d. Find the conditional UISuioumonl he 6.5 Variable speed limit control for free

Question

d. Find the conditional UISuioumonl he 6.5 Variable speed limit control for freeways. Refer to ti Canadian Journal of Civil Engineering (Jan. 2013) inves- tigation of the use of variable speed limits to control free- way traffic congestion, Exercise 4.9 (p. 140). Recall that the study site was an urban freeway divided into three sec- tions with variable speed limits posted in each section. The probability distribution of the optimal speed limit for each of the three sections was determined. One possible set of distributions is as follows (probabilities in parentheses). Section 1: 30 mph (.06), 40 mph (.24), 50 mph (24), 60 mph (.46); Section 2: 30 mph (.10), 40 mph (.24), 50 mph (.36), 60 mph (.30); Section 3: 30 mph (15), 40 mph (.18), 50 mph (.30), 60 mph (.37). Consider a randomly selected vehicle traveling through the study site at a randomly se- lected time. For this vehicle, let X represent the section and Y represent the speed limit at the time of selection. a. Which of the following probability distributions is rep resented by the given probabilities, p(x.y),pi(x), P2(y). PI(x|), or p2(y |x)? Explain.

Explanation / Answer

Given data is as

X : - Section and Y :- Speed limit

And probability matrrix is as

a) These probability table represent p2(y|x). Because , here we take probability for each individual x is 1 and then compute probability for every y for individual x variable.

We see that, the sum of each x (i.e. section 1, section 2, section 3 ) is 1. So it means that each cell represent the probability as p2(y|x).

b) If sections are as equal lenght then the p1(x) = 1 / 3 = 0.3333

Let if sections are as equal lenght then the probability of each cell is

P(x,y) = 1 / (total no of cells) = 1 / 12 = 0.0833

Then  p1(x) = sum of 1st raw probabilities = sum of probability of speed limit of section 1

= 0.0833 + 0.0833 + 0.0833 + 0.0833  

= 0.3333

So we satisfy the result

c) We use these 0.3333 probability to find Bivariate distribution as multipy each cell probability of original table(table 1) bye 0.3333

C)

E(x) = 1*0.333 + 2*0.333 + 3*0.333 = 1.9998 = 2

That is E(X) = Section 2

And E(Y) = 0.1*30 + 0.22*40 + 0.3*50 + 0.37*60 = 49 mph

Section Speed limit 30mph 40mph 50mph 60mph total Section 1 0.06 0.24 0.24 0.46 1 Section 2 0.10 0.24 0.36 0.30 1 Section 3 0.15 0.18 0.30 0.37 1

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