Last modified on 10 May 2011, at 18:32

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 DistributionsEdit

Section 4.1Edit

Exercise 1Edit

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 2Edit

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.Edit

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)

Ch04ex003a.png

  • 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.Edit

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}
  • 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}