$1$ . $5$ . $8$ . Given the cdf $\newline$ $F(x)=\left\{\begin{array}{ll} 0 & x<-1 \\ \frac{x+2}{4} & -1 \leq x<1 \\ 1 & 1 \leq x . \end{array}\right.$ $\newline$ Sketch the graph of $F(x)$ and then compute: $\newline$ (a) $P\left(-\frac{1}{2}<X \leq \frac{1}{2}\right)$ ; $\newline$ (b) $P(X=0)$ ; $\newline$ (c) $P(X=1)$ ; $\newline$ (d) $P(2<X \leq 3)$ .

Full solution

1

5

8

. Given the cdf

\newline

F(x)=\left\{\begin{array}{ll} 0 & x<-1 \\ \frac{x+2}{4} & -1 \leq x<1 \\ 1 & 1 \leq x . \end{array}\right.

\newline

Sketch the graph of

F(x)

and then compute:

\newline

(a)

P\left(-\frac{1}{2}<X \leq \frac{1}{2}\right)

;

\newline

(b)

P(X=0)

;

\newline

(c)

P(X=1)

;

\newline

(d)

P(2<X \leq 3)

Graph Sketch: Sketch the graph of $F(x)$ based on the given CDF. $\newline$ For $x < -1$ , $F(x) = 0$ , which is a horizontal line. $\newline$ For $-1 \leq x < 1$ , $F(x) = \frac{x+2}{4}$ , which is a line with a slope of $\frac{1}{4}$ starting at $F(-1) = \frac{1}{4}$ and ending at $F(1) = \frac{3}{4}$ . $\newline$ For $x \geq 1$ , $F(x) = 1$ , which is another horizontal line.
Calculate $P(-\frac{1}{2} < X \leq \frac{1}{2})$ : (a) Calculate $P(-\frac{1}{2} < X \leq \frac{1}{2})$ using $F(x)$ . $P(-\frac{1}{2} < X \leq \frac{1}{2}) = F(\frac{1}{2}) - F(-\frac{1}{2})$ . $F(\frac{1}{2}) = (\frac{1}{2} + 2)/4 = \frac{2.5}{4} = \frac{5}{8}$ . $F(-\frac{1}{2}) = (\frac{-1}{2} + 2)/4 = \frac{1.5}{4} = \frac{3}{8}$ . $P(-\frac{1}{2} < X \leq \frac{1}{2}) = \frac{5}{8} - \frac{3}{8} = \frac{2}{8} = \frac{1}{4}$ .
Calculate $P(X=0)$ : (b) Calculate $P(X=0)$ using $F(x)$ . $\newline$ Since $F(x)$ is a CDF of a continuous random variable, $P(X = a)$ for any specific value $a$ is $0$ . $\newline$ $P(X=0) = 0$ .
Calculate $P(X=1)$ : (c) Calculate $P(X=1)$ using $F(x)$ . $\newline$ Again, since $F(x)$ is a CDF of a continuous random variable, $P(X = a)$ for any specific value $a$ is $0$ . $\newline$ $P(X=1) = 0$ .
Calculate $P(2 < X \leq 3)$ : (d) Calculate $P(2 < X \leq 3)$ using $F(x)$ . $\newline$ Since $F(x)$ is $1$ for all $x \geq 1$ , $P(2 < X \leq 3) = F(3) - F(2)$ . $\newline$ $F(3) = 1$ and $F(2) = 1$ . $\newline$ $P(2 < X \leq 3) = 1 - 1 = 0$ .