Question:
The probability distribution of a random variable x is given as under:
$\mathrm{P}(\mathrm{X}=x)=\left\{\begin{array}{l}k x^{2} \text { for } x=1,2,3 \\ 2 k x \text { for } x=4,5,6 \\ 0 \quad \text { otherwise }\end{array}\right.$
where k is a constant. Calculate
(i) E(X)
(ii) E (3X2)
(iii) P(X ³ 4)
Solution:
The probability distribution of random variable X is given by:
(i) E(X) =
= 1 x k + 2 x 4k + 3 x 9k + 4 x 8k + 5 x 10k + 6 x 12k
= k + 8k + 27k + 32k + 50k + 72k = 190k
= 190 x 1/44 = 95/22 = 4.32 (approx.)
(ii) E(X2) = 3(k + 4 x 4k + 9 x 9k + 16 x 8k + 25 x 10k + 36 x 12k)
= 3(k + 16k + 81k + 128k + 250k + 432k) = 3(908k)
= 3 x 908 x 1/44 = 2724/44 = 61.9 (approx.)
(iii) P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)
= 8k + 10k + 12k = 30k
= 30 x 1/44 = 15/22