The C.D.F. (cumulative density function) of the random variable X is given by, F(x)= [0, x<0 x/2, 0<=x<1 2/3, 1<=x<2 11/12, 2<=x<3 1, 3<=x] a) Find P(X>1/2) b) Find P(2<X<=4) c) Find P(X=1) Does anyone know how to approach this?
Follow Math Help Forum on Facebook and Google+
Originally Posted by abc4616 The C.D.F. (cumulative density function) of the random variable X is given by, F(x)= [0, x<0 x/2, 0<=x<1 2/3, 1<=x<2 11/12, 2<=x<3 1, 3<=x] a) Find P(X>1/2) Definition of the cdf, tells you that: P(X>1/2)=F(1/2). b) Find P(2<X<=4) Again from the definition of the sdf: P(2<X<=4)=P(X<=4)-P(X<2)=F(4)-F(2) c) Find P(X=1) This is equal to the jump in the cdf at x=1, so its F(1)-F(1-)=2/3-1/2=1/6 RonL
View Tag Cloud