cumulative distribution function calculation 3

Question

given FX(X) = x^2, compute P(1/4 < X < 1/2).

sorry, I am new to here so don't really know how to type them more mathematically.

Answer 1

If the problem was carefully posed, it probably said that the cumulative distribution function $F_X(x)$ of a certain random variable $X$ is given by $F_X(x)=x^2$ for $0\le x\le 1$, and $F_X(x)=0$ for $x\lt 0$, $F_X(x)=1$ for $x\gt 1$.

Recall that by definition $F_X(x)$ is the probability that $X$ is $\le x$. In symbols, $F_X(x)=\Pr(X\le x)$. We have $$\Pr(1/4)\lt X\lt 1/2)=\Pr(X\lt 1/2)-\Pr(X\le 1/4).$$

Because $X$ has continuous distribution, $\Pr(X\lt 1/2)=\Pr(X\le 1/2)=F_X(1/2)=(1/2)^2=1/4$.

Also, $\Pr(X\le 1/4)=F_X(1/4)=1/16$.

It follows that our required probability is $1/4-1/16$, which is $3/16$.