Let the mean and the standard deviation of the observation 2,3,3,4,5,7, a, b be 4 and √2 respectively. Then the mean deviation about the mode of these observations is :

Solution

Maths sir · 4 steps

Measure the squared deviations from 4 and you will see that a and b have no choice: both must be 4.

1. Use the mean. There are 8 observations with mean 4, so their sum is 32. The six known values add to 24, so a + b = 8.
2. Use the variance. The variance is (√2)² = 2, so the squared deviations from the mean add to 8 × 2 = 16. For 2, 3, 3, 4, 5, 7 they are 4 + 1 + 1 + 0 + 1 + 9 = 16, which already uses up all of it.

   16 + (a-4)² + (b-4)² = 16 ⇒ (a-4)² + (b-4)² = 0

3. Find the mode. A sum of squares is zero only when each square is zero, so a = b = 4. The data are 2, 3, 3, 4, 4, 4, 5, 7, and the mode is 4.
4. So. The absolute deviations from 4 are 2, 1, 1, 0, 0, 0, 1, 3, which add to 8. The mean deviation about the mode is 8/8 = 1, option (c).

The idea to keep. Write the variance as the sum of (xᵢ - x̄)² divided by n. When the known values already account for the whole sum, every unknown must sit exactly on the mean.

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