Given the probability density function: f(x) | vedclass

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(A) Given the p.d.f. of a random variable $X$: $f(x) = \begin{cases} 3(1 - 2x^2), & 0 < x < 1 \ 0, & 	ext{otherwise} \end{cases}$ The probability is

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$\frac{179}{864}$

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(A) Given the p.d.f. of a random variable $X$: $f(x) = \begin{cases} 3(1 - 2x^2), & 0 The probability is calculated as: $P\left(\frac{1}{4} $= 3 \left[ x - \frac{2}{3}x^3 ight]_{1/4}^{1/3}$ $= 3 \left[ \left( \frac{1}{3} - \frac{2}{3} \cdot \frac{1}{27} ight) - \left( \frac{1}{4} - \frac{2}{3} \cdot \frac{1}{64} ight) ight]$ $= 3 \left[ \left( \frac{1}{3} - \frac{2}{81} ight) - \left( \frac{1}{4} - \frac{1}{96} ight) ight]$ $= 3 \left[ \frac{27-2}{81} - \frac{24-1}{96} ight] = 3 \left[ \frac{25}{81} - \frac{23}{96} ight]$ $= 3 \left[ \frac{25 	imes 32 - 23 	imes 27}{2592} ight] = 3 \left[ \frac{800 - 621}{2592} ight]$ $= 3 	imes \frac{179}{2592} = \frac{179}{864}$

$X$	$\alpha$	$1$	$0$	$-3$
$P(X)$	$\frac{1}{3}$	$K$	$\frac{1}{6}$	$\frac{1}{4}$

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = x)$	$k$	$3k$	$5k$	$7k$	$8k$	$k$

Given the probability density function: $f(x) = \begin{cases} 3(1 - 2x^2), & 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}$ The probability $P\left(\frac{1}{4} < X < \frac{1}{3}\right)$ is given by: $P\left(\frac{1}{4} < X < \frac{1}{3}\right) = \int_{1/4}^{1/3} 3(1 - 2x^2) \, dx$

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