5 students of a class have an average height | vedclass

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(C) Let the heights of $5$ students be $x_1, x_2, x_3, x_4, x_5$.Given mean $\bar{x} = \frac{\sum_{i=1}^5 x_i}{5} = 150 \implies \sum_{i=1}^5 x_i = 75

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) Let the heights of $5$ students be $x_1, x_2, x_3, x_4, x_5$.Given mean $\bar{x} = \frac{\sum_{i=1}^5 x_i}{5} = 150 \implies \sum_{i=1}^5 x_i = 750$.Given variance $\sigma^2 = \frac{\sum x_i^2}{5} - (\bar{x})^2 = 18$.$\frac{\sum x_i^2}{5} - (150)^2 = 18 \implies \frac{\sum x_i^2}{5} = 22500 + 18 = 22518$.$\sum_{i=1}^5 x_i^2 = 112590$.Now,a new student with height $x_6 = 156$ joins.The new sum of heights is $750 + 156 = 906$.The new mean $\bar{x}_{new} = \frac{906}{6} = 151$.The new sum of squares is $\sum_{i=1}^6 x_i^2 = 112590 + (156)^2 = 112590 + 24336 = 136926$.The new variance is $\frac{\sum_{i=1}^6 x_i^2}{6} - (\bar{x}_{new})^2 = \frac{136926}{6} - (151)^2$.$= 22821 - 22801 = 20 \, cm^2$.

$X$	$35$	$54$	$52$	$53$	$56$	$58$	$52$	$50$	$51$	$49$
$Y$	$108$	$107$	$105$	$105$	$106$	$107$	$104$	$103$	$104$	$101$

$5$ students of a class have an average height $150 \, cm$ and variance $18 \, cm^2$. $A$ new student,whose height is $156 \, cm$,joined them. The variance (in $cm^2$) of the height of these $6$ students is

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$X$ $35$ $54$ $52$ $53$ $56$ $58$ $52$ $50$ $51$ $49$

$Y$ $108$ $107$ $105$ $105$ $106$ $107$ $104$ $103$ $104$ $101$

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Find the mean and variance for the first $n$ natural numbers.

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