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The outcome of each of $30$ items was observed; $10$ items gave an outcome $\frac{1}{2} - d$ each, $10$ items gave outcome $\frac {1}{2}$ each and the remaining $10$ items gave outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac {4}{3}$ then $\left| d \right|$ equals
$\frac {2}{3}$
$2$
$\frac {\sqrt 5}{2}$
$\sqrt 2$
Solution
Variance remains some if same number is subracted from each observation. (subtract $10$ from each observation)
$\therefore \frac{{1{{\left( { – d} \right)}^2} + 10{{\left( 0 \right)}^2} + 10{{\left( d \right)}^2}}}{{30}} – {\left( {\frac{{10\left( { – d} \right) + 10\left( 0 \right) + 10\left( d \right)}}{{30}}} \right)^2} = \frac{4}{3}$
$\frac{{20{d^2}}}{{30}} = \frac{4}{3}$
$ \Rightarrow {d^2} = 2$
$\left( d \right) = \sqrt 2 $
Similar Questions
If the mean and variance of the frequency distribution
| $x_i$ | $2$ | $4$ | $6$ | $8$ | $10$ | $12$ | $14$ | $16$ |
| $f_i$ | $4$ | $4$ | $\alpha$ | $15$ | $8$ | $\beta$ | $4$ | $5$ |
are $9$ and $15.08$ respectively, then the value of $\alpha^2+\beta^2-\alpha \beta$ is $…………$.
