Let X be a continuous random variable with the probability density function
$ f(x) = \left\{ \begin{array}{ll} \frac{x+1}{2}, \quad -1 < x < 1 \\ 0, \quad \text{otherwise} \end{array} \right. $
Then $ P(\frac{1}{4}< X^2 <\frac{1}{2})$ is
I integrated $f(x)$ from $\frac{1}{4} $ to $ \frac{1}{2}$ and got $\frac{11}{64}$
But I don't know whether my answer is correct or not.
if my approach is wrong then what is the correct method to do this question?
Best Answer
Given
$$ P(\frac{1}{4}< X^2 <\frac{1}{2})$$ $$ P(\frac{1}{2}< X <\frac{1}{\sqrt{2}})$$
$$ P(\frac{1}{2}< X <\frac{1}{\sqrt{2}}) = \int_{\frac{1}{2}} ^{\frac{1}{\sqrt{2}}} \frac{x+1}{2} = 0.166 $$