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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).

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#### 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`