5 Sep 2017

(Question from Mathematics Standard Level for the IB Diploma)

The probability distribution of a random variable X is given by P(X=x) = kx(4-x), where x = 1, 2, 3.

a) Find the value of k.

We have 3 values in our sample space: 1, 2 and 3. The corresponding probability distribution is:

P(1) = k(1)(4-1) = 3k
P(2) = k(2)(4-2) = 4k
P(3) = k(3)(4-3) = 3k

Recall that all probabilities in a probability distribution needs to add to 1. So:

3k+4k+3k = 1
10k = 1
k = 1/10 = 0.1

b) Find E(X).

Recall,

$$E(X)=\mu = \sum^n_{i=1}x_ip_i \ \ \ \ \ \ \ \ \text{or} \ \ \sum^n_{i=1}x_iP(X=x_i)$$

Since k = 1/10, from part a) our probability distribution is:

P(1) = 3/10
P(2) = 4/10
P(3) = 3/10

So,

E(X) = (1)(3/10) + (2)(4/10) + (3)(3/10)
E(X) = 3/10 + 8/10 + 9/10
E(X) = 20/10 = 2

So E(X) = 2