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  • 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, 21 and 18.

7       The mean and standard deviation of a group of  \small 100 observations were found to be \small 20 and \small 3, respectively. Later on it was found that three observations were incorrect, which were recorded as \small 21,21  and  \small 18. Find the mean and standard deviation if the incorrect observations are omitted.

Answers (1)

best_answer

Given, 

Initial Number of observations, n = 100

\overline x =\frac{1}{n}\sum_{i=1}^nx_i

\dpi{100} \implies 20 =\frac{1}{100}\sum_{i=1}^{100}x_i  \implies \sum_{i=1}^{100}x_i = 2000

Thus, incorrect sum = 2000

Hence, New sum of observations = 2000 - 21-21-18 = 1940

New number of observation, n' = 100-3 =97

Therefore, New Mean = New Sum)/100

=\frac{1940}{97} 

= 20

Now, Standard Deviation,

\sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}

\\ \implies 3^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 9 + (\overline x)^2 \\ \implies \frac{1}{100}\sum_{i=1}^nx_i ^2 = 9 + 400 = 409 \\ \implies \sum_{i=1}^nx_i ^2 = 40900 ,which is the incorrect sum.

Thus,  New sum = Old (Incorrect) sum - (21x21) - (21x21) - (18x18)

= 40900 - 441 - 441 - 324

= 39694

Hence, Correct Standard Deviation = 

\sigma' =\sqrt{\frac{1}{n'}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{39694}{97} - (20)^2}

= \sqrt{108 - 104.04} = \sqrt{3.96}

= 3.036

Posted by

HARSH KANKARIA

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