 # cdf

A random variable X is continuous and bounded between zero and five, X:(0<=X<=5). The cumulative distribution function (cdf) for X is F(x)=x/5. Calculate P(2<=X<=4). Method1: P(2<=X<=4)=F(4)-F(2) F(4)=4-0/5-0=4/5, F(2)=2/5 F(4)-F(2)=0.4 but if Calculate in Method2: F(4)=p(<=4)=1/5+2/5+3/5+4/5 F(2)=p(<=2)=1/5+2/5 F(4)-F(2)=7/5 different answers? what’s wrong thanks

I’d go Method2. How could F(2)=2/5? It’s supposed to be cdf. Thus F(2) has to be 1/5 + 2/5

hyang Wrote: ------------------------------------------------------- > I’d go Method2. How could F(2)=2/5? It’s supposed > to be cdf. Thus F(2) has to be 1/5 + 2/5 no , in Method 1 ,F(2)=2/5 is actually from 2-0/5-0. the answer gives 0.4, which is Method 1. what wrong with Method 2?

A random variable X is >>>> continuous <<<< and bounded between zero and five. You cannot just add probs at discrete points. At discrete points, the prob will be zero for a continuous distribution.

F(x) = P(X <= x) so obviously it is some number <= 1 for any x. They gave you the formula for F(x) as F9X0 = x/5. Method 2 uses some other bogus method for changing the definition of F(x) that they gave you. I’d like to tell you whats wrong with it, but I don’t even know where you came up with it.

Method 1 looks right. Method 2 is wrong because you can’t simply add discrete numbers to determine the probability for a continuous function. The key here is the function is continuous. What about F(2.3544…?)