Question

The probability distribution of X is as follows:

x	0	1	2	3	4
P[X = x]	0.1	k	2k	2k	k

Find

1. k
2. P(X < 2)
3. P[1 ≤ X < 4]

Answer 1

i. Since P[X = x] is a probability distribution of x,

`sum_(x  =  0)^4 P[X = x] = 1`

∴ P[X = 0] + P[X = 1] + P[X = 2] + P[X = 3] + P[X = 4] = 1

0.1 + k + 2k + 2k + k = 1

6k = 1 – 0.1 = 0.9

k = `0.9/6`

= 0.15

ii. P(X < 2) = P[X = 0] + P[X = 1]

= 0.1 + k

= 0.1 + 0.15   ...[∵ k = 0.15]

= 0.25

iii. P[1 ≤ X < 4] = P[X = 1] + P[X = 2] + P[X = 3]

= k + 2k + 2k

= 5k

= 5(0.15)   ...[∵ k = 0.15]

= 0.75