2 Marks Questions

Question 12 Marks

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 cases of it is replaced by 12 .

Answer

Here we are given that, n = 20, $\bar x = 10$ and $\sigma = 2$
$\therefore \;\bar x = \frac{1}{n}\Sigma {x_i} \Rightarrow n \times \bar x = \Sigma {x_i }\\ \Rightarrow\Sigma {x_i}=20\times10=200$
Therefore Incorrect $\Sigma {x_i} = 200$
Now $\frac{1}{n}\Sigma x_i^2 - {(\bar x)^2} =\sigma^2$
$\Rightarrow \frac{1}{{20}}\Sigma x_i^2 - {(10)^2} = 4 \Rightarrow \Sigma x_i^2 = 2080$
If it is replaced by 12,
When wrong item 8 is replaced by 12
Therefore, Correct $\Sigma {x_i}$ = Incorrect $\Sigma {x_i}$ - 8 + 12
= 200 - 8 + 12 = 204
$\therefore$ Correct mean = $\frac{{204}}{{20}}$ = 10.2
Also correct $\Sigma x_i^2$ = Incorrect $\Sigma x_i^2$ - (8)² + (12)²
= 2080 - 64 + 144 = 2160
$\therefore$ Correct variance $= \frac{1}{{20}}(correct\;\Sigma x_1^2)$ - (correct mean)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

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Question 22 Marks

The mean and standard deviation of 20 observation is found to be 10 and 2 respectively. On rechecking, it was found that observation 8 was incorrect. Calculate the correct mean and standard deviation in cases of the wrong items is omitted.

Answer

Here n = 20, $\bar x = 10$ and $\sigma = 2$
$\therefore \;\bar x = \frac{1}{n}\Sigma {x_i} \Rightarrow n \times \bar x = \Sigma {x_i }\\ \Rightarrow\Sigma {x_i}=20\times10=200$
Therefore Incorrect $\Sigma {x_i} = 200$
Now $\frac{1}{n}\Sigma x_i^2 - {(\bar x)^2} =\sigma^2$
$\Rightarrow \frac{1}{{20}}\Sigma x_i^2 - {(10)^2} = 4 \Rightarrow \Sigma x_i^2 = 2080$

If wrong item is omitted.
When wrong item 8 is omitted from the data then we have 19 observations.
Therefore Correct $\Sigma {x_i}$ = Incorrect $\Sigma {x_i} - 8$
Correct $\Sigma {x_i} = 200 - 8 = 192$
Therefore Correct mean = $\frac{{192}}{{19}}$ = 10.1
Also correct $ \Sigma x_i^2$ = Incorrect $ \Sigma x_i^2 - {(8)^2}$
$\Rightarrow$ Correct $\Sigma x_i^2 = 2080 - 64$ = 2016
Hence Correct variance $ = \frac{1}{{19}}\left( {correct\;\Sigma x_i^2} \right)$ - (correct mean)²
$= \frac{1}{{19}} \times 2016 - {\left( {\frac{{192}}{{19}}} \right)^2}$
$= \frac{{2016}}{{19}} - \frac{{36864}}{{361}} = \frac{{38304 - 36884}}{{361}} = \frac{{1440}}{{361}}$
Correct S.D. $ = \sqrt {\frac{{1440}}{{361}}} = \sqrt {3.99}$ = 1.997

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