Let X be a continuous random variable with probability distribution

Evaluate k and find p(1 $\lt$ x $\leq$ 2)

Since the total probability is 1

$∫_{-∞}^∞p(x)$ = 1

$∫_0^3 \frac{x}{6}+k) \ dx = [\frac{x^2}{12}+kx]_0^3 = \frac{3}{4} +3k $ = 1

∴ 3k = 1 - $\frac{3}{4}$ = $\frac{1}{4}$ or k = $\frac{1}{12}$

∴ p(x) = $ \begin{cases} \frac{x}{6}+\frac{1}{12} \,\,\,\,\,\, if \ 0 \leq x \leq 3 \\[2ex] 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, elsewhere \end{cases} $

∴ $p (1 \lt x ≤2) = ∫_1^2 (\frac{x}{6}+ \frac{1}{12}) \ dx =[\frac{x^2}{12}+ \frac{x}{12}]_1^2 = \frac{1}{3}$