Question

If the standard deviation of the numbers $ - 1,0,1,k$ is \[\sqrt 5 \] where \[k > 0\], then $k$ is equal to?
A. $2\sqrt {\dfrac{{10}}{3}} $
B. $2\sqrt 6 $
C. $4\sqrt {\dfrac{5}{3}} $
D. $\sqrt 6 $

Answer 1

Hint: Recall the formula of standard deviation, \[S.D. = \sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}} - {{\left( {\dfrac{1}{n}\sum\limits_{i = 1}^n {\left( {{x_i}} \right)} } \right)}^2}} \]. We will substitute the given values in the formula of standard deviation and then solve the equation for the value of $k$.

Complete step-by-step answer:
We know that the standard deviation of the data is given as \[S.D. = \sqrt {\dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i}} \right)}^2}} - {{\left( {\dfrac{1}{n}\sum\limits_{i = 1}^n {\left( {{x_i}} \right)} } \right)}^2}} \], where $\overline x $ is the mean of the data and $n$ is the number of observations.
Now, substitute all the given values in the formula of standard deviation.
$
   \Rightarrow S.D. = \sqrt {\dfrac{{{{\left( { - 1} \right)}^2} + {0^2} + {1^2} + {k^2}}}{4} - {{\left( {\dfrac{{ - 1 + 0 + 1 + k}}{4}} \right)}^2}} \\
   \Rightarrow \sqrt 5 = \sqrt {\dfrac{{2 + {k^2}}}{4} - {{\left( {\dfrac{k}{4}} \right)}^2}} \\
   \Rightarrow \sqrt 5 = \sqrt {\dfrac{{2 + {k^2}}}{4} - \dfrac{{{k^2}}}{{16}}} \\
$
Squaring on both sides,
$
   \Rightarrow 5 = \dfrac{{2 + {k^2}}}{4} - \dfrac{{{k^2}}}{{16}} \\
   \Rightarrow 5 = \dfrac{{8 + 4{k^2} - {k^2}}}{{16}} \\
   \Rightarrow 72 = 3{k^2} \\
$
Divide both sides by 3
$
   \Rightarrow {k^2} = 24 \\
   \Rightarrow k = \pm \sqrt {24} \\
$
But, we are given that \[k > 0\].
Hence, the value of $k$ is $\sqrt {24} = 2\sqrt 6 $.
Thus, option B is correct.

Note: One must know the formula of standard deviation. We have both negative and positive value of $k$ but we have taken only positive value because we are given in the question that $k > 0$