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, then the correct standard deviation is

• A. $29.1$
• B. $10.24$
• C. $30.1$
• D. $27.1$

Answer 1

Answer: (B)

We have, $n=100,\bar{x}=40,\sigma =10$
Mean $(\bar{x})=\frac{\Sigma x_i}{n}=40$
$\Rightarrow \frac{\Sigma x_i}{100}=40\Rightarrow \Sigma x_i=4000$
Corrected $\Sigma x_i=4000-30-70+3+27$
$=4030-100=3930$
Corrected mean $=\frac{3930}{100}=39.3$
Variance $\left({\sigma }^2\right)=\frac{\Sigma x_i^2}{n}-(40{)}^2$
$\Rightarrow 100=\frac{\Sigma x_i^2}{100}-1600\Rightarrow \Sigma x_i^2=170000$
Corrected $\Sigma x_i^2=170000-(30{)}^2-(70{)}^2+(3{)}^2+(27{)}^2$
$=170000-900-4900+9+729$
$=164938$
Corrected standard deviation ( $\sigma )$
$=\sqrt{\frac{164938}{100}-(39.3{)}^2}=\sqrt{1649.38-1544.49}$
$=\sqrt{104.48}=10.24$