Question

The mean and S.D. of 20 observations are 10 and 2 respectively. Later on, it was noticed that item 8 was incorrect Calculate A.M. and S.D. if (i) the incorrect item is omitted (ii) the incorrect item is replaced by 12.

Answer 1

As we know that

$Mean= 10 = > \frac{\rm^{20}\Sigma_ {i=1}x_{i}}{20} = 10 \\ \\ = > \rm^{20}\Sigma_ {i=1}x_{i} = 200 \\ \\ SD = 2 = > {\sigma}^{2} = 4 \\ \\ = > \frac{\rm^{20}\Sigma_ {i=1}x_{i}^{2}}{20} - ( {10)}^{2} = 4 \\ \\ = > \rm^{20}\Sigma_ {i=1}x_{i}^{2} = 2080$

Thus incorrect
$\rm^{20}\Sigma_ {i=1}x_{i}^{2} = 200 \\ \\ and \: incorrect\:\:\rm^{20}\Sigma_ {i=1}x_{i}^{2} = 2080$

(i) the incorrect item is omitted:

On omitting 8,we are left with 19 observations

since correct
$\rm^{19}\Sigma_ {i=1}x_{i} = incorrect (\rm^{20}\Sigma_ {i=1}x_{i}) - 8 \\ \\ = 200 - 8 = 192 \\ \\ thus \: correct \: \rm^{19}\Sigma_ {i=1}x_{i}= 192$
and correct mean
$= \frac{192}{19} = 10.105$
Also correct
$\rm^{19}\Sigma_ {i=1}x_{i}^{2} = incorrect \: (\rm^{20}\Sigma_ {i=1}x_{i}^{2} ) - 64 \\ \\ = 2080 - 64 = 2016 \\ \\ correct \: variance \: = \frac{1}{19}( correct \: \rm^{19}\Sigma_ {i=1}x_{i}^{2}) - {(correct \: mean)}^{2} \\ \\ = ( \frac{2016}{19} ) - ( { \frac{192}{19} })^{2} = \frac{1440}{361} \\ \\ correct \: SD = \sqrt{ \frac{1440}{361} } = 1.997$
correct mean = 10.105

correct SD = 1.997

(ii) the incorrect item is replaced by 12

Incorrect
$\rm^{20}\Sigma_ {i=1}x_{i}= 200 \\ \\ correct \: \rm^{20}\Sigma_ {i=1}x_{i}= 200 - 8 + 12 = 204 \\ \\ correct \: mean = \frac{204}{20} = 10.2 \\ \\ also \: incorrect \: \rm^{20}\Sigma_ {i=1}x_{i}^{2} = 2080 \\ \\ correct \: \rm^{20}\Sigma_ {i=1}x_{i}^{2} = 2080 - {8}^{2} + {12}^{2} = 2160 \\ \\ correct \: variance = \frac{2160}{20} - ( {10.2)}^{2} \\ \\ = 3.96 \\ \\ = > correct \: SD = \sqrt{3.96} \\ \\ = 1.989$
Thus for this case mean 10.2 and SD =1.989

Hope it helps you.