Question

The mean and standard deviation of 20 observation are found to be 10 and 2 respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases :

(i) If wrong item is omitted

(ii) If it is replaced by 12.

Answer 1

Here $n=20,\overline{x}=10and\sigma =2$

$\therefore \overline{x}=\frac{1}{n}\sum x_i\Rightarrow \sum x_i=n\times \overline{x}=20\times 10=200$

$\therefore Incorrect\sum x_i=200$

Now $\frac{1}{n}\sum x_i^2-\left({\overline{x}}^2\right)=4$

$\Rightarrow \frac{1}{20}\sum x_i^2-(10{)}^2=4\Rightarrow \sum x_i^2=2080$

(i) If wrong item is omitted.

When wrong item 8 is omitted from the data then we have 19 observations.

$\therefore$ Correct $\sum x_i=Incorrect\sum x_i-8$

Correct $\sum x_i=200-8=192$

$\therefore$ Correct mean = $\frac{192}{19}=10.1$

Also correct $\sum x_i^2$ = Incorrect $\sum x_i^2-(8{)}^2$

$\Rightarrow \sum x_i^2$ = 2080 - 64 = 2016

$\therefore$ Correct variance

= $\frac{1}{19}\times 2016-{\left(\frac{192}{19}\right)}^2=\frac{2016}{19}-\frac{36864}{361}$

= $\frac{38304-36864}{361}=\frac{1440}{361}$

Correct S.D. = $\sqrt{\frac{1440}{361}}=\sqrt{3.99}=1.997$

(ii) If it is replaced by 12

When wrong item 8 is replaced by 12

$\therefore$ Correct $\sum x_i$ = Incorrect $\sum x_i-8+12$

= 200 - 8 + 12 = 204

$\therefore$ Correct mean = $\frac{204}{20}=10.2$

Also correct $\sum x_i^2$ = Incorrect $\sum x_i^2-(8{)}^2+(12{)}^2$

= 2080 - 64 + 144 = 2160

$\therefore$ Correct variance

= $\frac{1}{20}\left(correct\right(\sum x_i^2)-(correctmean{)}^2$

= $\frac{2160}{20}-{\left(\frac{204}{20}\right)}^2=\frac{2160}{20}-\frac{41616}{400}$

= $\frac{43200-41616}{400}=\frac{1584}{400}$

Correct S.D. = $\sqrt{\frac{1584}{400}}=\sqrt{3.96}=1.989$