Let α, β ∈ R. Let the mean and the variance of 6 observations -3,4,7,-6, α, β be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :

Solution

Maths sir · 5 steps

Use the mean for α + β and the variance for α² + β², then pin down both numbers.

1. Use the mean. Mean 2 means the six values add to 12. The four known values add to -3 + 4 + 7 - 6 = 2, so α + β = 10.
2. Use the variance. With variance 23 and mean 2, the average of the squares is 23 + 2² = 27, so the squares add to 162. The known squares give 9 + 16 + 49 + 36 = 110, so α² + β² = 52.

   2αβ = (α+β)² - (α²+β²) = 100 - 52 = 48 ⇒ αβ = 24

3. Find alpha and beta. Two numbers with sum 10 and product 24 are the roots of t² - 10t + 24 = 0, which are 4 and 6. So the observations are -3, 4, 7, -6, 4, 6.
4. Deviations from 2. The values |xᵢ - 2| are 5, 2, 5, 8, 2, 4. They add to 26.
5. So. Mean deviation about the mean = 26/6 = 13/3, which is option (d).

The idea to keep. When two observations are unknown, the mean gives their sum and the variance gives their sum of squares. Together these fix their product, so you can find each value.

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