Question

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

• A. 9
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (D)

2

Answer 2

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