InterviewSolution
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InterviewSolution
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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 are recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted. |
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Answer» Incorrect mean is `20`. There were `100` observations. So, sum of all onservations `= 20**100 = 2000` When in correct observations omiited then sum of observations `= 2000-21-21-18 = 1940` `:.` Correct mean `= 1940/(100-3) = 1940/97 = 20` Now, we will find the correct standard deviation. We know, `sigma = sqrt(1/N sum (x_i)^2-(bar x)^2)` When incorrect observations were present, then standard deviation was `9`. `:. 3 = sqrt(1/100 sum (x_i)^2 - (20)^2)` `=>9 = 1/100 sum (x_i)^2 - 400` `=>40900 = sum (x_i)^2` This is the sum when observations were incorrect. `:.` Correct sum, when observations are omitted `= 40900 - 21^2-21^2-18^2 = 39694 ` `:.` Correct `sum (x_i)^2 = 39694` `:. sigma = sqrt(1/97(39694) - 400) = sqrt(409.16-400) = sqrt(9.16) = 3.03` So, the correct standard deviation is `3.03.` |
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