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, 22 and 18. Find the mean and standard deviation if the incoorect observations were omitted.

Answer 1

Given n = 100 , $\overline{x}=20,\sigma =3$

$\because \overline{x}=20$

$\Rightarrow \frac{\sum x}{100}=20$

$\Rightarrow \sum x_i=100\times 20$

$\Rightarrow \sum x=2000$

Now, incorrect observations 21, 21 and 18 are omitted, then correct sum is

$\sum x_i=2000-21-21-18=2000-60$

= 1940

Now, correct mean of remaining 97 observations is

$\overline{x}=\frac{1940}{97}=20$

Again, $\sigma =3\Rightarrow \sqrt{\frac{\sum x_i^2}{n}-(\overline{x}{)}^2}$

$\Rightarrow \frac{\sum x_i^2}{100}-(20{)}^2=9$

$\Rightarrow \frac{\sum x^2}{100}=9+400$

= $409\times 100=40900$

Now, correct $\sum x^2=40900-(21{)}^2-(21{)}^2-(18{)}^2=40900-441-441-324$

= 40900 - 1206=39694

Now, correct SD for remaining 97 observations is

$\sigma =\sqrt{\frac{39694}{97}-{\left(\frac{1940}{97}\right)}^2}=\sqrt{4092-(20{)}^2}=\sqrt{409.2-400}=\sqrt{9.2}=3.03$