The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and a and b are respectively mean and variance of remaining 6 observation, then a+3 b-5 is equal to ___________.

Solution

Maths sir · 4 steps

Turn the mean and variance back into a total and a sum of squares, take out 14, then rebuild.

1. Recover the totals. With 7 observations and mean 8, the sum is Σ x = 7 × 8 = 56. Use the variance formula to recover the sum of squares.

   σ² = (Σ x²)/n - x̄² ⇒ 16 = (Σ x²)/7 - 64 ⇒ Σ x² = 560

2. Remove 14. For the remaining 6 observations, Σ x = 56 - 14 = 42 and Σ x² = 560 - 196 = 364. So the new mean is a = 42/6 = 7.
3. New variance. b = 364/6 - 7² = 182/3 - 49 = 35/3. So 3b = 35.
4. So. a + 3b - 5 = 7 + 35 - 5 = 37.

The common slip: -8. Subtracting the old x̄² = 64 instead of the new 7² = 49 makes b = -10/3, a negative variance, and the final answer becomes -8.

The idea to keep. Mean and variance cannot be edited directly when data is added or removed. Convert them to Σ x and Σ x², adjust those, then recompute both.

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