Let mu be the mean and sigma be the standard | vedclass

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Question

$\sum f _{ i }=62$

$3 k ^2+16 k -12 k -64=0$

$k =	ext { or }-\frac{16}{3}(	ext { rejected) }$

$\mu=\frac{\sum f _{ i } x _{ i }}{\sum f _{

Vedclass · Accepted Answer

$8$

Vedclass · Answer

$\sum f _{ i }=62$

$3 k ^2+16 k -12 k -64=0$

$k =	ext { or }-\frac{16}{3}(	ext { rejected) }$

$\mu=\frac{\sum f _{ i } x _{ i }}{\sum f _{ i }}$

$\mu=\frac{8+2(15)+3(15)+4(17)+5}{62}=\frac{156}{62}$

$\sigma^2=\sum f _{ i } x _{ i }^2-\left(\sum f _{ i } x _{ i }ight)^2$

$=\frac{8 	imes 1^2+15 	imes 13+17 	imes 16+25}{62}-\left(\frac{156}{62}ight)^2$

$\sigma^2=\frac{500}{62}-\left(\frac{156}{62}ight)^2$

$\sigma^2+\mu^2=\frac{500}{62}$

${\left[\sigma^2+\mu^2ight]=8}$

$X_i$	$0$	$1$	$2$	$3$	$4$	$5$
$f_i$	$k+2$	$2k$	$K^{2}-1$	$K^{2}-1$	$K^{2}-1$	$k-3$

Height in cms	$70-75$	$75-80$	$80-85$	$85-90$	$90-95$	$95-100$	$100-105$	$105-110$	$110-115$
No. of children	$3$	$4$	$7$	$7$	$15$	$9$	$6$	$6$	$3$

Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution

$X_i$ $0$ $1$ $2$ $3$ $4$ $5$

$f_i$ $k+2$ $2k$ $K^{2}-1$ $K^{2}-1$ $K^{2}-1$ $k-3$

where $\sum f_i=62$. if $[x]$ denotes the greatest integer $\leq x$, then $\left[\mu^2+\sigma^2\right]$ is equal $.........$.

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