7.3 Variance and Standard Deviation

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 ¯ = fx 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 ¯ = fx 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