$i f x_{i}^{'} s a r e t h e m i d p o i n t s o f t h e c l a s s i n t e r v a l s o f a g r o u p e d d a t a, f_{i}^{'} s a r e t h e c o r r e s p o n d i n g f r e q u e n c i e s a n d ¯ x i s t h e m e a n t h e n \sum f_{i} (x_{i} - ¯ x) = ?$

(a) 1 (b) 0 (c) -1 (d) 2

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Solution

option b

Mean (x) = $\frac{S u m o f a l l t h e o b s e r v a t i o n s}{N u m b e r o f o b s e r v a t i o n s}$

x = $\frac{(f 1 x 1 + f 2 x 2 + \dots . . + f n x n)}{f 1 + f 2 + . . . . . . + f n}$
x = Σfixi / Σfi , Σfi = n
x = Σfixi / n
n x = Σfixi ……………(1)

Σ(fixi - x) = ( f1x1 - x) + ( f2x2 - x)+ …..+(fnxn - x)
Σ(fixi - x) = (f1x1 + f2x2 + …..+ fnxn) - (x +x +....n times)
Σ(fixi - x) = Σfixi -nx
Σ(fixi - x) = nx - nx ( From eq1)
Σ(fixi - x) = 0

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if x′is are the midpoints of the class intervals of a grouped data,f′is are the corresponding frequencies and ¯x is the mean then ∑fi(xi−¯x)=? (a) 1 (b) 0 (c) -1 (d) 2

$i f x_{i}^{'} s a r e t h e m i d p o i n t s o f t h e c l a s s i n t e r v a l s o f a g r o u p e d d a t a, f_{i}^{'} s a r e t h e c o r r e s p o n d i n g f r e q u e n c i e s a n d ¯ x i s t h e m e a n t h e n \sum f_{i} (x_{i} - ¯ x) = ?$

(a) 1 (b) 0 (c) -1 (d) 2