The standard deviation of the scores 505, 510 | vedclass

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(C) The given scores are $505, 510, 515, \ldots, 595$. This is an arithmetic progression with $a = 505$,$d = 5$,and $n = 19$ terms.The mean $\bar{X} =

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$5 \sqrt{30}$

Vedclass · Answer

(C) The given scores are $505, 510, 515, \ldots, 595$. This is an arithmetic progression with $a = 505$,$d = 5$,and $n = 19$ terms.The mean $\bar{X} = \frac{505 + 595}{2} = 550$.The standard deviation $\sigma$ is given by $\sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{X})^2}$.Let $x_i = 550 + 5k$,where $k$ ranges from $-9$ to $9$.Then $(x_i - \bar{X})^2 = (5k)^2 = 25k^2$.$\sum (x_i - \bar{X})^2 = 25 \sum_{k=-9}^{9} k^2 = 25 	imes 2 	imes \sum_{k=1}^{9} k^2$.Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get $\sum_{k=1}^{9} k^2 = \frac{9(10)(19)}{6} = 285$.So,$\sum (x_i - \bar{X})^2 = 50 	imes 285 = 14250$.$\sigma = \sqrt{\frac{14250}{19}} = \sqrt{750} = \sqrt{25 	imes 30} = 5 \sqrt{30}$.

${x_i}$	$3$	$8$	$13$	$18$	$23$
${f_i}$	$7$	$10$	$15$	$10$	$6$

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

The standard deviation of the scores $505, 510, 515, 520, \ldots, 595$ is

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