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 :

Solution

Maths sir · 3 steps

Recover the sum and the sum of squares, fix both, then rebuild the variance.

1. Recover the totals. From the mean, Σ x = 10 × 20 = 200. From the variance 8² = 64, we get (Σ x²)/10 - 20² = 64, so Σ x² = 10 × 464 = 4640.
2. Correct both totals. Take out 50 and put in 40. The corrected sum is 200 - 50 + 40 = 190, so the correct mean is 19. The corrected sum of squares is 4640 - 2500 + 1600 = 3740.

   σ² = (Σ x²)/n - x̄² = 3740/10 - 19² = 374 - 361 = 13

3. So. The correct variance is 13, option (c).

The idea to keep. When a wrong observation is replaced, the mean and SD cannot be fixed directly. Go back to Σ x and Σ x², swap the value in both, then use the new mean.

Answer from the NTA final answer key. Solution by PrepWiser.