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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: (ii)

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:

 (ii) If it is replaced by \small 12.

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 + 12 = 204

Therefore, Correct Mean = (Correct Sum)/20

\dpi{100} =\frac{204}{20} 

= 10.2

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) + (12x12)

= 2080 – 64 + 144

= 2160

Hence, Correct Standard Deviation = 

\sigma' =\sqrt{\frac{1}{n}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{2160}{20} - (10.2)^2}

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

= 1.98

Posted by

HARSH KANKARIA

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