The p.d.f. of a continuous random variable X | vedclass

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(C) We are given the probability density function $f(x) = \frac{x+2}{18}$ for $-2 < x < 4$. We need to find $P[|x| < 1]$. The condition $|x| < 1$ is e

Vedclass · Accepted Answer

$\frac{2}{9}$

Vedclass · Answer

(C) We are given the probability density function $f(x) = \frac{x+2}{18}$ for $-2 We need to find $P[|x| The condition $|x| Thus,$P[|x| Evaluating the integral: $P = \frac{1}{18} \int_{-1}^{1} (x+2) \, dx = \frac{1}{18} \left[ \frac{x^2}{2} + 2x \right]_{-1}^{1}$. Substituting the limits: $P = \frac{1}{18} \left[ (\frac{1}{2} + 2) - (\frac{(-1)^2}{2} + 2(-1)) \right]$. $P = \frac{1}{18} \left[ (\frac{1}{2} + 2) - (\frac{1}{2} - 2) \right]$. $P = \frac{1}{18} [ \frac{1}{2} + 2 - \frac{1}{2} + 2 ] = \frac{1}{18} [4] = \frac{4}{18} = \frac{2}{9}$. Therefore,the correct option is $C$.

$X$	$30$	$10$	$-10$
$P(X)$	$\frac{1}{5}$	$A$	$B$

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

The p.d.f. of a continuous random variable $X$ is given by $f(x) = \frac{x+2}{18}$ for $-2 < x < 4$ and $f(x) = 0$ otherwise. Then $P[|x| < 1] = $

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