If $\mathrm{f}(x)= \begin{cases}3\left(1-2 x^2 ight) & ; 0

Question

If $\mathrm{f}(x)= \begin{cases}3\left(1-2 x^2ight) & ; 0 < x < 1 \ 0 & ; 	ext { otherwise }\end{cases}$ is a probability density function of $\mathrm{X}$, then $\mathrm{P}\left(\frac{1}{4} < x < \frac{1}{3}ight)$ is

Aicharya · Accepted Answer

$\begin{aligned} \mathrm{P}\left(\frac{1}{4} < x < \frac{1}{3}ight)= & \int_{\frac{1}{4}}^{\frac{1}{3}} \mathrm{f}(x) \mathrm{d} x=\int_{\frac{1}{4}}^{\frac{1}{3}} 3\left(1-2 x^2ight) \mathrm{d} x \ & =\left[3 x-2 x^3ight]_{\frac{1}{4}}^{\frac{1}{3}} \ & =\left(1-\frac{2}{27}ight)-\left(\frac{3}{4}-\frac{1}{32}ight) \ & =\frac{1}{4}+\frac{1}{32}-\frac{2}{27} \ & =\frac{179}{864}\end{aligned}$

If $\mathrm{f}(x)= \begin{cases}3\left(1-2 x^2\right) & ; 0 < x < 1 \\ 0 & ; \text { otherwise }\end{cases}$…

Solution