If the variance of 10 natural numbers 1,1,1, | vedclass

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Question

$\sigma^{2}=\frac{\Sigma x ^{2}}{ n }-\left(\frac{\Sigma x }{ n }\right)^{2}$

$=\frac{9+ k ^{2}}{10}-\left(\frac{9+ k }{10}\right)^{2}<10$

$9

Vedclass · Accepted Answer

$11$

Vedclass · Answer

$\sigma^{2}=\frac{\Sigma x ^{2}}{ n }-\left(\frac{\Sigma x }{ n }\right)^{2}$

$=\frac{9+ k ^{2}}{10}-\left(\frac{9+ k }{10}\right)^{2}<10$

$90+10 k^{2}-81-k^{2}-18 k < 1000$

$9 k ^{2}-18 k -991 < 0$

$k^{2}-2 k < \frac{991}{9}$

$( k -1)^{2} < \frac{1000}{9}$

$\frac{-10 \sqrt{10}}{3} <  k -1 < \frac{10 \sqrt{10}}{3}$

$k < \frac{10 \sqrt{10}}{3}+1$

$k \leq 11$

Maximum value of $k$ is $11 .$

Classes	$0-30$	$30-60$	$60-90$	$90-120$	$120-150$	$50-180$	$180-210$
$f_i$	$2$	$3$	$5$	$10$	$3$	$5$	$2$

If the variance of $10$ natural numbers $1,1,1, \ldots ., 1, k$ is less than $10 ,$ then the maximum possible value of $k$ is ...... .

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