The variance of the numbers 8, 21, 34, 47, ld | vedclass

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(D) The given sequence is an arithmetic progression with first term $a = 8$ and common difference $d = 13$. Let $n$ be the number of terms. The $n$-th

Vedclass · Accepted Answer

$8788$

Vedclass · Answer

(D) The given sequence is an arithmetic progression with first term $a = 8$ and common difference $d = 13$. Let $n$ be the number of terms. The $n$-th term is $a_n = a + (n-1)d = 320$. $8 + (n-1)13 = 320 \implies 13(n-1) = 312 \implies n-1 = 24 \implies n = 25$. The mean $\bar{x} = \frac{8 + 320}{2} = \frac{328}{2} = 164$. The variance $\sigma^2$ of an arithmetic progression with $n$ terms is given by $\sigma^2 = \frac{(n^2-1)d^2}{12}$. Substituting the values: $\sigma^2 = \frac{(25^2 - 1) 	imes 13^2}{12} = \frac{(625 - 1) 	imes 169}{12} = \frac{624 	imes 169}{12}$. $\sigma^2 = 52 	imes 169 = 8788$.

$x_i$	$92$	$93$	$97$	$98$	$102$	$104$	$109$
$f_i$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

The variance of the numbers $8, 21, 34, 47, \ldots, 320$ is . . . . . . .

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