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જો સંખ્યા $-1, 0, 1, k$ નો પ્રમાણિત વિચલન $\sqrt 5$ હોય તો $k$ = ............... ( જ્યાં $k > 0,$)
A
$4\sqrt {\frac {5}{3}}$
B
$\sqrt 6$
C
$2\sqrt 6$
D
$2\sqrt {\frac {10}{3}}$
(JEE MAIN-2019)
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 $
Standard 11
Mathematics
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નીચે આપેલ માહિતી માટે પ્રમાણિત વિચલન શોધો :
${x_i}$ | $3$ | $8$ | $13$ | $18$ | $25$ |
${f_i}$ | $7$ | $10$ | $15$ | $10$ | $6$ |
medium