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

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JEE Mains Previous Paper 1 (Held On: 09 Apr 2019 Shift 1)
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  1. \(2\sqrt 6\)
  2. \(2\sqrt {\frac{{10}}{3}}\)
  3. \(4\sqrt {\frac{5}{3}}\)
  4. \(\sqrt 6\)

Answer (Detailed Solution Below)

Option 1 : \(2\sqrt 6\)
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JEE Main 04 April 2024 Shift 1
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Detailed Solution

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Mean of given observation = \(\frac{k}{4}\)

Given, standard deviation, \(\sigma = \sqrt 5\) 

Since, Standard deviation is the square root of variance.

Variance, σ2 = 5      -----(i)

Formula for variance, \({\sigma ^2} = \frac{1}{n}\Sigma {\left( {{x_i} - \bar x} \right)^2}\) 

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

\(= \frac{{{{\left( {\frac{k}{4} + 1} \right)}^2} + {{\left( {\frac{k}{4}} \right)}^2} + {{\left( {\frac{k}{4} - 1} \right)}^2} + {{\left( {\frac{{ - 3k}}{4}} \right)}^2}}}{4}\)

\(= \frac{{\left( {\frac{{{{\rm{k}}^2}}}{{16}} + \frac{{2k}}{4} + 1} \right) + \frac{{{{\rm{k}}^2}}}{{16}} + \left( {\frac{{{{\rm{k}}^2}}}{{16}} - \frac{{2k}}{4} + 1} \right) + \frac{{9{{\rm{k}}^2}}}{{16}}}}{4}\)

\(= \frac{{\left( {\frac{{{{\rm{k}}^2}}}{{16}}} \right) + \frac{{{{\rm{k}}^2}}}{{16}} + \left( {\frac{{{{\rm{k}}^2}}}{{16}}} \right) + \frac{{9{{\rm{k}}^2}}}{{16}} + 2}}{4}\)

\({\sigma ^2} = \frac{{\frac{{12{k^2}}}{{16}} + 2}}{4}\)       -----(ii)

Equating (i) and (ii), we get

\(\frac{{\frac{{12{k^2}}}{{16}} + 2}}{4} = 5\)

\(\Rightarrow \frac{{12{k^2}}}{{16}} + 2 = 20\)

\(\Rightarrow \frac{{12{k^2}}}{{16}} = 20 - 2\;\)

⇒ 12k2 = 18 × 16

⇒ 12k2 = 288

\(\Rightarrow {k^2} = \frac{{288}}{{12}} = 24\)

\(\Rightarrow k = \sqrt {24}\)

\(\Rightarrow k = 2\sqrt 6\)

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