If f(x) = begin{cases} 3(1 - 2x^2) & ; 0

Question

(C) We are given the probability density function $f(x) = 3(1 - 2x^2)$ for $0 < x < 1$.To find $P\left(\frac{1}{4} < x < \frac{1}{3}\right)$,we integr

Vedclass · Accepted Answer

$\frac{179}{864}$

Vedclass · Answer

(C) We are given the probability density function $f(x) = 3(1 - 2x^2)$ for $0 To find $P\left(\frac{1}{4} $P\left(\frac{1}{4} $= \int_{\frac{1}{4}}^{\frac{1}{3}} (3 - 6x^2) dx$$= [3x - 2x^3]_{\frac{1}{4}}^{\frac{1}{3}}$$= \left(3\left(\frac{1}{3}ight) - 2\left(\frac{1}{3}ight)^3ight) - \left(3\left(\frac{1}{4}ight) - 2\left(\frac{1}{4}ight)^3ight)$$= \left(1 - \frac{2}{27}ight) - \left(\frac{3}{4} - \frac{2}{64}ight)$$= \left(\frac{25}{27}ight) - \left(\frac{3}{4} - \frac{1}{32}ight)$$= \frac{25}{27} - \frac{3}{4} + \frac{1}{32}$Finding a common denominator $(864)$:$= \frac{25 	imes 32 - 3 	imes 216 + 1 	imes 27}{864}$$= \frac{800 - 648 + 27}{864} = \frac{179}{864}$

$x$	$0$	$1$	$2$	$3$
$P(x)$	$0.2$	$k$	$k$	$2k$

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

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