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# While calculating the mean and variance of $10$ readings, a student wrongly used the reading $52$ for the correct reading $25$, the obtained the mean and variance as $45$ and $16$ respectively . Find the correct mean and the variance.

$\begin{array}{1 1}(A)\; 42.3,43.81\\(B)\;67,45\\(C)\;16,10\\(D)\;6,45\end{array}$

Can you answer this question?

Toolbox:
• The formulae used to calculate mean and variance are $Mean (\bar{x}) = \large\frac{\sum x}{n}$
• Variance $( \sigma^2)=\bigg[\large\frac{\sum x^2}{n}- \bigg(\frac{\sum x}{n}\bigg)^2\bigg]$
Mean $\bar{x}=45$
Variance $= (\sigma^2)=16$
$n=10$
Mean $= \large\frac{\sum x}{n}$
=> $45 = \large\frac{\sum x}{10}$
$\sum x=450$
Now replacing the wrong reading 52 by 25
$\therefore \sum x= 450-52+25=423$
The correct mean $\bar{x}= \large\frac{ \sum x}{n} =\frac{423}{10}$
$\qquad = 42.3$
Mean = 42.3
Given variance= 16
$\therefore \large\frac{\sum x^2}{n} - \bigg( \large\frac{\sum x}{n} \bigg)^2 $$=16 \large\frac{\sum x^2}{10} - \bigg( \large\frac{450}{10} \bigg)^2$$=16$
$\large\frac{ \sum x^2}{10 }$$= 16 +45^2 \large\frac{\sum x^2}{10}$$=16 +2025$
$\sum x^2=2041 \times 10$
$\qquad= 20410$
now replacing the wrong reading 52 by 25
$\sum x^2 =20410 -52^2 +25^2$
$\qquad= 20410-2704+625$
$\qquad= 18331$
Correct variance $\sigma ^2 =\large\frac{\sum x^2}{n} - \bigg( \large\frac{\sum x}{n}\bigg)^2$
$\qquad= \large\frac{18331}{10} - \bigg( \large\frac{423}{10}\bigg)^2$
$\qquad= 1833.1 -(42.3^2)$
$\qquad= 1833.1 -1789.29$
$\qquad= 43.81$
Hence A is the correct answer.
answered Jul 7, 2014 by