Question

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

(1) \(\sqrt{3.86}\)

(2) \(1.8\)

(3) \(\sqrt{3.96}\)

(4) \(1.94\)

Answer 1

Correct option is (3) \(\sqrt{3.96}\) 

Mean \( (\overline{\mathrm{x}})=10\)

\(\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}=10\)

\(\Sigma \mathrm{x}_{\mathrm{i}}=10 \times 20=200\)

If 8 is replaced by 12, then ∑x_i = 200 - 8 + 12 + 204

∴ Correct mean\((\overline{\mathrm{x}})=\frac{\sum \mathrm{x}_{\mathrm{i}}}{20}\)

\(=\frac{204}{20}=10.2\)

∵ Standard deviation = 2  

∴ Variance = (S.D.)² = 2² = 4

\(\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^{2}}{20}-\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}\right)^{2}=4\)

\(\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^{2}}{20}-(10)^{2}=4\)

\(\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^{2}}{20}=104\)

\(\Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^{2}=2080\)

Now, replaced '8' observations by '12' 

Then, \(\Sigma x_{\mathrm{i}}^{2}=2080-8^{2}+12^{2}=2160\)

∴ Variance of removing observations 

\(\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^{2}}{20}-\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}\right)^{2}\)

\( \Rightarrow \frac{2160}{20}-(10.2)^{2}\)

\( \Rightarrow 108-104.04\)

\(\Rightarrow 3.96\)

Correct standard deviation

\(=\sqrt{3.96}\)