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

$\text{Mean}=40\Rightarrow \Rightarrow \frac{{\sum }_{i=1}^{100}{x}_i}{100}=40$

$\Rightarrow {\sum }_{i=1}^{100}{x}_i=4000$

$SD=5.1\Rightarrow \Rightarrow {\sigma }^2=(5.1{)}^2$

$\Rightarrow \frac{{\sum }_{i=1}^{100}{x}_1^2}{100}-(40{)}^2=26.01$

$\Rightarrow {\sum }_{i=1}^{100}{x}_i^2=162601$

Thus, incorrect $\left({\sum }_{i=1}^{100}{x}_i\right)=4000$ and incorrect $\left({\sum }_{i=1}^{100}{x}_i^2\right)=162601$

Now, incorrect $\left({\sum }_{i=1}^{100}{x}_i\right)=4000$

$\Rightarrow \text{correct}\left({\sum }_{i=1}^{100}{x}_i^2\right)=\{162601-(50{)}^2+\left(40{)}^2\right\}=161701$

$\Rightarrow$ correct variance $=\frac{\text{correct}\left({\sum }_{i=1}^{100}{x}_i^2\right)}{100}-\text{(correct mean)}{}^2$

$=\{\frac{161701}{100}-(39.9{)}^2\}=\{1617.01-(40-0.1{)}^2\}$

= (1617.01) - {1600 + 0.01 - 8}

= (1617.01 - 1592.01) = 25

$\Rightarrow$ correct SD $=\sqrt{25}=5$

Hence, correct mean = 39.9 and correct SD = 5.