Question

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

• A. 9
• B. 4
• C. 3
• D. 2
• E. 1

Answer 1

Answer: (D)

2

Given, $\sum _{i=1}^n=(x_i-5)=9$
and $\sum _{i=1}^9(x_i-5{)}^2=45$
$\therefore$ Standard deviation
$=\sqrt{\frac{{\left[\sum _{i=1}^9(x_i-5)\right]}^2-\sum _{i=1}^9(x_i-5{)}^2}{9}}$
$=\sqrt{\frac{(9{)}^2-45}{9}}=\sqrt{\frac{81-45}{9}}$
$=\sqrt{\frac{36}{9}}=\sqrt{4}=2$