If the variance of the numbers 9, 15, 21, ldo | vedclass

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(C) The given sequence is $9, 15, 21, \ldots, (6n+3)$. This is an Arithmetic Progression with first term $a = 9$ and common difference $d = 6$. The va

Vedclass · Accepted Answer

$\frac{P}{9}$

Vedclass · Answer

(C) The given sequence is $9, 15, 21, \ldots, (6n+3)$. This is an Arithmetic Progression with first term $a = 9$ and common difference $d = 6$. The variance of an $AP$ with $n$ terms is given by $\sigma^2 = \frac{(n^2-1)d^2}{12}$. Substituting $d = 6$,we get $P = \frac{(n^2-1) 	imes 6^2}{12} = \frac{(n^2-1) 	imes 36}{12} = 3(n^2-1)$. The first $n$ even numbers are $2, 4, 6, \ldots, 2n$. This is an $AP$ with $a = 2$ and $d = 2$. The variance of these $n$ even numbers is $\sigma_{even}^2 = \frac{(n^2-1) 	imes 2^2}{12} = \frac{(n^2-1) 	imes 4}{12} = \frac{n^2-1}{3}$. From the first equation,$n^2-1 = \frac{P}{3}$. Substituting this into the second variance,we get $\sigma_{even}^2 = \frac{1}{3} 	imes \frac{P}{3} = \frac{P}{9}$.

$C$.$I$.	$75$-$175$	$175$-$275$	$275$-$375$	$375$-$475$	$475$-$575$	$575$-$675$	$675$-$775$
$f_i$	$3$	$2$	$1$	$0$	$1$	$2$	$3$

If the variance of the numbers $9, 15, 21, \ldots, (6n+3)$ is $P$,then the variance of the first $n$ even numbers is

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$C$.$I$. $75$-$175$ $175$-$275$ $275$-$375$ $375$-$475$ $475$-$575$ $575$-$675$ $675$-$775$
$f_i$ $3$ $2$ $1$ $0$ $1$ $2$ $3$
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