The density function of a random variable X is given by f(x) = kx (2-x), 0≤x≤2. [7 Find (i) k (ii) mean (iii) variance

Answer 1

To find k, we need to make sure that the total area under the density function is equal to 1. We can do this by integrating the density function over the interval [0,2] and setting the result equal to 1:

∫[0,2] kx(2-x) dx = 1
k ∫[0,2] (2x - x^2) dx = 1
k [(x^2) - (x^3/3)] [0,2] = 1
k [(2^2) - (2^3/3)] - [0 - (0/3)] = 1
k (4 - 8/3) = 1
k (12/3 - 8/3) = 1
k = 3/2

(i) k = 3/2

To find the mean (expected value) of X, we use the formula:

E(X) = ∫[0,2] x*f(x) dx
E(X) = ∫[0,2] x*(3/2)*x*(2-x) dx
E(X) = ∫[0,2] (3/2)x^2*(2-x) dx
E(X) = (3/2) ∫[0,2] (2x^2 - x^3) dx
E(X) = (3/2) [(2(2^3)/3 - (2^4/4)) - (0)]
E(X) = (3/2) [(16/3 - 16) - 0]
E(X) = (3/2) [16/3 - 48/3]
E(X) = (3/2) [-32/3]
E(X) = -16

(ii) The mean of X is -16.

To find the variance of X, we use the formula:

Var(X) = E(X^2) - [E(X)]^2

First, we find E(X^2):

E(X^2) = ∫[0,2] x^2*f(x) dx
E(X^2) = ∫[0,2] x^2*(3/2)*x*(2-x) dx
E(X^2) = (3/2) ∫[0,2] x^3*(2-x) dx
E(X^2) = (3/2) [(2(2^4)/4 - (2^5)/5) - 0]
E(X^2) = (3/2) [(16 - 32/5) - 0]
E(X^2) = (3/2) [(80/5 - 32/5)]
E(X^2) = (3/2) [48/5]
E(X^2) = 72/5

Then we find the variance:

Var(X) = E(X^2) - [E(X)]^2
Var(X) = 72/5 - (-16)^2
Var(X) = 72/5 - 256
Var(X) = 72/5 - 1280/5
Var(X) = -1208/5

(iii) The variance of X is -1208/5.