The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50…

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The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is

MARKS Content Team · Accepted Answer

Correct answer: (B) 13. Incorrect mean is (∑i=110x_i)/10=20 ⇒∑i=110x_i=200 ....i So, corrected mean is (∑i=1)/(10x_i-50+4010)=(200-10)/10=19 Incorrect variance is (∑i=1)/10xi210-400=64 ⇒∑i=110xi2=4640 Now, corrected variance is =(∑i=1)/(9xi2+40^2-50^210)-19^2 =(4640-900)/10-19^2=374-361=13

The mean and standard deviation of $10$ observations are $20$ and $8$ respectively. Later on, it was observed that one observation was recorded as $50$ instead of $40$ . Then the correct variance is

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