The average of n numbers x_1, x_2, x_3, ..., | vedclass

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(B) The initial average $M$ is given by $M = \frac{x_1 + x_2 + x_3 + ... + x_n}{n}$.This implies the sum of the $n$ numbers is $nM = x_1 + x_2 + x_3 +

Vedclass · Accepted Answer

$\frac{nM - x_n + x'}{n}$

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(B) The initial average $M$ is given by $M = \frac{x_1 + x_2 + x_3 + ... + x_n}{n}$.This implies the sum of the $n$ numbers is $nM = x_1 + x_2 + x_3 + ... + x_n$.When $x_n$ is replaced by $x'$,the new sum of the numbers becomes $S' = (x_1 + x_2 + x_3 + ... + x_{n-1} + x_n) - x_n + x'$.Substituting the sum $nM$ for the original numbers,we get $S' = nM - x_n + x'$.The new average is the new sum divided by $n$,which is $\frac{nM - x_n + x'}{n}$.

Size of Item	$1$	$2$	$3$	$4$	$5$	$6$	$7$
Frequency	$2$	$4$	$5$	$8$	$7$	$3$	$2$

$x_i$	$5$	$8$	$11$	$14$	$17$
$f_i$	$4$	$5$	$6$	$10$	$20$

The average of $n$ numbers $x_1, x_2, x_3, ..., x_n$ is $M$. If $x_n$ is replaced by $x'$,then the new average is:

Similar Questions

The upper quartile for the following distribution is given by the size of:
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Frequency $2$ $4$ $5$ $8$ $7$ $3$ $2$

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$f_i$ $4$ $5$ $6$ $10$ $20$

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