Question

Solve the following problem :

The following is the c.d.f of a r.v.X.

x	– 3	– 2	– 1	0	1	2	3	4
F (x)	0.1	0.3	0.5	0.65	0.75	0.85	0.9	1

Find the probability distribution of X and P(–1 ≤ X ≤ 2).

Answer 1

P(X = –3) = F(–3) = 0.1
P(X = –2) = F(–2) – F(–3) = 0.3 – 0.1 = 0.2
P(X = –1) = F(–1) – F(–2) = 0.5 – 0.3 = 0.2
P(X = 0) = F(0) – F(–1) = 0.65 – 0.5 = 0.15
P(X = 1) = F(1) – F(0) = 0.75 – 0.65 = 0.1
P(X = 2) = F(2) – F(1) = 0.85 – 0.75 = 0.1
P(X = 3) = F(3) – F(2) = 0.9 – 0.85 = 0.05
P(X = 4) = F(4) – F(3) = 1 – 0.9 = 0.1
∴ The probability distribution of X is as follows:

X = x	– 3	– 2	– 1	0	1	2	3	4
P(X = x)	0.1	0.2	0.2	0.15	0.1	0.1	0.05	0.1

P(–1 ≤ X ≤ 2)
= P(X = –1 or X = 0 or X = 1 or X = 2)
= P(X = –1) + P(X = 0) + P(X = 1) + P(X = 2)
= 0.2 + 0.15 + 0.1 + 0.1
= 0.55