Question

A random variable X has the following probability distribution.

| X = x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X = x) | 0 | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |

Find (i) k (ii) the mean and (iii) P(0 < X < 5).

Answer 1

Sum of the probabilities = 1

0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1

⇒ 10k2 + 9k = 1

⇒ 10k2 + 9k – 1 = 0

⇒ 10k2 + 10k – k – 1 = 0

⇒ 10k(k + 1) – 1 (k + 1) = 0

⇒ (10k – 1) (k + 1) = 0

Mean (µ) = 0(0) + 1(k) + 2(2k) + 3(2k) + + 4(3k) + 5(k2) + 6(2k2) + 7(7k2 + k)

= 0 + k + 4k + 6k + 12k + 5k2 + 12k2 + 49k2 + 7k

= 66k2 + 30k