Solutions To Mathematics Textbooks/Probability and Statistics for Engineering and the Sciences (7th ed) (ISBN-10: 0-495-38217-5)/Chapter 4

Chapter 4 - Continuous Random Variables and Probability Distributions

Section 4.1

Exercise 1

Given the density function $f(x) = \begin{cases} 0.5x & 0 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}$

Part a. Find $P(X \leq 1)$

$\begin{array}{rl} P(X \leq 1) &= \int_{-\infty}^1 f(x) dx \\ &= \int_{0}^1 0.5 x dx = \left. \frac{x^2}{4} \right|_0^1 \\ &= \frac{1}{4} \end{array}$

Part b.

$\begin{array}{rl} P(0.5 \leq X \leq 1.5) &= \int_{0.5}^{1.5} f(x) dx = \left. \frac{x^2}{4} \right|_{0.5}^{1.5} \\ &= \frac{1}{2} \end{array}$

Part c.

$\begin{array}{rl} P(1.5 < X ) &= \int_{0.5}^{2} f(x) dx = \left. \frac{x^2}{4} \right|_{0.5}^{2} \\ &= 0.438 \end{array}$

Exercise 2

Let $X \text{ be Uniform}(-5,5)$

Part a.

$\begin{array}{rl} P(X < 0) &= \int_{-5}^0 \frac{1}{5- -5} dx \\ \\ &= \left. \frac{x}{10} \right|_{-5}{0} \\ \\ &= \frac{1}{2} \end{array}$

Part b.

$P(-2.5 < X < 2.5) = \int_{-2.5}^{2.5} \frac{1}{10} dx = \frac{1}{2}$

Part c.

$P(-2 < X < 3) = \int_{-2}^{3} \frac{1}{10} dx = \frac{1}{2}$

Part d. For $-5 < k < k + 4 < 5$ , compute

$\begin{array}{rl} P(k < X < k+4) &= \int_{k}^{k+4} \frac{1}{10} dx \\ \\ &= \left. \frac{x}{10} \right|_{k}^{k+4} \\ \\ &= \frac{k+4}{10} - \frac{k}{10} \\ \\ &= \frac{4}{10} = \frac{2}{5} \end{array}$

Exercise 3.

Let $f(x)$ be a probability density function.

$f(x) = \begin{cases} 0.09375(4-x^2) & -2 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}$

Part a. graph $f(x)$

Part b.

$P(X > 0) = \int_{0}^{2} f(x) dx = 0.5$

Part c.

$P(-1 < X < 1) = \int_{-1}^{1} f(x) dx = 0.6875$

Part d.

$P(X < -0.5 \text{ or } X > 0.5) = 1 - P(-5 \leq X \leq 0.5) = 1 - \int_{-0.5}^{0.5} f(x) dx = 1 - 0.3671888 = 0.632812$

Exercise 4.

Let $X$ have the Rayleigh distribution with the probability density function

$f(x) = \begin{cases} \frac{x}{\theta^2} e^{-x^2/\left(2\theta^2\right)} & x > 0 \\ & \\ 0 & \text{otherwise} \end{cases}$

Part a.

Verify that $f(x)$ is a pdf.

- First notice that $f(x) \geq 0$ for all $x$
- Next show the integral over the whole number line equals one:

$\begin{array}{rll} \int_{-\infty}^{\infty}f(x)dx &= \int_{0}^{\infty} \frac{x}{\theta^2} e^{-x^2/\left(2\theta^2\right)} dx & \text{let } y = x/\theta, dy = 1/\theta dx \\ \\ &= \int_{0}^{\infty} y e^{-y^2/2} dy & \\ \\ &= \left. e^(-y^2/2) \right|_{0}^{\infty} \\ \\ &= 0 - - 1 \\ &= 1 \end{array}$