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 =

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 $$ P(X \leq 1) $$

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


 * Part b.

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


 * Part c.

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

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

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

Verify that $$ f(x) $$ is a pdf.
 * Part a.
 * 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} $$


 * Part b. Let $$ \theta = 100$$.
 * Probability $$ X $$ is at most 200


 * $$ P(X \leq 200) = \int_{0}^{200} f(x) dx = 1 - \frac{1}{e^2} \approx 0.864665 $$


 * Probability $$ X $$ is less than 200


 * $$ P(X < 200) = \int_{0}^{200} f(x) dx = 1 - \frac{1}{e^2} \approx 0.864665 $$


 * Probability $$ X $$ is at least 200


 * $$ P(X \geq 200) = 1 - P(X < 200) = \frac{1}{e^2} \approx 0.135335 $$


 * The probability $$ X $$ is between 100 and 200 assuming $$ \theta = 100 $$.


 * $$ P(100 < X < 200) = \int_{100}^{200} f(x)dx = \frac{e^{3/2} - 1}{e^2} \approx 0.471195 $$


 * Give an expression for $$ P(X \leq x) $$, i.e., define the cumulative distribution function.


 * $$ P(X \leq x) = \int_{-\infty}^{x} f(y)dy = \begin{cases} 0 & x \leq 0 \\ 1 - \exp\left( \frac{-x^2}{2\theta^2} \right) & x > 0 \end{cases} $$