The time (in minutes) for a lab assistant to prepare the equipment for a certain experiment is a random variable taking values between 25 and 35 minutes with p.d.f. `f(x)={{:((1)/(10)","25lexle35),(0", otherwise"):}`, then F (x) =

The time (in minutes) for a lab assistant to prepare the equipment for a certain experiment is a random variable taking values between 25 and 35 minutes with p.d.f. f(x)={{:((1)/(10)","25lexle35),(0", otherwise"):} , then the probability that preparation time exceeds 33 minutes is

The time (in minutes) for a lab assistant to prepare the equipment for a certain experiment is a random variable taking values between 25 and 35 minutes with p.d.f f(x) = {{:((1)/(10), 25 le x le 35),(0, "otherwise"):} What is the probability that preparation time exceeds 33 minutes ? . Also find the c.d.f of X.

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 continuous r.v. X is f(x)={{:((1)/(10)","-5lexle5),(0", otherwise"):} , then P(X lt0) =

The p.d.f. of a continuous r.v. X is f(x)={{:((1)/(10)","-5lexle5),(0", otherwise"):} , then P(X lt0) =

The p.d.f. of a r.v. X is f(x)={{:((1)/(x^(2))","1ltxltoo),(0", otherwise"):} , then F (x) =

The p.d.f. of a r.v. X is f(x)={{:(3(1-2x^(2))","0ltxlt1),(0", otherwise"):} , then F (x) =

The p.d.f. of a continuous r.v. X is f(x)={{:((x)/(8)", "0ltxlt4),(0", otherwise"):} , then F (x) =

The p.d.f. of a r.v. X is f(x)={{:(kx","0ltxlt2),(0", otherwise"):} , then k =

The p.d.f. of a r.v. X is f(x)={{:(kx", "0ltxlt2),(0", otherwise"):} , then k =