How to calculate the following probability from density function?

Question

Suppose that the variable $X$ takes values on the interval $[0,2]$.

Its Probability density function here is:

$f(x) =c(2-x)^3.$

Question : How can I calculate the following possibility:

$P(1 < X < 4) $ ?

After integrated the density I got that:

$\int^2_0 f(x) =[−c(x−2)^4/4]_0^2 = 4c =1$

From that I got:

$ c = 0.25$

So the Cumulative distribution function should be : $F(x) = -0.25 * (x-2)^4/4$

Now I know that $P(1 < X < 4) = F(4) - F(1)$, but I cannot get the given solution ($0.06$)

I get $F(4) =-1$
and $F(1) = -1/16$

Any help would be appreciated!

Answer 1

A distribution function cannot take negative values. You did not integrate properly. $F(x)=-c\frac {(2-t)^{4}} 4|_0^{x}=4c-c\frac {(2-x)^{4}} 4$. Since $X$ takes values in $[0,2]$ we have $X <4$ with probability $1$. Hence $P(1<X<4)=1-F(1)=\frac 1 {16}$.

$0.06$ is only approximate value for this probability.