The mean and standard deviation of 10 observations are 20 and 8 respectively.

Question

The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is :

Accepted Answer

1. Calculate the sum of the original observations:

$\frac{{x}_{1}+{x}_{2}+\ldots +{x}_{9}+50}{10}=20$

${x}_{1}+{x}_{2}+\ldots +{x}_{9}=150$

2. Calculate the sum of the squares of the original observations using the original variance:

${\sigma }^{2}={8}^{2}=64$

$64=\frac{{x}_{2}^{1}+{x}_{2}^{2}+\ldots +{x}_{2}^{9}+2500}{10}-400$

${x}_{2}^{1}+{x}_{2}^{2}+\ldots +{x}_{2}^{9}=2140$

3. Calculate the new mean after correcting the error:

$=\frac{150+40}{10}=19$

4. Calculate the new variance using the corrected sum of observations and the corrected sum of squares of observations:

${\sigma }^{2}=\frac{2140+1600}{10}-(19{)}^{2}$

$\Rightarrow$ ${\sigma }^{2}=13$

The correct variance after correcting the error is 13 (Option C).

Solutions · 1