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Measures of Variability
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Measures of Variability
Why to Measure of Variability?
Information given by various measures of central tendency is too limited; and
A statistical value that indicates the degree or extent to which the observations in a set spread around the central tendency.
Set A: 80, 82, 86, 89 and 93
86xSet B: 83, 85, 86, 87 and 89
Set C: 84, 85, 86, 87 and 88 86x
86x
The spread of the scores in each sets are different
Set A has the most spread of score, while Set C has the least
Computing for the degree of dispersion of the scores from the average will describe a set of distribution adequately
Sets of Data
Range
Easiest and simplest to determine, depends pair of extreme values
Unstable, easily fluctuates with the change of highest and lowest score
Most unreliable measure, does not give the dispersion or spread of the scores in between two extreme values
Characteristics
Range
The difference between the highest and th lowest score
Difference between the exact lower limit of the lowest score and the exact upper limit of the highest score
Exclusive Range
Inclusive Range
Difference between the highest and the lowest score
Interquartile Range (IR) & Semi-Interquartile Range or Quartile Deviation (QD)
distance from the first quartile to the third quartile
one half the distance from the first quartile to the third quartile
IR = Q3 – Q1
2.. 13 QQ
DQ
Illustrative Example:
Daily Allowances
No. Of Students
P80 – P8990 – 99
100 – 109110 – 119120 – 129130 – 139140 - 149
591012302014
Below is a distribution of the daily allowances of 100 freshman college students.
Table 1
Range
IR = Q3 – Q1
= 149.5 – 79.5
= P705
1424366686
100
= 134 – 110.33
= P 23.67
<cf
1
1
1
41
.1 QQ
B
Q Cf
CfnLQ
1012
2425 109.5
33.110P
3
3
3
43
.3 QQ
B
Q Cf
CfnLQ
1020
6675 129.5
00.134P
213 QQ
QD
2
33.110134
84.11P
Mean or Average Deviation
measures the average deviation of the values from the arithmetic mean, ignoring the algebraic sign of each deviation.
Ungrouped Data Grouped Data
n
xxfMD
k
iii
1
n
xxMD
n
ii
1
Where: xi = class mark fi = class frequency n = total frequency
The life expectancies (in months) of a hypothetical species of birds in captivity are : 20, 24, 32, 36, 40 and 46. Solve for the mean deviation.
Illustrative Example MD Ungrouped Data
xi
202432364046
xxi
n
xxMD
n
ii
1
Solution:
6
46
= 7.67mos.
139137
13
46198
n
xx
n
ii
1
6
198
= 33mos.
Find the mean deviation of the data in Table Y
Daily Allowances
fi
P 80 – P 8990 – 99
100 – 109110 – 119120 – 129130 – 139140 – 149
59
1012302014
Illustrative Example MD Grouped Data
ix xxi xxf ii
Table Y
1, 356.8
n
xxfMD
k
iii
1
100
0.360,1
60.13P
84.594.5
104.5114.5124.5134.5144.5
36.926.916.9
6.93.1
13.123.1
184.5242.1169.0
82.893.0
262.0323.4
Variance and Standard Deviation
similar to mean deviation since it is computed based on the difference of each value form the mean, with two expectations – the deviations are squared and the average of the deviations is found using n-1 as divisor instead of n. variance will be denoted by s2.
Formula Variance Ungrouped Data
1
2
12
n
xxs
n
ii
Variance
Standard Deviation
1
1
2
n
xxs
n
ii
Formula Variance Ungrouped Data
N
xn
ii
1
2
Sample mean is given Population mean is given
Square root of the variance is more frequently used of dispersion or variability.
Solve for the variance and the standard deviation of the life expectancy of six birds in captivity. The mean is 33 months.
202432364046
Illustrative Example:
ix xxi
478
2xxi 1
2
12
n
xxs
n
ii
5
4782 s
6.952 s
6.95s
moss 78.9
16981
19
49169
-13-9-137
13
Variance and Standard Deviation of Grouped Data
)1(1
2
1
2
nn
xfxfn
s
k
i
k
iiiii
1
1
2
2
n
xxfs
k
iii
)1(1
2
1
2
2
nn
xfxfn
s
k
i
k
iiiii
1
1
2
n
xxfs
k
iii
Where: xi = class mark fi = class frequency
n = total frequency
Compute the variance and the standard deviation of the data in Table Y. The mean is P 121.40.
Daily Allowance
6808.056512.492856.10571.32288.30
3432.207471.54
ix xxi if
Table Y
27939.00 100
1361.61723.61285.61
47.619.61
171.61533.61
-36.9-26.9-16.9-6.93.1
13.123.1
84.594.5
104.5114.5124.5134.5144.5
59
1012302014
P 80 – P 89 90 – 99100 – 109110 – 119120 – 129130 – 139140 – 149
2xxi 2xxf ii
Illustrative example:
Solution
1
1
2
2
n
XXifs
k
ii
99
27939
21.282s
s = P16.80
The Coding Formula for Variance and for Standard Deviation
ClassP80 – P8990 – 99100 – 109
110 – 119120 – 129130 – 139140 – 149
59
1012302014
2iiUfiiUfiUif
2
2
11
2
2
)1(c
nn
UfUfn
s
k
iii
k
iii
cnn
UfUfn
s
k
iii
k
iii
)1(
2
11
2
-3-2-1
-15-18-10
0304042
4536100
3080
126
123
0
100 69 327
2
2
11
2
2
)1(c
nn
UfUfn
s
k
iii
k
iii
22
2 10)99(100
)69()327(100
s
21.2822 Ps
21.282s
s = P 16.80
Relative Dispersion
amount of variability relative to an average
100.. X
SVC
10013
13
QQCQD
A. Coefficient of Variation expresses the standard deviation as a percentage of the mean.
B. Coefficient of Quartile Deviation makes use of the first and third quartiles.
100.. X
SVC
10013
13
QQCQD
Illustrative examples: For the data in Table Y
10040.121
80.16
%84.13
10033.110134
33.110134
%69.9
Standard Score
A student scored 43 in a long test in Chemistry wherein
the mean and the standard deviation of the scores were
38 and 6, respectively. In a long test in Physics, the
student scored 45. the mean and standard deviation in
Physics were respectively 42 and 10. in which long test
was the student’s relative standing higher?
s
xxZ
Illustrative example:
The Z scores of the student are:
6
3843 Z
10
4245 Z
In Chemistry,
In Physics,
The student’s relative performance was higher in
Chemistry than in Physics.
Solution:
83.0
30.0
SKEWNESS
curve trails off to the left, then the distribution is negatively skewed
s
xxSk
)~(3
Sk > 0 (skewed to the right)
Sk = 0 (symmetric distribution)
Sk < 0 (skewed to
the left)
frequency curve
curve trails off to the right, then the distribution is positively skewed
Illustrative Example
s
xxSk
)~(3
For a distribution whose mean is 50.5, median is 51.6 and the standard deviation is 4.2, the Pearsonian coefficient of skewness is:
Skewness
42
)6.515.50(3
79.0
Solution:
KURTOSIS
Symmetrical curves vary in shape because they do not have the same peakedness
41
4)(
ns
xxfKur
k
iii
for ungrouped data
41
4)(
ns
xxKur
k
ii
for grouped data
k > 3, the curve
is leptokurtic
k < 3, the curve is platykurtic
k = 3, the curve
is mesokurtic
Class fi
18 – 2627 – 3536 - 4445 – 5354 – 6263 – 7172 – 8081 – 8990 - 98
2356
10111283
4)( xxf ii
41
4)(
ns
xxKur
k
ii
xxi
2)4.329(60
137,797,15
43.2
Table XYZ
60
7.63x-41.7-32.7-23.7-14.7
5.73.3
12.321.330.3
6,047,4773,430,1431,577,478
280,16910,556
1,305247,664
1,646,6772,528,668
15,797,137
Illustrative Example
223140495867768594
xi
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