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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Statistics

BITSAT 2015 · Mathematics · Statistics

The mean square deviation of a set of n observation x_ 1 , x_ 2^ 2...

The mean square deviation of a set of n observation x1,x22,…xn\mathrm{x}_{1}, \mathrm{x}_{2^{2}}, \ldots \mathrm{x}_{\mathrm{n}} about a point c\mathrm{c} is defined as $\frac{1}{n} \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

Options

  1. A3
  2. B2
  3. C1
  4. DNone of these

Correct answer

A. 3

Step-by-step solution

We have 1n∑i=1n\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}+2\right)^{2}=18andand1n∑i=1n(xi−2)2=10⇒∑i=1n(xi+2)2=18n and \begin{array}{l}\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=10 \\ \Rightarrow \sum_{i=1}^{n}\left(x_{i}+2\right)^{2}=18 n \text{ and } \end{array}$\begin{array}{l} \sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=10 n \\ \Rightarrow \sum_{i=1}^{n}\left(x_{i}+2\right)^{2}+\sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=28 n \end{array} and ∑i=1n(xi+2)2−∑i=1n(xi−2)2=8n\sum_{i=1}^{n}\left(x_{i}+2\right)^{2}-\sum_{i=1}^{n}\left(x_{i}-2\right)^{2}=8 n⇒2∑i=1n(xi+4)2=28n and 2∑i=1n4xi=8n⇒∑i=1nx12+4n=14n and ∑i=1nxi=n⇒∑i=1nxi2=10n and ∑i=1nxi=n∴σ=1n∑i=1nxi2−(1n∑i=1′nxi)2=10nn−(nn)2=3\begin{array}{l} \Rightarrow2 \sum_{i=1}^{n}\left(x_{i}+4\right)^{2}=28 n \text{ and }2 \sum_{i=1}^{n} 4 x_{i}=8 n \\ \Rightarrow \sum_{i=1}^{n} x_{1}^{2}+4 n=14 n \text{ and } \sum_{i=1}^{n}x_{i}=n \\ \Rightarrow \sum_{i=1}^{n} x_i^{2}=10 n \text{ and } \sum_{i=1}^{n}x_{i}=n \\ \therefore \sigma=\sqrt{\frac{1}{n} \sum_{i=1}^{n} x_i^{2}-\left(\frac{1}{n} \sum_{i=1^{\prime}}^{n} x_{i}\right)^{2}}=\sqrt{\frac{10 n}{n}-\left(\frac{n}{n}\right)^{2}}=3 \end{array}$
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