Maharashtra State BoardHSC Arts 12th Board Exam
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In the p.m.f. of r.v. X X 1 2 3 4 5 P (X) 120 320 a 2a 120 Find a and obtain c.d.f. of X. - Mathematics and Statistics

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Sum

In the p.m.f. of r.v. X

X 1 2 3 4 5
P (X) `1/20` `3/20` a 2a `1/20`

Find a and obtain c.d.f. of X.

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Solution

For p.m.f. of a r.v. X

`sum_("i" = 1)^5` P(X = x) = 1

∴ P(X = 1) +  P(X = 2) + P(X = 3) +  P(X = 4) + P(X = 5) = 1

∴ `1/20 + 3/20+ "a" + 2"a" + 1/20 = 1`

∴ 3a = `1 - 5/20`

∴ 3a = `1 - 1/4`

∴ 3a =`3/4`

∴ a = `1/4`

∴ The p.m.f. of the r.v. X is

X 1 2 3 4 5
P(X = x) `1/20` `3/20` `5/20` `10/20` `1/20`

Let F(x) be the c.d.f. of X.

Then F(x) = P(X ≤ x)

∴ F(1) = P(X ≤ 1) =  P(X = 1) = `1/20`

F(2) = P(X≤ 2) =  P(X = 1) +  P(X = 2)

= `1/20 + 3/20`

= `4/20`

= `1/5`

F(3) = P(X ≤ 3) =  P(X = 1) +  P(X = 2) + P(X = 3)

= `1/20 + 3/20 + 5/20`

= `9/20`

F(4) = P(X ≤ 4) =  P(X = 1) +  P(X = 2) + P(X = 3) + P(X = 4)

= `1/20 + 3/20 + 5/20 + 10/20`

= `19/20`

F(5) = P(X ≤ 5) =  P(X = 1) +  P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) 

= `1/20 + 3/20 + 5/20 + 10/20 + 1/20`

= `20/20`

= 1

Hence, the c.d.f. of the random variable X is as follows :

xi 1 2 3 4 5
F(xi) `1/20` `1/5` `9/20` `19/20` 1
Concept: Types of Random Variables
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