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The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is:
(1) 3.99 (2) 4.01 (3) 4.02 (4) 3.98
Actual $\sum x_i^2 = 2120$. Variance $= \frac{2120}{20} - (10.1)^2 = 3.99$.
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