for probability that most extreme cases will be excluded from a
smaller sample.
2. Makes the sample more representative of the population for every small
sample.
3. Reduces the denominator to a larger extent (If n=5 they we have a 20%
reduction in the denominator). But in large samples the n-1 correction does
not have as large effect.
Standard Deviation (s) or (sd):
Is defined as a positive square root of variance.
   2 ……………. (Population)
s  s 2 ……………… (Sample)
The mean squared difference from the sample mean will, on average,
underestimate the population variance. In some samples it will overestimate
it, but most of the time it will underestimate it, if the formula is modified so
that the sum of squared deviations is divided by n-1 rather than N, then the
tendency to underestimate the population variance is eliminated.
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Coefficient of Variation (C.V):
Is defined as the ratio of the standard deviation to the mean. It is
independent of the units employed.
C.V 
C .V 


s
__
or
C.V 

100%

or
C.V 
s
X
__
……………. (Population)
100% ……………… (Sample)
X
Note: For biological data coefficient of variation (C.V) of the order of 10% to
15% are common. For homogenous material this figure my be reduced to 5%
while a coefficient of variation of 25% would indicate very considerable
variability.
Example: The data below present the technician from two different
laboratories, all making the same specific blood chemistry
determination using a solution with a know concentration (5 mg/ml).
Laboratories
1
2
C.V 
Technician
5, 8, 6
6, 4, 9
Mean
6.333
6.333
1.528
100%  24%
6.333
C.V 
Standard deviation
1.528
2.517
2.517
100%  39.7%
6.333
Lab. 1 gives most accurate result.
Example: A set of data (4, 6, 3, 4, 5 and 2) compute: The range, the variance,
the standard deviation and the coefficient of variation?
Solution:
R = XL - XS ................................ R= 6-2=4
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2

 n
 
  Xi  

1  n
2
2
S 
Xi   i 1  



n  1 i 1
n




1  2
(4  6  3  4  5  2)2 
2
2
2
2
2
S 
(4  6  3  4  5  2 ) 
2
6 1 
6

2
s  s 2 …................................. s  2 = 1.414
n
__
X
 Xi
i 1
N
C .V 
s
__
__
…................................ X 
4 63 45 2
4
6
…................................ C.V 
X
1.414
100%  35.35%
4
Example: The following shows the hemoglobin values (g/100ml) of 30 children
receiving treatment for hemolytic anemia, compute: The variance,
the standard deviation and the coefficient of variation?
Hemoglobin
6.5 – 7.5
7.5 – 8.5
8.5 – 9.5
9.5 – 10.5
10.5 – 11.5
11.5 – 12.5
Midpoint Frequency
(mi)
(fi)
7
8
9
10
11
12
20
1
5
11
9
3
1
 fi =30
Solution:
2

 k
 
  mifi  
k
1
2
2
 

mi fi   i 1 k

Variance for grouped data ………… S  k


fi  1  i 1
fi 


i 1
i 1


1  2
(7 *1  8 * 5  ...  12 *1) 2 
2
2
S 
(7 *1  8 * 5  ...  12 *1) 
  1.199
30  1 
30

2
s  s 2 …....................................... s  1.199 =1.095
k
__
X 
 mifi
i 1
k
 fi
………………………..
(7 *1  8 * 5  9 *11  10 * 9  11* 3  12 *1)
 9.367
30
i 1
C.V 
1.095
100%  11.69%
9.367
Home Work 3:
Q1: The following table gives the results of a survey to study the ages and
hemoglobin levels of patients of a certain clinic.
Age (Years)
Hemoglobin level (g/dl)
Mean
30
60
Standard Deviation
6
10
Determine whether hemoglobin levels of the patients are more variable than
ages?
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Q2: The weights (in kg) of 8 pregnant women gave the following results:
8
x
i 1
i
8
x
 495
i 1
Find: (a) The mean.
2
i
 30659
(b) The variance.
(c) The standard deviation.
(d) The coefficient of variation.
Q3: The following are the glucose levels (g/100ml) of a sample of 50 children.
126
114
115
115
110
117
115
88
116
119
101
113
113
109
115
100
112
90
108
83
116
113
89
122
109
111
132
106
123
117
120
130
104
149
118
138
128
126
140
110
118
122
127
121
108
108
121
111
137
134
Prepare: 1. Grouped data by frequency distribution.
2. Find the mean, median and mode.
3. Computed the R, S2, S and C.V.?
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