Question

The mean square deviation of a set of $n$ observation $x_{1}, x_{2},... x_{n}$ about a point $c$ is defined as $\frac{1}{n} \displaystyle\sum_{i=1}^{n}\left(x_{i}-c\right)^{2}$.The mean square deviations about $- 2$ and $2$ are $18$ and $10$ respectively, the standard deviation of this set of observations is

• A. 3
• B. 2
• C. 1
• D. None of these

Answer 1

The Correct Option is A

We have $\frac{1}{n} \displaystyle\sum_{i=1}^{n}\left(x_{i}+2\right)^{2}=18$ and
$\frac{1}{n} \displaystyle\sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=10$
$\Rightarrow \displaystyle\sum_{i=1}^{n}\left(x_{i}+2\right)^{2}=18 n$ and
$\displaystyle\sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=10 n$
$\Rightarrow \displaystyle\sum_{i=1}^{n}\left(x_{i}+2\right)^{2}+\displaystyle\sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=28 n$
and $\displaystyle\sum_{i=1}^{n}\left(x_{i}+2\right)^{2}-\displaystyle\sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=8 n$
$\Rightarrow 2 \displaystyle\sum_{i=1}^{n}\left(x_{i}+4\right)^{2}=28 n 2 \sum_{i=1}^{n} 4 x_{i}=8 n$
$\Rightarrow \displaystyle\sum_{i=1}^{n} x_{i}^{2}+4 n=14 n \displaystyle\sum_{i=1}^{n} x_{i}=n$
$\Rightarrow \displaystyle\sum_{i=1}^{n} x_{i}^{2}=10 n \displaystyle\sum_{i=1}^{n} x_{i}=n$
$\therefore \sigma=\sqrt{\frac{1}{n} \displaystyle\sum_{i=1}^{n} x_{i}^{2}-\left(\frac{1}{n} \displaystyle\sum_{i=1}^{n} x_{i}\right)^{2}}$
$=\sqrt{\frac{10 n}{n}-\left(\frac{n}{n}\right)^{2}}=3$