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Statistics Part 2

Section: Mathematics

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Question : 28 of 50

Marks: +1, -0

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On respectively, it was found that an observation by mistake was taken 8 instead of 12 . The correct standard deviation is

[6 Apr 2024 Shift 1]

$\sqrt{3.86}$
1.8
$\sqrt{3.96}$
1.94

Solution: 👈: Video Solution

Mean

(\overline{x})=10

\;\Rightarrow\;\frac{\Sigma x_i}{20}=10

\;\Sigma x_i =10 \times 20=200

If 8 is replaced by 12 , then

\Sigma x_1=200-8+12=204

\;\therefore\;\text{ Correct mean }(\overline{x})=\;\frac{\Sigma x_i}{20}

\;=\;\frac{204}{20}=10.2

\;\because\;\text{ Standard deviation }\;=2

\;\therefore\;\text{ Variance }\;=(\;\text{ S.D. }\;)^2=2^2=4

\;\Rightarrow\;\frac{\Sigma x_i^2}{20}- \left(\frac{\Sigma x_i}{20}\right)^2=4

\;\Rightarrow\;\frac{\Sigma x_i^2}{20}-(10)^2=4

\;\Rightarrow\;\frac{\Sigma x_i^2}{20}=104

\;\Rightarrow\;\Sigma x_i^2=2080

Now, replaced ' 8 ' observations by ' 12 '

Then,

\Sigma x_i^2=2080-8^2+12^2=2160

\therefore

Variance of removing observations

\;\Rightarrow\;\frac{\Sigma x_i^2}{20}- \left(\frac{\Sigma x_i}{20}\right)^2

\;\Rightarrow\;\frac{2160}{20}-(10.2)^2

\;\Rightarrow\;108-104.04

\;\Rightarrow\;3.96

Correct standard deviation

=\sqrt{3.96}

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