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 $n \bar{x} - \bar{x} (n) = n \bar{x} - n \bar{x} = 0$.

Marks	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Number of students	$26$	$26$	$x$	$110$	$84$	$y$	$36$	$32$

Class	$200-299$	$300-399$	$400-499$	$500-599$	$600-699$	$700-799$	$800-899$
Frequency	$3$	$61$	$118$	$139$	$126$	$151$	$2$

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

Similar Questions

In the formula $Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$ for the mode,$f_{2} = \ldots \ldots \ldots$

The frequency distribution of marks obtained by $400$ students is given below. If the mean of the frequency distribution is $41.2$,find the missing frequencies $x$ and $y$.
Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Number of students $26$ $26$ $x$ $110$ $84$ $y$ $36$ $32$

For the calculations of mode of a grouped data,the maximum frequency of a class is denoted by $\ldots \ldots \ldots . .$

Find the mean of the following data:

Class $200-299$ $300-399$ $400-499$ $500-599$ $600-699$ $700-799$ $800-899$

Frequency $3$ $61$ $118$ $139$ $126$ $151$ $2$

If $Z+M=88$ and $Z-M=2$,then $M=\ldots \ldots \ldots \ldots . .$

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