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If the standard deviation of the numbers $-1, 0, 1, k$ is $\sqrt 5$ where $k > 0,$ then $k$ is equal to
$4\sqrt {\frac {5}{3}}$
$\sqrt 6$
$2\sqrt 6$
$2\sqrt {\frac {10}{3}}$
Solution
$S.D. = \sqrt {\frac{{\sum {{{\left( {x – \bar x} \right)}^2}} }}{n}} $
$\bar x = \frac{{\sum x }}{4} = \frac{{ – 1 + 0 + 1 + k}}{4} = \frac{k}{4}$
Now $\sqrt 5 = \sqrt {\frac{{{{\left( { – 1 – \frac{k}{4}} \right)}^2} + {{\left( {0 – \frac{k}{4}} \right)}^2} + {{\left( {1 – \frac{k}{4}} \right)}^2} + {{\left( {k – \frac{k}{4}} \right)}^2}}}{4}} $
$ \Rightarrow 5 \times 4 = 2{\left( {1 + \frac{k}{{16}}} \right)^2} + \frac{{5{k^2}}}{8}$
$ \Rightarrow 18 = \frac{{3{k^2}}}{4}$
$ \Rightarrow {k^2} = 24$
$ \Rightarrow k = 2\sqrt 6 $
Similar Questions
If the variance of the frequency distribution is $160$ , then the value of $\mathrm{c} \in \mathrm{N}$ is
| $X$ | $c$ | $2c$ | $3c$ | $4c$ | $5c$ | $6c$ |
| $f$ | $2$ | $1$ | $1$ | $1$ | $1$ | $1$ |
