Let the mean and variance of 12 observations be 9/2 and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is m/n, where m and n are coprime, then m+n is equal to :

Solution

Maths sir · 3 steps

Variance cannot be patched directly, so rebuild Σ x and Σ x², fix them, then recompute.

1. Recover the totals. With 12 observations and mean 9/2, the sum is Σ x = 12 × 9/2 = 54. From σ² = (Σ x²)/12 - x̄² we get (Σ x²)/12 = 4 + 81/4 = 97/4, so Σ x² = 291.
2. Correct the totals. Remove 9 and 10, then add 7 and 14. The new sum is 54 - 19 + 21 = 56. The new sum of squares is 291 - (81 + 100) + (49 + 196) = 355.

   σ² = 355/12 - (56/12)² = 355/12 - 196/9 = (1065 - 784)/36 = 281/36

3. So. 281 is prime, so 281/36 is already in lowest terms. Then m = 281, n = 36 and m + n = 317, which is option (a).

The idea to keep. For a corrected-observation problem, always go back to Σ x and Σ x². Fix both totals, and remember the mean changes too, so recompute it before subtracting x̄².

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