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). - Mathematics and Statistics

Sum

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 0.5 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

Concept: Probability Distribution of Discrete Random Variables
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Chapter 7: Probability Distributions - Miscellaneous Exercise 2 [Page 244]

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Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 7 Probability Distributions
Miscellaneous Exercise 2 | Q 9.1 | Page 244
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