91影视

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

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

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

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

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

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

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

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

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

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

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

Therefore,

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