# 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

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.

#### 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
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 7 Probability Distributions
Miscellaneous Exercise | Q 5 | Page 242