Question

A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure being 40. Find the correct mean and S.D.

Answer 1

$$\begin{gathered} {\overline{x}}_o=40,{\mathrm{\sigma }}_{\mathrm{o}}=5.1,\mathrm{and}\mathrm{n}=100 \\ \Rightarrow {\mathrm{\Sigma x}}_{\mathrm{o}}=4000 \\ \mathrm{Corrected}\mathrm{sum}\mathrm{of}\mathrm{observations},{\mathrm{\Sigma x}}_{\mathrm{n}}=4000-50+40=3990 \\ \mathrm{n}=100 \\ \Rightarrow {\overline{\mathrm{x}}}_{\mathrm{n}}=\frac{{\mathrm{\Sigma x}}_{\mathrm{n}}}{\mathrm{n}}=39.90 \\ \mathrm{Also},\mathrm{SD},{\mathrm{\sigma }}_{\mathrm{o}}=5.1 \\ \Rightarrow \mathrm{\Sigma }{\left({\mathrm{x}}_{\mathrm{i}}-{\overline{\mathrm{x}}}_{\mathrm{o}}\right)}^2=2601 \\ \Rightarrow \mathrm{\Sigma }{\left({\mathrm{x}}_{\mathrm{i}}-{\overline{\mathrm{x}}}_{\mathrm{n}}\right)}^2=2601-{\left(50-40\right)}^2+{\left(40-39.90\right)}^2=2601-100+0.01=2501.01 \\ \Rightarrow {\sigma }_n=\sqrt{\frac{\mathrm{\Sigma }{\left({\mathrm{x}}_{\mathrm{i}}-{\overline{\mathrm{x}}}_{\mathrm{n}}\right)}^2}{n}}=\sqrt{\frac{2501.01}{100}}=5 \end{gathered}$$