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. Find the mean and standard deviation if the incorrect observations are omitted.

Asked by Abhisek | 11 months ago |  145

1 Answer

Solution :-

Number of observations (n) = 100

Incorrect mean (x̅)= 20

Incorrect standard deviation (σ) = 3

Incorrect sum of observations = 2000

Correct sum of observations = 2000-21-21-18 = 2000 - 60 = 1940

correct mean = \( \dfrac{Correct\;sum}{100-3}\)

\( \dfrac{1940}{97}\) = 20

\( (σ)= \sqrt{\dfrac{1}{n} \displaystyle\sum_{i=1}^{n} X_1-\dfrac{1}{n^2} (\displaystyle\sum_{i=1}^{n}X)^2}\)

 \( 3= \sqrt{\dfrac{1}{n} \displaystyle\sum_{i=1}^{n} X_1^2-\overline(X)^2}\)

\(3= \sqrt{\dfrac{1}{100}\times Incorrect \displaystyle\sum X_1^2-(20)^2}\)

\( Incorrect \displaystyle\sum X_1^2=100(9+400)\)

\( Incorrect \displaystyle\sum X_1^2=40900\)

\(correct \displaystyle\sum_{i=1}^{n}X_1^2 \) = 

\( Incorrect \displaystyle\sum X_1^2-(21)^2-(21)^2-(18)^2\)

= 40900-441-441-324

= 40900 - 1206

= 39694

Hence, correct standard deviation

 \(3= \sqrt{\dfrac{correct \displaystyle\sum X_1^2}{n}-(correct\;mean)^2}\)

 \( =3= \sqrt{\dfrac{39694}{97}-(20)^2}\)

\( \sqrt{409.216-400}\)

= 3.036

Answered by Pragya Singh | 11 months ago

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