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 $P(X\leq 1)$

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

• 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

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

• 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:

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

• 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.

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