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)

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