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Let $x_1,x_2,.........,x_{100}$ are $100$ observations such that $\sum {{x_i} = 0,\,\sum\limits_{1 \leqslant i \leqslant j \leqslant 100} {\left| {{x_i}{x_j}} \right|} } = 80000\,\& $ mean deviation from their mean is $5,$ then their standard deviation, is-
$10$
$30$
$40$
$50$
Solution
$\bar{x}=\frac{\sum x_{i}}{100}=0$ and
$\frac{\sum\left|x_{i}-\bar{x}\right|}{100}=5 \Rightarrow \sum\left|x_{i}\right|=500$
$ \Rightarrow \sum {x_i^2} + 2\sum\limits_{1 \le i < j \le 100} {\left| {{x_i}{x_j}} \right|} = {(500)^2}$
$\Rightarrow \frac{\sum x_{i}^{2}}{100}=\frac{(500)^{2}-2 \sum\left|x_{i} x_{j}\right|}{100}=2500-1600$
$S. D.=\sqrt{\frac{\sum\left(x_{i}-\bar{x}\right)^{2}}{100}}=\sqrt{900}=30$
Similar Questions
If the variance of the frequency distribution is $160$ , then the value of $\mathrm{c} \in \mathrm{N}$ is
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