# The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively.

Question:

The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Solution:

Number of observations (n) = 100

Incorrect mean $(\bar{x})=20$

Incorrect standard deviation $(\sigma)=3$

$\Rightarrow 20=\frac{1}{100} \sum_{i=1}^{100} x_{i}$

$\Rightarrow \sum_{i=1}^{100} x_{j}=20 \times 100=2000$

$\therefore$ Incorrect sum of observations $=2000$

$\Rightarrow$ Correct sum of observations $=2000-21-21-18=2000-60=1940$

$\therefore$ Correct mean $=\frac{\text { Correct sum }}{100-3}=\frac{1940}{97}=20$

Standard deviation $(\sigma)=\sqrt{\frac{1}{n} \sum_{i=1}^{n} x_{i}-\frac{1}{n^{2}}\left(\sum_{i=1}^{n} x_{i}\right)^{2}}=\sqrt{\frac{1}{n} \sum_{j=1}^{n} x_{i}^{2}-(\bar{x})^{2}}$

$\Rightarrow 3=\sqrt{\frac{1}{100} \times \operatorname{Incorrect} \sum x_{j}{ }^{2}-(20)^{2}}$

$\Rightarrow$ Incorrect $\sum x_{i}{ }^{2}=100(9+400)=40900$

Correct $\sum_{i=1}^{n} x_{i}^{2}=\operatorname{Incorrect} \sum_{i=1}^{n} x_{i}^{2}-(21)^{2}-(21)^{2}-(18)^{2}$

$=40900-441-441-324$

$=39694$

$\therefore$ Correct standard deviation $=\sqrt{\frac{\text { Correct } \sum x_{i}^{2}}{n}-(\text { Correct mean })^{2}}$

$=\sqrt{\frac{39694}{97}-(20)^{2}}$

$=\sqrt{409.216-400}$

$=\sqrt{9.216}$

$=3.036$