If the probability density function of a continuous random variab… - Vidyadip

MCQ

If the probability density function of a continuous random variable X is
$\begin{aligned}\mathrm{f}(x) & =3\left(1-2 x^2\right) ; & & 0<x<1 \\& =0 ; & & \text { otherwise }\end{aligned}$
Then $\mathrm{P}\left(\frac{1}{4}<\mathrm{X}<\frac{1}{3}\right)=$

A
$\frac{128}{752}$
B
$\frac{331}{752}$
C
$\frac{165}{864}$
✓
$\frac{179}{864}$

✓

Answer

Correct option: D.

$\frac{179}{864}$

(D)
$\mathrm{P}\left(\frac{1}{4}<\mathrm{X}<\frac{1}{3}\right)=\int_{\frac{1}{4}}^{\frac{1}{3}} \mathrm{f}(x) \mathrm{d} x=\int_{\frac{1}{4}}^{\frac{1}{3}} 3\left(1-2 x^2\right) \mathrm{d} x$
$ =\left[3 x-2 x^3\right]_{\frac{1}{4}}^{\frac{1}{3}}$
$=\left(1-\frac{2}{27}\right)-\left(\frac{3}{4}-\frac{1}{32}\right)$
$=\frac{1}{4}+\frac{1}{32}-\frac{2}{27}$
$=\frac{179}{864}$

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