Question

The mean and standard deviation of $20$ observations were calculated as $10$ and $2.5$ respectively. It was found that by mistake one data value was taken as $25$ instead of $35 .$ If $\alpha$ and $\sqrt{\beta}$ are the mean and standard deviation respectively for correct data, then $(\alpha, \beta)$ is :

• A. $(11,26)$
• B. $(10.5,25)$
• C. $(11,25)$
• D. $(10.5,26)$

Answer 1

Answer: (D)

Given:
Mean $(\bar{x})=\frac{\sum x_{i}}{20}=10$
or $\Sigma x _{ i }=200$ (incorrect)
or $200-25+35=210=\Sigma X _{ i }$ (Correct)
Now correct $\overline{ x }=\frac{210}{20}=10.5$
again given $S.D =2.5(\sigma)$
$\sigma^{2}=\frac{\Sigma x_{i}^{2}}{20}-(10)^{2}=(2.5)^{2}$
or $\Sigma x _{ i }^{2}=2125$ (incorrect)
or $\Sigma x _{ i }^{2}=2125-25^{2}+35^{2}$
$=2725$ (Correct)
$\therefore $ correct $\sigma^{2}=\frac{2725}{20}-(10.5)^{2}$
$\underline{\sigma}^{2}=26$
or $\sigma=26$
$\therefore \underline{\alpha}=10.5, \beta=26$