The mean and standard deviation of 100 observations are 40 and 5.1...

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

Options

A445
B451
C447
D449

Correct answer

D. 449

Step-by-step solution

\begin{aligned}& \text{ 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 \times\left(40^2\right)+100(5.1)^2 \\ & \sum \mathrm{x}_{\mathrm{i}^2=16 \times 10^4+(5.1)^2 \times 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 \times 44.9=449\end{aligned}

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