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  • The mean and standard deviation of 20 observations 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)

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

 (i) If wrong item is omitted.

Answers (1)

best_answer

Given, 

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

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

\implies 10 =\frac{1}{20}\sum_{i=1}^{20}x_i  \implies \sum_{i=1}^{20}x_i = 200

Thus, incorrect sum = 200

Hence, correct sum of observations = 200 – 8 = 192

Therefore, Correct Mean = (Correct Sum)/19

=\frac{192}{19} 

= 10.1

Now, Standard Deviation,

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

\\ \implies 2^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 4 + (\overline x)^2 \\ \implies \frac{1}{20}\sum_{i=1}^nx_i ^2 = 4 + 100 = 104 \\ \implies \sum_{i=1}^nx_i ^2 = 2080 ,which is the incorrect sum.

Thus,  New sum = Old sum - (8x8)

= 2080 – 64

= 2016

Hence, Correct Standard Deviation = 

\sigma' =\sqrt{\frac{1}{n'}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{2016}{19} - (10.1)^2}

\dpi{100} = \sqrt{106.1 - 102.01} = \sqrt{4.09}

=2.02

Posted by

HARSH KANKARIA

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