# Chapter 4 - Continuous Random Variables and Probability Distributions

## Section 4.1

### Exercise 1

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

• Part a. Find ${\displaystyle P(X\leq 1)}$

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

• Part b.

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

• Part c.

${\displaystyle {\begin{array}{rl}P(1.5

### Exercise 2

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

• Part a.

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

• Part b.

${\displaystyle P(-2.5

• Part c.

${\displaystyle P(-2

• Part d. For ${\displaystyle -5, compute

${\displaystyle {\begin{array}{rl}P(k

### Exercise 3.

Let ${\displaystyle f(x)}$ be a probability density function.

${\displaystyle f(x)={\begin{cases}0.09375(4-x^{2})&-2\leq x\leq 2\\0&{\text{otherwise}}\end{cases}}}$
• Part a. graph ${\displaystyle f(x)}$

• Part b.

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

• Part c.
${\displaystyle P(-1
• Part d.
${\displaystyle P(X<-0.5{\text{ or }}X>0.5)=1-P(-5\leq X\leq 0.5)=1-\int _{-0.5}^{0.5}f(x)dx=1-0.3671888=0.632812}$

### Exercise 4.

Let ${\displaystyle X}$ have the Rayleigh distribution with the probability density function

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

• Part a.

Verify that ${\displaystyle f(x)}$ is a pdf.

• First notice that ${\displaystyle f(x)\geq 0}$ for all ${\displaystyle x}$
• Next show the integral over the whole number line equals one:
${\displaystyle {\begin{array}{rll}\int _{-\infty }^{\infty }f(x)dx&=\int _{0}^{\infty }{\frac {x}{\theta ^{2}}}e^{-x^{2}/\left(2\theta ^{2}\right)}dx&{\text{let }}y=x/\theta ,dy=1/\theta dx\\\\&=\int _{0}^{\infty }ye^{-y^{2}/2}dy&\\\\&=\left.e^{(}-y^{2}/2)\right|_{0}^{\infty }\\\\&=0--1\\&=1\end{array}}}$
• Part b. Let ${\displaystyle \theta =100}$.
• Probability ${\displaystyle X}$ is at most 200
${\displaystyle P(X\leq 200)=\int _{0}^{200}f(x)dx=1-{\frac {1}{e^{2}}}\approx 0.864665}$
• Probability ${\displaystyle X}$ is less than 200
${\displaystyle P(X<200)=\int _{0}^{200}f(x)dx=1-{\frac {1}{e^{2}}}\approx 0.864665}$
• Probability ${\displaystyle X}$ is at least 200
${\displaystyle P(X\geq 200)=1-P(X<200)={\frac {1}{e^{2}}}\approx 0.135335}$

• The probability ${\displaystyle X}$ is between 100 and 200 assuming ${\displaystyle \theta =100}$.
${\displaystyle P(100
• Give an expression for ${\displaystyle P(X\leq x)}$, i.e., define the cumulative distribution function.
${\displaystyle P(X\leq x)=\int _{-\infty }^{x}f(y)dy={\begin{cases}0&x\leq 0\\1-\exp \left({\frac {-x^{2}}{2\theta ^{2}}}\right)&x>0\end{cases}}}$