Question

Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, Find the correct standard deviation.

Answer 1

Mean = 40

SD= 10

n = 100

$\sum x_i=40\times 100=4000$

Corrected Sum = 4000- 30-70+30+27 = 3930

Corrected Mean = $\frac{3930}{100}=39.3$

Variance = 100

100 = $\frac{\sum x_i^2}{100}-(40{)}^2$

Incorrect $\sum x_i^2=170000$

Corrected $\sum x_i^2=$

Incorrect $\sum x_i^2-$ (Sum of squares of incorrect values) + (Sum of squares of corrected values )

Corrected $\sum x_i^2=170000-(900+4900)+(9+729)$

Corrected $\sum x_i^2=164938$

Corrected $\sigma =\sqrt{\frac{Corrected\sum x_i^2}{n}-(CorrectedMean{)}^2}$

Corrected $\sigma =\sqrt{\frac{164938}{100}-(39.3{)}^2}=10.24$