Question

If $F\left(x\right)$ is the c.d.f. associated with the p.d.f. $f\left(x\right)$ and if:
$f\left(x\right)\begin{array}{cc}=\frac{3}{2}(1-x^2),& 0<x<1\\ =0,& otherwise\end{array}$ then $P(\frac{1}{4}<x<\frac{1}{2})=?$

• A. $\frac{46}{128}$
• B. $\frac{48}{128}$
• C. $\frac{41}{128}$
• D. $\frac{58}{128}$

Answer 1

The correct option is C $\frac{41}{128}$

Given $f\left(x\right)=\{\frac{3}{2}(1-x^2),0<x<1\}$

$\{0,otherwise\}$

for $f\left(x\right)$

If $x$ is in internal $(-\infty ,0)$

$\Rightarrow f\left(x\right)={\int }_{-\infty }^{x}f\left(x\right)dx$

$=0$

If $x$ is in internal $[0,1],$ then

$\Rightarrow f\left(x\right)={\int }_{-\infty }^{x}f\left(x\right)dx$

$={\int }_{-\infty }^{0}0dx+{\int }_0^x\frac{3}{2}(1-x^2)dx$

$={\int }_0^x\frac{3}{2}dx-{\int }_0^x\frac{3}{2}{x}^{2}dx$

$={\left[\frac{3}{2}x\right]}_0^x-{\left[\frac{x^3}{2}\right]}_0^x$

$=\frac{3}{2}x-\frac{x^3}{2}$

$=\frac{x}{2}(3-x^2)$

If $x$ is in internal $(1,\infty ),$ then

$f\left(x\right)=1$

Now, $p(\frac{1}{4}<x<\frac{1}{2})=f\left(\frac{1}{2}\right)-f\left(\frac{1}{4}\right)$

$=\frac{1}{4}(3-\frac{1}{4})-\frac{1}{8}(3-\frac{1}{16})$

$=\frac{1}{4}\left(\frac{11}{4}\right)-\frac{1}{8}\left(\frac{47}{16}\right)$

$=\frac{11}{16}-\frac{47}{126}$

$=\frac{88-47}{128}$

$=\frac{41}{128}$

Hence, the answer is $\frac{41}{128}.$