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# Engineering

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

  • Option 1)

    2\sqrt{6}              

  • Option 2)

    2\sqrt{\frac{10}{3}}

  • Option 3)

    4\sqrt{\frac{5}{3}}

  • Option 4)

    \sqrt{6}

 

Answers (1)
S solutionqc

S.D=\sqrt{\frac{\Sigma xi^{2}}{n}-\left ( \frac{\Sigma xi}{n} \right )^{2}}=\sqrt{5}

         \Rightarrow \frac{\Sigma xi^{2}}{n}-\left ( \frac{\Sigma xi}{n} \right )^{2}=5

       \Rightarrow\frac{(-1^{2}+0^{2}+1^{2}+k^{2})}{4}-\left ( \frac{-1+0+1+k}{4} \right )^{2}=5

         \Rightarrow\frac{2+k^{2}}{4}-\frac{k^{2}}{16}=5

                 8+4k^{2}-k^{2}=80

                  3k^{2}=72

                   k=2\sqrt{6}


Option 1)

2\sqrt{6}              

Option 2)

2\sqrt{\frac{10}{3}}

Option 3)

4\sqrt{\frac{5}{3}}

Option 4)

\sqrt{6}

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