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}$
$=\sqrt{104.48}=10.24$