Question

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 ______.

Options

• –27

• 7

• –7

• 9

Answer 1

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 –7.

Explanation:

Y = ax + b

X = {x ∈ N: 1 ≤ x ≤ 17}

Mean, μ = `(sum_(n = 1)^17 Y_i)/n`

= `(sum_(n = 1)^17 ax_i + b)/n`

= `(asum_(n = 1)^17x_i + sum_(n = 1)^17b_i)/n` ...`(sumx_i = sumn = (n(n + 1))/2)`

= `(a xx (17 xx 18)/2  + b xx 17)/17`

17 = 9a + b  (μ = 17 given) ...(i)

Variance = `(sum_(i = 1)^17(Y_i - μ)^2)/n`

= `(sum_(i = 1)^17 (ax_i + b - 17)^2)/n`

= `(sum_(i = 1)^17 (ax_i + 17 - 9a - 17)^2)/n`  From equation (i)

216 = `(sum_(i = 1)^17 a^2(x_i - 9)^2)/17` (Variance = 216 given)

216 = `(sum_(i = 1)^17 (a^2x_i^2 - 18a^2x_i + 81a^2))/17`

= `(a^2(17 xx 18 xx 35)/6 - (18a^2 xx 17 xx 18)/2 + 81a^2 xx 17)/17`

216 = 105a² – 162a² + 81a²

= 216 = 24a²

a = ±3  ...(∵ a > 0)

⇒ a = 3

17 = 9a + b

⇒ b = –10

Then a + b = –7