Question

If $\sum _{i=1}^9\left(x_i-5\right)=9\mathrm{and}\sum _{i=1}^9{\left(x_i-5\right)}^2=45,$then the standard deviation of the 9 times $x_1,x_2,\dots \dots x_9$ is

• A. $9$
• B. $4$
• C. $3$
• D. $2$
• E. $1$

Answer 1

The correct option is D

$2$

Explanation for the correct option:

Step 1:Calculate variance.

$\begin{array}{rcl}\therefore Var& =& \left(\frac{1}{n}\right)\underset{i=1}{\overset{n}{\sum (}}{x_i-5)}^2–{\left[\left(\frac{1}{n}\right)\sum _{i=1}^n(x_i-5)\right]}^2\\ & =& \left(\frac{1}{9}\right)\times 45-{\left[\left(\frac{1}{9}\right)\times 9\right]}^2\\ & =& 5-1\\ & =& 4\end{array}$

Here, $\mathrm{Var}$ is 4

Step 2: Calculate the standard deviation.

Standard Deviation $=\sqrt{\mathrm{Variance}}$

Therefore, the Standard Deviation$\begin{array}{rcl}& =& \sqrt{4}\end{array}$

$=2$

Hence, (D) is the correct option.