Question

The mean and standard deviation of $10$ observations $x_1,x_2,x_3.....x_{10}$ are $\bar{x}$ and $\sigma$ respectively. Let $10$ is added to $x_1,x_2.....x_9$ and $90$ is subtracted from $x_{10}.$ If still, the standard deviation is the same, then $x_{10}-\bar{x}$ is equal to

• A. $35$
• B. $45$
• C. $55$
• D. $50$

Answer 1

Answer: (B)

$\sum _{i=1}^{10}\frac{{\left(x_i-\bar{x}\right)}^2}{10}={\left(\sigma \right)}^2....\left(1\right)$
$\frac{\sum _{i=1}^9{\left(x_i+10-\bar{x}\right)}^2+{\left(x_{10}-90-\bar{x}\right)}^2}{10}={\left(\sigma \right)}^2....\left(2\right)$
From $\left(1\right)$ and $(2)$
$\frac{\sum _{i=1}^{10}{\left(x_i-\bar{x}\right)}^2}{10}=\frac{\sum _{i=1}^{10}{\left(x_i-\bar{x}\right)}^2+20\sum _{i=1}^9\left(x_i-\bar{x}\right)-180\left(x_{10}-\bar{x}\right)+100\times 9+8100}{10}$
$2\times \sum _{i=1}^9\left(x_i-\bar{x}\right)+2\left(x_{10}-\bar{x}\right)-20\left(x_{10}-\bar{x}\right)+90+810=0$
$\Rightarrow 900=20\left(x_{10}-\bar{x}\right)\Rightarrow x_{10}-\bar{x}=45$