If x_{i} are the midpoints of the class inter | vedclass

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(A) The mean $\bar{x}$ of grouped data is given by the formula $\bar{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$,where $n = \sum f_{i}$.This can be rewr

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(A) The mean $\bar{x}$ of grouped data is given by the formula $\bar{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$,where $n = \sum f_{i}$.This can be rewritten as $\sum f_{i} x_{i} = n \bar{x}$.We need to evaluate the sum $\sum (f_{i} x_{i} - f_{i} \bar{x})$.Using the properties of summation,we get $\sum f_{i} x_{i} - \sum f_{i} \bar{x}$.Since $\bar{x}$ is a constant,we can take it out of the summation: $\sum f_{i} x_{i} - \bar{x} \sum f_{i}$.Substituting $\sum f_{i} x_{i} = n \bar{x}$ and $\sum f_{i} = n$,we get $n \bar{x} - \bar{x} (n) = n \bar{x} - n \bar{x} = 0$.

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
Frequency	$5$	$15$	$20$	$18$	$2$

Number of pages written per day	$16-18$	$19-21$	$22-24$	$25-27$	$28-30$
Number of days	$1$	$3$	$4$	$9$	$13$

If $x_{i}$ are the midpoints of the class intervals of grouped data,$f_{i}$ are the corresponding frequencies,and $\bar{x}$ is the mean,then $\sum (f_{i} x_{i} - f_{i} \bar{x})$ is equal to

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