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Question

Mathematics Question on Measures of Dispersion

The mean square deviation of a set of nn observation x1,x2,...xnx_{1}, x_{2},... x_{n} about a point cc is defined as 1ni=1n(xic)2\frac{1}{n} \displaystyle\sum_{i=1}^{n}\left(x_{i}-c\right)^{2}.The mean square deviations about 2- 2 and 22 are 1818 and 1010 respectively, the standard deviation of this set of observations is

A

3

B

2

C

1

D

None of these

Answer

3

Explanation

Solution

We have 1ni=1n(xi+2)2=18\frac{1}{n} \displaystyle\sum_{i=1}^{n}\left(x_{i}+2\right)^{2}=18 and
1ni=1n(xi2)2=10\frac{1}{n} \displaystyle\sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=10
i=1n(xi+2)2=18n\Rightarrow \displaystyle\sum_{i=1}^{n}\left(x_{i}+2\right)^{2}=18 n and
i=1n(xi2)2=10n\displaystyle\sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=10 n
i=1n(xi+2)2+i=1n(xi2)2=28n\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 i=1n(xi+2)2i=1n(xi2)2=8n\displaystyle\sum_{i=1}^{n}\left(x_{i}+2\right)^{2}-\displaystyle\sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=8 n
2i=1n(xi+4)2=28n2i=1n4xi=8n\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
i=1nxi2+4n=14ni=1nxi=n\Rightarrow \displaystyle\sum_{i=1}^{n} x_{i}^{2}+4 n=14 n \displaystyle\sum_{i=1}^{n} x_{i}=n
i=1nxi2=10ni=1nxi=n\Rightarrow \displaystyle\sum_{i=1}^{n} x_{i}^{2}=10 n \displaystyle\sum_{i=1}^{n} x_{i}=n
σ=1ni=1nxi2(1ni=1nxi)2\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}}
=10nn(nn)2=3=\sqrt{\frac{10 n}{n}-\left(\frac{n}{n}\right)^{2}}=3