JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer If the standard deviation of the numbers -1, 0, 1, k is \[\sqrt{5}\]where \[k>0,\]then k is equal to [JEE Main 9-4-2019 Morning]

    A) \[2\sqrt{\frac{10}{3}}\]                     

    B) \[2\sqrt{6}\]

    C) \[4\sqrt{\frac{5}{3}}\]                       

    D) \[\sqrt{6}\]

    Correct Answer: B

    Solution :

    \[S.D=\sqrt{\frac{\Sigma {{\left( x-\overline{x} \right)}^{2}}}{n}}\]           \[\overline{x}=\frac{\Sigma x}{4}=\frac{-1+0+1+k}{4}=\frac{k}{4}\]   Now\[\sqrt{5}=\sqrt{\frac{{{\left( -1-\frac{k}{4} \right)}^{2}}+{{\left( 0-\frac{k}{4} \right)}^{2}}+{{\left( 1-\frac{k}{4} \right)}^{2}}+{{\left( k-\frac{k}{4} \right)}^{2}}}{4}}\]           \[\Rightarrow 5\times 4=2{{\left( 1+\frac{k}{16} \right)}^{2}}+\frac{5{{k}^{2}}}{8}\]           \[\Rightarrow \]\[18=\frac{3{{k}^{2}}}{4}\]\[\Rightarrow \]\[{{k}^{2}}=24\]\[\Rightarrow \]\[k=2\sqrt{6}\]

You need to login to perform this action.
You will be redirected in 3 sec spinner