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(C) Step $1$: Use the property of the p.d.f. that $\int_{-\infty}^{\infty} f(x) dx = 1$.Step $2$: Calculate the integral: $\int_{0}^{1} kx dx + \int_{

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$1/4$

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(C) Step $1$: Use the property of the p.d.f. that $\int_{-\infty}^{\infty} f(x) dx = 1$.Step $2$: Calculate the integral: $\int_{0}^{1} kx dx + \int_{1}^{2} k dx + \int_{2}^{3} (-kx + 3k) dx = 1$.Step $3$: Evaluate the integrals: $[k \frac{x^2}{2}]_{0}^{1} + [kx]_{1}^{2} + [-k \frac{x^2}{2} + 3kx]_{2}^{3} = 1$.Step $4$: $\frac{k}{2} + k + [(- \frac{9k}{2} + 9k) - (- \frac{4k}{2} + 6k)] = 1$.Step $5$: $\frac{k}{2} + k + [\frac{9k}{2} - 4k] = 1 \implies \frac{k}{2} + k + \frac{k}{2} = 1 \implies 2k = 1 \implies k = 1/2$.Step $6$: Calculate $P(2 Step $7$: Substitute $k = 1/2$: $[-\frac{x^2}{4} + \frac{3x}{2}]_{2}^{3} = (-\frac{9}{4} + \frac{9}{2}) - (-1 + 3) = \frac{9}{4} - 2 = 1/4$.

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

Let $X$ be a continuous random variable with the probability density function (p.d.f.) given by $f(x) = \begin{cases} kx, & 0 \leq x < 1 \\ k, & 1 \leq x < 2 \\ -kx + 3k, & 2 \leq x < 3 \\ 0, & \text{otherwise} \end{cases}$. Find $P(2 < X \leq 3)$.

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