Refer to the Canadian Journal of Civil Engineering (January 2013) study of variable speed limits, Exercise 4.23 (p. 225). Recall that a portion of an urban freeway was divided into three sections of equal length, and variable speed limits were posted (independently) in each section. The probability distribution of the optimal speed limit for each section follows (Probabilities in parentheses). Section 1: 30 mph (.05),40 mph (.25), 50 mph (.25), 60 mph (.45);

Section 2 30 mph (.10), 40 mph (.25), 50 mph (.35), 60 mph (.30); Section 3: 30 mph (.15), 40 mph (.20), 50 mph (.30), 60 mph (.35). A vehicle adhering to the speed limit will travel through the three sections of the freeway at a steady (fixed) speed. Let x represents this speed.

a.List the possible values of x.

b.Find P(x) = 30. [Hint: The event {x = 30} is the union of the events {x = 30 in Section 1}, {x = 30 in Section 2}, and {x = 30 in Section 3}. Also, P1x = 30

in Section 1) = P1x = 30 | Section 1) * P (Section 1), where P (Section 1) =$\frac{\mathbf{1}}{\mathbf{3}}$ , since the sections are of equal length.]

c.Find the probability distribution for x.

d.What is the probability that the vehicle can travel at least 50 mph through all three freeway sections?

From the given information urban freeway is divided into three sections of equal length, and the variable speed limits were posted independently in each section.

The probability distribution of the optimal speed limit for each section follows:

Section 1-30 mph$\left(0.05\right)$

$40$mph$\left(0.25\right)$

50 mph$\left(0.25\right)$

60 mph$\left(0.45\right)$

Section 2 - 30 mph$\left(0.10\right)$

40 mph$\left(0.25\right)$

50 mph$\left(0.35\right)$

60 mph$\left(0.30\right)$

Section 3-30 mph$\left(0.15\right)$

40 mph$\left(0.20\right)$

50 mph$\left(0.30\right)$

60 mph$\left(0.35\right)$

To find the possible values of by, arranged in a table to get:

Here we take only the highest values of .

For$p\left(x=30\right)$we get that

$\begin{array}{rcl}p\left(x=30\right)& =& p\left[x=30\left(\text{section}1\right),x=30\left(\text{section}2\right),x=30\left(\text{section}3\right)\right]\\ & =& p\left[x=30\left(\text{section}1\right)+x=30\left(\text{section}2\right)+x=30\left(\text{section}3\right)\right]\\ & & \end{array}$....(1)

Then

$\begin{array}{rcl}p\left(x=30\left(\text{section}1\right)\right)& =& p\left(x=30\left|\text{section}1\right.\right)\times p\left(\text{section}1\right)\\ & =& 0.05\times \frac{1}{3}\\ & =& 0.001\\ & & \end{array}$

Where $p\left(\text{section}1\right)=\frac{1}{3}$

Similarly, we define for section 2 and section3 , we get

$\begin{array}{rcl}p\left(x=30\left(\text{section}2\right)\right)& =& p\left(x=30\left|\text{section}2\right.\right)\times p\left(\text{section}2\right)\\ & =& 0.10\times \frac{1}{3}\\ & =& 0.033\\ & & \end{array}$

$\begin{array}{rcl}p\left(x=30\left(\text{section}3\right)\right)& =& p\left(x=30\left|\text{section}3\right.\right)\times p\left(\text{section}3\right)\\ & =& 0.15\times \frac{1}{3}\\ & =& 0.05\\ & & \end{array}$

Therefore, $\left(1\right)$to get

$\begin{array}{rcl}p\left(x=30\right)& =& 0.001+0.033+0.05\\ & =& 0.084\\ & & \end{array}$

The distribution of follows the Binomial distribution. Because if we take the speed-independent Bernoulli trials. The trials are divided into three categories. Then the all-independentBernoulli trials follow a binomial distribution.

For$p\left(x=50\right)$we get that

$\begin{array}{rcl}p\left(x=50\right)& =& p\left[x=50\left(\text{section}1\right),x=50\left(\sec tion2\right),x=50\left(\text{section}3\right)\right]\\ & =& p\left[x=50\left(\text{section}1\right)+x=50\left(\sec tion2\right)+x=50\left(\text{section}3\right)\right]\\ & & \end{array}$

Then follow the step-2 method to get

$\begin{array}{rcl}p\left(x=50\left(\sec tion1\right)\right)& =& p\left(x=50\left|\text{section}1\right.\right)\times p\left(\text{section}1\right)\\ & =& 0.25\times \frac{1}{3}\\ & =& 0.083\\ & & \end{array}$

Similarly, we define for section 2 and section3, we get

$\begin{array}{rcl}p\left(x=50\left(\text{section}2\right)\right)& =& p\left(x=50\left|\text{section}2\right.\right)\times p\left(\text{section}2\right)\\ & =& 0.35\times \frac{1}{3}\\ & =& 0.116\\ & & \end{array}$

$\begin{array}{rcl}p\left(x=50\left(\text{section}3\right)\right)& =& p\left(x=50\left|\text{section}3\right.\right)\times p\left(\text{section}3\right)\\ & =& 0.30\times \frac{1}{3}\\ & =& 0.10\\ & & \end{array}$

Therefore,

$\begin{array}{rcl}p\left(x=50\right)& =& 0.083+0.116+0.10\\ & =& 0.299\\ & & \end{array}$