A discrete random variable X has the probability function.P (X = x) =(k(1 − x)2 if x = −1, 0, 1 and 20 otherwisewhere k is a constant.(a) Show that k = 16 .(b) Find E(X).(c) Show that E(X2) = 43(d) Find Var(1 − 3X)

(a) The sum of all probabilities in a probability distribution must equal 1. Therefore, we can set up the following equation:

P(X = -1) + P(X = 0) + P(X = 1) = 1

Substituting the given probability function:

k(1 - (-1))^2 + k(1 - 0)^2 + k(1 - 1)^2 = 1 k(4) + k(1) + k(0) = 1 4k + k = 1 5k = 1 k = 1/5 k = 0.2

So, k = 0.2, not 16.

(b) The expected value E(X) of a discrete random variable X is given by the sum of the product of each outcome and its probability. Therefore,

E(X) = Σ [x * P(X = x)]

Substituting the given values and probabilities:

E(X) = -1 * k(1 - (-1))^2 + 0 * k(1 - 0)^2 + 1 * k(1 - 1)^2 E(X) = -1 * 0.2(4) + 0 + 0 E(X) = -0.8

So, E(X) = -0.8.

(c) The expected value E(X^2) of a discrete random variable X is given by the sum of the product of the square of each outcome and its probability. Therefore,

E(X^2) = Σ [x^2 * P(X = x)]

Substituting the given values and probabilities:

E(X^2) = (-1)^2 * k(1 - (-1))^2 + 0^2 * k(1 - 0)^2 + 1^2 * k(1 - 1)^2 E(X^2) = 1 * 0.2(4) + 0 + 0 E(X^2) = 0.8

So, E(X^2) = 0.8, not 43.

(d) The variance Var(aX + b) of a linear transformation of a random variable X is given by a^2 * Var(X), where a and b are constants. Therefore,

Var(1 - 3X) = (-3)^2 * Var(X)

To find Var(X), we use the formula Var(X) = E(X^2) - [E(X)]^2:

Var(X) = E(X^2) - [E(X)]^2 Var(X) = 0.8 - (-0.8)^2 Var(X) = 0.8 - 0.64 Var(X) = 0.16

Substituting this into the variance of the transformation:

Var(1 - 3X) = (-3)^2 * 0.16 Var(1 - 3X) = 9 * 0.16 Var(1 - 3X) = 1.44

So, Var(1 - 3X) = 1.44.