Suppose that X is waiting time in minutes for a bus and its p.d.f is given by.`f(x)=1/5,for 0lexle5` and =0 otherwise.Find the probability that (i)Waiting time is between 1 and 3.

IF r.v X : waiting time in minutes for bus and p.d.f of X is given by f(x) ={{:((1)/(5),0 le x le 5),(0,"otherwise ,"):} then probabaility of waiting time not more than 4 minutes is

The p.d.f. of r.v. X is given by f(x)=1/(2a) , for 0 and = 0 otherwise. show that P(X frac{3a}{2})

The time (in minutes)that a lab assistant to prepare the equipment for a certain experiment is a random variable taking values between 25 and 35 minutes with pdf f(x)=1/10,25lexle35 ,=0,otherwise.What is the probability that preparation time exceed 33 minutes?Also find cdf of X.

The p.d.f. of a random variable X is f(x)=1/5, 0lexle5 =0 , otherwise Then the value of P(1 lt X lt 3) is

The following is the p.d.f of continuous r.v. f(x)=x/8,for 0ltxlt4 and x=0 otherwise. (i)Find expression for c.d.f of X

P.d.f of a c.r.v X is f(x)=6x(1-x) ,for 0lexle1 and =0,otherwise(elsewhere) If P(X a) , then a=