The mean and standard deviation of 20 observations are found to be...

The mean and standard deviation of 20 observations are found to be 10 and 2 . respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

Options

A1.8
B1.94
C $\sqrt{3.96}$
D $\sqrt{3.86}$

Correct answer

C. $\sqrt{3.96}$

Step-by-step solution

\begin{aligned} & \text{ Mean }(\overline{\mathrm{x}})=10 \\ & \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}=10 \\ & \Sigma \mathrm{x}_{\mathrm{i}}=10 \times 20=200 \end{aligned}

If 8 is replaced by 12, then

\Sigma x_i=200-8+12=204 \therefore

Correct mean

(\overline{\mathrm{x}})=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20} =\frac{204}{20}=10.2 \because

Standard deviation

=2 \therefore

Variance

=(\text{ S.D. })^2=2^2=4 \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}^2}{20}-\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}\right)^2=4 \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}^2}{20}-(10)^2=4 \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}^2}{20}=104 \Rightarrow \Sigma x_i^2=2080

Now, replaced ' 8 ' observations by ' 12 ' Then,

\Sigma \mathrm{x}_{\mathrm{i}^2=2080-8^2+12^2=2160 \therefore

Variance of removing observations

\begin{aligned} & \Rightarrow \frac{\Sigma x_i^2}{20}-\left(\frac{\Sigma x_i}{20}\right)^2 \\ & \Rightarrow \frac{2160}{20}-(10.2)^2 \\ & \Rightarrow 108-104.04 \\ & \Rightarrow 3.96 \end{aligned}

Correct standard deviation

=\sqrt{3.96}

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