# A random variable X has the following probability distribution - Mathematics and Statistics

A random variable X has the following probability distribution :

 X=x 0 1 2 3 4 5 6 P[X=x] k 3k 5k 7k 9k 11k 13k

(a) Find k
(b) find P(O <X< 4)
(c) Obtain cumulative distribution function (c. d. f.) of X.

#### Solution

sum P(x)=1

P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1

K+3x +5k +7k +9k +11k +13k=1

49k=1

k=1/49

P(0<x<4)=P(1)+P(2)+p(3)

=3k+5k+7k

=15k

=15/49

F(0)=P(0)=1/49

F(1)=P(0)+P(1)=1/49+3/49=4/49

F(2)=P(0)+P(1)+P(2)=1/49+3/49+5/49=9/49

F(3)=P(0)+P(1)+P(2)+P(3)=1/49+3/49+5/49+7/49=16/49

F(4)=P(0)+P(1)+P(2)+P(3)+P(4)=1/49+3/49+5/49+7/49+9/49=25/49

F(5)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)=1/49+3/49+5/49+7/49+9/49+11/49=36/49

F(6)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1/49+3/49+5/49+7/49+9/49+11/49+13/49=49/49

 Cummulative districbution function (c.d.f.) of x

 X 0 1 2 3 4 5 6 F(x) 1/49 4/49 9/49 16/46 25/49 36/49 1
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2013-2014 (March)

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