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Measures of Dispersion (Part 3)

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Measures of Dispersion (Part 3)
7.3 Variance and Standard Deviation

1. The variance is a measure of the mean for the square of the deviations from the mean.

2. The standard deviation refers to the square root for the variance.

(A) Ungrouped Data




Example 1:
Find the variance and standard deviation of the following data.
15, 17, 21, 24 and 31

Solution:
Variance,  σ 2 = ∑ x 2 N − x ¯ 2 σ 2 = 15 2 + 17 2 + 21 2 + 24 2 + 31 2 5 − ( 15 + 17 + 21 + 24 + 31 5 ) 2 σ 2 = 2492 5 − 21.6 2 σ 2 = 31.84 Standard deviation,  σ  =  variance σ  =  31.84 σ  =  5.642


(B) Grouped Data (without Class Interval)




Example 2:
The data below shows the numbers of children of 30 families:

Number of child
2
3
4
5
6
7
8
Frequency
6
8
5
3
3
3
2




Find the variance and standard deviation of the data.
Solution:
Mean  x ¯ = ∑ f x ∑ f = ( 6 ) ( 2 ) + ( 8 ) ( 3 ) + ( 5 ) ( 4 ) + ( 3 ) ( 5 ) + ( 3 ) ( 6 ) + ( 3 ) ( 7 ) + ( 2 ) ( 8 ) 6 + 8 + 5 + 3 + 3 + 3 + 2 = 126 30 = 4.2 ∑ f x 2 ∑ f = ( 6 ) ( 2 ) 2 + ( 8 ) ( 3 ) 2 + ( 5 ) ( 4 ) 2 + ( 3 ) ( 5 ) 2 + ( 3 ) ( 6 ) 2 + ( 3 ) ( 7 ) 2 + ( 2 ) ( 8 ) 2 6 + 8 + 5 + 3 + 3 + 3 + 2 = 634 30 = 21.13 Variance,  σ 2 = ∑ f x 2 ∑ f − x ¯ 2 σ 2 = 21.133 − 4.2 2 σ 2 = 3.493 Standard deviation,  σ  =  variance σ  =  3.493 σ  = 1 .869


(C) Grouped Data (with Class Interval)


Example 3:

Daily Salary(RM)
Number of workers
10 – 14
40
15 – 19
25
20 – 24
15
25 – 29
12
30 – 34
8
Find the mean of daily salary and its standard deviation.

Solution:

Daily Salary (RM)
Number of workers, f
Midpoint, x
fx
fx2
10 – 14
40
12
480
5760
15 – 19
25
17
425
7225
20 – 24
15
22
330
7260
25 – 29
12
27
324
8748
30 – 34
8
32
256
8192
Total
100

1815
37185
Mean  x ¯ = ∑ f x ∑ f Mean of daily salary = 1815 100 = 18.15 Variance,  σ 2 = ∑ f x 2 ∑ f − x ¯ 2 Standard deviation,  σ  =  variance σ 2 = 37185 100 − 18.15 2 σ 2 = 42.43 σ  =  42.43 σ  = 6 .514