Question

If \(\displaystyle\sum_{i=1}^{n}(x_i-a)=n\) and \(\displaystyle\sum_{i=1}^{n}(x_i-a)^2=na\) \((n, a>1)\) then the standard deviation of n observations \(x_1, x_2, \ldots, x_n\) is :

• A. \(a-1\)
• B. \(n\sqrt{a-1}\)
• C. \(\sqrt{n(a-1)}\)
• D. \(\sqrt{a-1}\)

Hint:

Standard deviation does not change if every observation is shifted by \(a\); use \(y_i=x_i-a\).

Answer 1

The Correct Option is D

Let \(y_i=x_i-a\). Then \(\sum y_i=n\) and \(\sum y_i^2=na\).
Standard deviation is unchanged by shifting each observation by a constant, so SD of \(x_i\) = SD of \(y_i\).
Variance = \(\dfrac{\sum y_i^2}{n}-\left(\dfrac{\sum y_i}{n}\right)^2=\dfrac{na}{n}-\left(\dfrac nn\right)^2=a-1\).
Standard deviation = \(\sqrt{a-1}\).
Answer: (4) \(\sqrt{a-1}\)