Homework SourseThe cdf of the random variable X is given by Fxx 13 23 x12
The cdf of the random variable X is given by
Fx(x) = 1/3 + (2/3) (x+1)2 -1 x 0
0 x<-1
Find the probability of the events A = {X> 1/3 }, B= {|X- 1/3 | <1 } and C= {X<0}
Fx(x) = 1/3 + (2/3) (x+1)2 -1 x 0
0 x<-1
Find the probability of the events A = {X> 1/3 }, B= {|X- 1/3 | <1 } and C= {X<0}
Solution
Given
Fx (x) = 1/3 +(2/3)/(x+1)2 -1<=x<=0
=0 x< -1
Therefore it is like saying x has a probability to be onlybetween -1 and 0.
P(x> 1/3) = 0 because x doesn’t exceed 0 .
P(B) = P(|x-1/3|<1)
=P(-2/3< x< 4/3) = P(x<4/3)- P(x<-2/3) =1-1/3+(2/3)/(2/3+1)2 = 90.67%
P(c ) = P(x<0) = Fx (0) = 1/3+2/3 = 1
Hope this helps you. Feel free to ask for any clarifications
