Question

The discrete random variable X has the probability function.

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

Evaluate p(x < 6), p(x ≥ 6) and p(0 < x < 5)

Answer 1

P(x < 6) = P(x = 0) + P(x = 1) + P(x = 2)+ P(x = 3) + p(x = 4) + P(x = 5)

= 0 + k + 2k +2k + 3k + k²

= 8k + k² 

= `8(1/10) +  (1/10)^2`

= `8/10 + 1/100`

= `(80 + 1)/100`

P(x < 6) = `81/100`

P(x ≥ 6) = P(x = 6) + p(x = 7)

= 2k² + 7k² + k

= 9k² + k

= `9(1/10)^2 + 1/10`

= `9/100 + 1/10`

= `(9 + 10)/100`

= `19/100`

∴ P(x ≥ 6) = 19/100`

P(0 < x < 5) = P(x = 1) + P(x = 2) + p(x = 3) + P(x = 4)

= k + 2k + 2k + 3k

= 8k

= `8(1/10)`

∴ P(0 < x < 5) = `8/10`