Question

The probability distribution of a random variable x is given as under:
P(X = x) = `{{:("k"x^2,  "for"  x = 1"," 2"," 3),(2"k"x,  "for"  x = 4"," 5"," 6),(0,  "otherwise"):}`
where k is a constant. Calculate E(X)

Answer 1

Given that: P(X = x) = `{{:("k"x^2,  "for"  x = 1"," 2"," 3),(2"k"x,  "for"  x = 4"," 5"," 6),(0,  "otherwise"):}`

∴ Probability distribution of random variable X is

X	1	2	3	4	5	6	otherwise
P(X)	k	4k	9k	8k	10k	12k	0

We know that `sum_("i" = 1)^"n" "P"("X"_"i")` = 1

∴ k + 4k + 9k + 8k + 10k + 12k = 1

⇒ 44k = 1

⇒ k = `1/44`

E(X) = `sum_("i" = 1)^"n" "P"_"i""X"_"i"`

= 1 × k + 2 × 4k + 3 × 9k + 4 × 8k + 5 × 10k + 6 × 12k

= k + 8k + 27k + 32k + 50k + 72k

= 190k

= `190 xx 1/44`

= `95/22`

= 4.32  ......(Approx)