The mean of 5 observations is 15 and variance | vedclass

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(D) Let the observations be $x_1, x_2, x_3, x_4, x_5$. The mean $\bar{x} = \frac{\sum_{i=1}^{5} x_i}{5} = 15$,so $\sum x_i = 75$. Given variance $\sig

Vedclass · Accepted Answer

$\frac{2659}{49}$

Vedclass · Answer

(D) Let the observations be $x_1, x_2, x_3, x_4, x_5$. The mean $\bar{x} = \frac{\sum_{i=1}^{5} x_i}{5} = 15$,so $\sum x_i = 75$. Given variance $\sigma^2 = \frac{\sum x_i^2}{5} - (\bar{x})^2 = 9$. $\Rightarrow \frac{\sum x_i^2}{5} - 225 = 9$ $\Rightarrow \sum x_i^2 = 5(234) = 1170$. Now,two new observations $-5$ and $13$ are added. The new sum of observations is $75 - 5 + 13 = 83$. The new sum of squares is $1170 + (-5)^2 + (13)^2 = 1170 + 25 + 169 = 1364$. The new mean is $\bar{x}' = \frac{83}{7}$. The new variance is $\sigma'^2 = \frac{\sum x_i^2 + 25 + 169}{7} - (\bar{x}')^2 = \frac{1364}{7} - (\frac{83}{7})^2$. $\sigma'^2 = \frac{1364 	imes 7 - 6889}{49} = \frac{9548 - 6889}{49} = \frac{2659}{49}$. Thus,the correct option is $D$.

Measure	Height	Weight
Mean	$162.6 \ cm$	$52.36 \ kg$
Variance	$127.69 \ cm^2$	$23.1361 \ kg^2$

The mean of $5$ observations is $15$ and variance is $9$. If two observations having values $-5$ and $13$ are combined with these observations,then what will be the new variance?

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