Q.The following is the p.d.f. of continuous r.v. f

Integrate the pdf to get the cdf $F(x)=\int_{-1}^x f(t)\,dt$, then read off the required probabilities.

$f(x) = \dfrac{x^2}{3}$ for $-1<x<2$, and $0$ otherwise.

$F(x) = \int_{-1}^{x} \dfrac{t^2}{3}\,dt = \left[\dfrac{t^3}{9}\right]_{-1}^{x} = \dfrac{x^3}{9} - \dfrac{(-1)^3}{9} = \dfrac{x^3+1}{9}$

So the cdf is:

$F(x) = \begin{cases} 0, & x\le -1 \\ \dfrac{x^3+1}{9}, & -1<x<2 \\ 1, & x\ge2 \end{cases}$

(Check: $F(2)=\dfrac{8+1}{9}=1$ ✓)

$P(X<1) = F(1) = \dfrac{1^3+1}{9} = \dfrac29$

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Integrate the p.d.f. from $0$ to $x$ to get the c.d.f., then evaluate at the two given points (noting $x=9$ lies outside the support).

$f(x)=\frac{x}{32},\quad 0<x<8;\qquad f(x)=0 \text{ otherwise}$

a. c.d.f.:

For $x\le0$: $F(x)=0$.

For $0<x<8$:

$F(x)=\int_0^x\frac{t}{32}\,dt=\frac{1}{32}\left[\frac{t^2}{2}\right]_0^x=\frac{x^2}{64}$

For $x\ge8$: $F(x)=1$.

$F(x)=\begin{cases}0, & x\le0\\[4pt] \dfrac{x^2}{64}, & 0<x<8\\[4pt] 1, & x\ge8\end{cases}$

b. Evaluate:

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