## VARIANCE FOR GROUPED AND UNGROUPED DATA

Question 1 :

Calculate the variance of the following table.

x    2    4    6     8    10    12    14    16

f    4    4    5    15     8      5      4    5

Solution :

 x246810121416 f445158545 d = x-82-8 = -64-8 = -46-8 = -28-8 = 010-8 = 212-8 = 414-8 = 616-8 = 8 d23616404163664 fd2144642003280144320

Σf  =  50 and Σfd²  =  804

σ2  =  √(Σfd²/Σf)

=  √(804/50)

=  16.08

Question 2 :

Find the variance of the following distribution

 Class interval20-2425-2930-3435-3940-4445-49 Frequency15252812128

Solution :

Here we consider the first data that is class interval as (x) and no of frequency as (f).

 x222732374247 f15252812128 d = x-3222-32 = -1027-32 = -532-32 = 037-32 = 542-32 = 1047-32 = 15 d210025025100225 fd21500625030012001800

Σfd² =  5425 and Σf  =  100

σ  =  √(Σfd²/Σf)

=  √(5425/100)

=  √54.25

=   7.37

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