If the standard deviation of the numbers -1, | vedclass

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Question

$S.D. = \sqrt {\frac{{\sum {{{\left( {x - \bar x} \right)}^2}} }}{n}} $

$\bar x = \frac{{\sum x }}{4} = \frac{{ - 1 + 0 + 1 + k}}{4} = \frac{k}{4}$

Vedclass · Accepted Answer

$2\sqrt 6$

Vedclass · Answer

$S.D. = \sqrt {\frac{{\sum {{{\left( {x - \bar x} ight)}^2}} }}{n}} $

$\bar x = \frac{{\sum x }}{4} = \frac{{ - 1 + 0 + 1 + k}}{4} = \frac{k}{4}$

Now $\sqrt 5 = \sqrt {\frac{{{{\left( { - 1 - \frac{k}{4}} ight)}^2} + {{\left( {0 - \frac{k}{4}} ight)}^2} + {{\left( {1 - \frac{k}{4}} ight)}^2} + {{\left( {k - \frac{k}{4}} ight)}^2}}}{4}} $

$ \Rightarrow 5 	imes 4 = 2{\left( {1 + \frac{k}{{16}}} ight)^2} + \frac{{5{k^2}}}{8}$

$ \Rightarrow 18 = \frac{{3{k^2}}}{4}$

$ \Rightarrow {k^2} = 24$

$ \Rightarrow k = 2\sqrt 6 $

Class	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
$f_i$	$3$	$7$	$12$	$15$	$8$	$3$	$2$

If the standard deviation of the numbers $-1, 0, 1, k$ is $\sqrt 5$ where $k > 0,$ then $k$ is equal to

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