The mean and standard deviation of 100 observations are 40 and 5.1 , respectively, By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are $\mu$ and $\sigma$ respectively, then $10(\mu+\sigma)$ is equal to

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Actual means }=\mu=\frac{100(40)-50+40}{100} \ & \mu=40-\frac{1}{10}=39.9 \end{aligned}$ Incorrect variance $\begin{aligned} & (5.1)^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{100}-(\overline{\mathrm{x}})^2 \ & \sum \mathrm{x}_{\mathrm{i}}^2=100 	imes\left(40^2ight)+100(5.1)^2 \ & \sum \mathrm{x}_{\mathrm{i}}^2=16 	imes 10^4+(5.1)^2 	imes 100=162601 \ & \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2-50^2+40^2}{100}-(\mu)^2 \ & \sigma^2=1617.01-(39.9)^2=25 \ & \sigma=5 \ & 10(\mu+\sigma)=10(39.9+5) \ & =10 	imes 44.9=449\end{aligned}$ *

The mean and standard deviation of 100 observations are 40 and 5.1 , respectively, By mistake one…

Solution