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Question
The following is the c.d.f. of 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 p.m.f. of X.
i. P(–1 ≤ X ≤ 2)
ii. P(X ≤ 3 / X > 0).
Solution
From the given table
F(−3)= 0.1, F(−2) = 0.3, F(−1) = 0.5
F(0) = 0.65, f(1) = 0.75, F(2) = 0.85
F(3) = 0.9, F(4) = 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 | 00.5 | 0.1 |
i. 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
ii. P(X ≤ 3 / X > 0)
= `("P"("X" = 1 "or" "X" = 2 "or" "X" + 3))/("P"("X" = 1 "or" "X" = 2 "or" "X" = 3 "or" "X" = 4)` ......`[("Using conditional probability"),("P"("A"/"B") = ("P"("A" ∩ "B"))/("P"("B")))]`
= `("P"("X" = 1) + "P"("X" = 2) + "P"("X" = 3))/("P"("X" = 1) + "P"("X" = 2) + "P"("X" = 3) + "P"("X" = 4))`
= `(0.1 + 0.1 + 0.05)/(0.1 + 0.1 + 0.05 + 0.1)`
= `0.25/0.35`
= `5/7`
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