Question

The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?

Answer 1

$$\begin{gathered} n=100 \\ \mathrm{Mean}=\overline{X}=40 \\ \sigma =\mathrm{SD}=5.1 \\ \frac{1}{n}\sum x_{\mathrm{i}}=\overline{X}\mathit{} \\ \Rightarrow \sum x_i=100\times 40=4000\left(\mathrm{This}\mathrm{is}\mathrm{an}\mathrm{incorrect}\mathrm{reading}\mathrm{due}\mathrm{to}\mathrm{misread}\mathrm{values}.\right) \\ \mathrm{Corrected}\mathrm{sum},\sum x_i=4000-50+40 \\ =3990 \\ \Rightarrow \mathrm{Corrected}\mathrm{mean}=\frac{\mathrm{Corrected}\mathrm{sum}}{100} \\ =\frac{3990}{100} \\ =39.9...\left(1\right) \end{gathered}$$

To find the corrected SD:

$$\begin{gathered} \sqrt{\mathrm{Variance}}=\mathrm{\sigma } \\ \Rightarrow {\sigma }^2={\left(5.1\right)}^2=\mathrm{Variance} \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{formula}, \\ \frac{1}{n}\sum _{}{x_i}^2-{\left(\overline{X}\right)}^2=\mathrm{Variance} \\ \Rightarrow \frac{1}{100}\sum _{}{x_{\mathrm{i}}}^2-{\left(40\right)}^2=26.01 \\ \Rightarrow \frac{1}{100}\sum _{}{x_{\mathrm{i}}}^2-1600=26.01 \\ \Rightarrow \frac{1}{100}\sum _{}{x_{\mathrm{i}}}^2=1626.01 \\ \Rightarrow \sum _{}{x_{\mathrm{i}}}^2=162601\left(\mathrm{But},\mathrm{this}\mathrm{is}\mathrm{incorrect}\mathrm{due}t\mathrm{o}\mathrm{misread}\mathrm{values}\right) \\ \Rightarrow \mathrm{Corrected}\sum _{}{x_i}^2=162601-{50}^2+{40}^2 \\ =161701....\left(2\right) \\ \mathrm{Corrected}\mathrm{variance}=\frac{1}{100}\mathrm{Corrected}\sum _{}{x_i}^2-{\left(\mathrm{Corrected}\mathrm{mean}\right)}^2 \\ =\frac{161701}{100}-{\left(39.9\right)}^2\left[\mathrm{using}\mathrm{equations}\left(1\right)\mathrm{and}\left(2\right)\right] \\ =1617.01-1592.01 \\ =25 \\ \mathrm{Corrected}\mathrm{SD}=\sqrt{\mathrm{Corrected}\mathrm{variance}} \\ =\sqrt{25} \\ =5 \end{gathered}$$

Corrected mean = 39.9
Corrected standard deviation = 5