Question

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

• A. 9
• B. 4
• C. 3
• D. 2
• E. 1

Answer 1

Answer: (D)

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