What is the standard deviation of the numbers | vedclass

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(A) The given numbers $31, 32, 33, \dots, 47$ form an arithmetic progression with $n+1$ terms.Here,the number of terms $N = 47 - 31 + 1 = 17$.The form

Vedclass · Accepted Answer

$2\sqrt{6}$

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(A) The given numbers $31, 32, 33, \dots, 47$ form an arithmetic progression with $n+1$ terms.Here,the number of terms $N = 47 - 31 + 1 = 17$.The formula for the standard deviation of $N$ consecutive integers is $\sigma = \sqrt{\frac{N^2 - 1}{12}}$.Substituting $N = 17$:$\sigma = \sqrt{\frac{17^2 - 1}{12}} = \sqrt{\frac{289 - 1}{12}} = \sqrt{\frac{288}{12}} = \sqrt{24} = 2\sqrt{6}$.Thus,the correct option is $A$.

$x_i$	$6$	$10$	$14$	$18$	$24$	$28$	$30$
$f_i$	$2$	$4$	$7$	$12$	$8$	$4$	$3$

What is the standard deviation of the numbers $31, 32, 33, \dots, 47$?

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