Let X = x ∈ N : 1 ≤ x ≤ 17 and
Y = ax + b: x ∈ X and a, b ∈ R, a > 0. If mean
and variance of elements of Y are 17 and 216
respectively then a + b is equal to :

Solution

Maths sir · 4 steps

Adding b shifts the data and leaves the spread alone, so let the variance give you a first.

1. Know X first. X is the first 17 natural numbers. Their mean is (17+1)/2 = 9, and their variance is (n²-1)/12 = (289-1)/12 = 24.

   ȳ = ax̄ + b, σ_y² = a² σₓ²

2. Variance gives a. 24a² = 216, so a² = 9. Since a > 0, a = 3.
3. Mean gives b. 3 × 9 + b = 17, so b = 17 - 27 = -10.
4. So. a + b = 3 + (-10) = -7, which is option (c).

The idea to keep. Under y = ax + b, the mean becomes ax̄ + b but the variance becomes only a²σ². Use the variance to find a, then the mean to find b.

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