C\" www.webassign.net/web/Student/Assignment-Responses/last?dep-17845191 The joi
ID: 3061018 • Letter: C
Question
C" www.webassign.net/web/Student/Assignment-Responses/last?dep-17845191 The joit probablity distribution of the number of cars and the number Y of buses per signal cycle at a proposed left-turn lane is displayed in the accompanylng jeint probabity table 10.050 00300.020 3 0.3500090 060 0.300 0060 040 0050 0.030020 (a) what is the probbity thet there is exactly one car and exactly one bus during a cle b) What is the probability that there is at mout one car and at most one bus duringcycle (c) at-the P Plexactly one car) Plexactly one bus)" y that there exactly ene danng a evoe, txactly one bus, ) Suppose the lh-turn lane is to have a capacity of five cars and one bus is equlvalent so theee cars. What is the probabity of an overflow during a cycle? (e) Are X and Y independent rv's? Explein Need Help?Explanation / Answer
a) P(X=1 and Y=1) = 0.030
b) P(X<=1 and Y<=1) = P(X=1,Y=1) + P(X=0,Y=1) + P(X=1,Y=0) + P(X=0,Y=0)
= 0.030 + 0.015 + 0.050 + 0.025 = 0.12
c) P(Exactly one car) = P(X=1) = 0.050 + 0.030 + 0.020 = 0.10
P(Exactly one Bus) = P(Y=1) = 0.015 + 0.030 + 0.075 + 0.090 +0.060 +0.030 =0.3
d) Suppose the left trun lane is to have acapcaity of five cars and one bus is equivelent to three cars. What is the probability of an overflow during a cycle
Overflow = more than 5 cars or the equivalent in the left turn lane.
From the problem statement one bus is equivalent to three cars. The equation is
in terms of cars.
P(X + 3Y > 5) = 1 – P(X +3Y < 5)
So find all p(x,y) that make P(X + 3Y < 5) true.
p(0,0) = p(0 + 3(0) < 5) TRUE p(0,0) = .025
p(1,0) = p(1 + 3(0) < 5) TRUE p(1,0) = .05
p(2,0) = p(2 + 3(0) < 5) TRUE p(2,0) = .125
p(3,0) = p(3 + 3(0) < 5) TRUE p(3,0) = .05
p(4,0) = p(4 + 3(0) < 5) TRUE p(4,0) = .1
p(5,0) = p(5 + 3(0) < 5) TRUE p(5,0) = .05
p(0,1) = p(0 + 3(1) < 5) TRUE p(0,1) = .015
p(0,2) = p(0 + 3(2) < 5) FALSE
p(1,1) = p(1 + 3(1) < 5) TRUE p(1,1) = .03
p(1,2) = p(1 + 3(2) < 5) TRUE p(1,2) = .075
p(1,3) = p(1 + 3(3) < 5) FALSE
p(1,2) and p(2,2) and p(3,2) and p(4,2) and p(5,2) and p(4,1) and p(5,1) are also FALSE
Sum all the true statements and use the equation above:
1 – P(X + 3Y £ 5) = 1 – [p(0,0) + p(1,0) + p(2,0) + p(3,0) + p(4,0) + p(5,0) +
p(0,1) + p(1,1) + p(1,2) = 1 - .620 =
Therefore, P(X + 3Y > 5) = 1 – P(X + 3Y < 5) = .380
e)
Using Table and the definition of independence,
px(0) = p(0,0) + p(0,1) + p(0,2) = .05
py(0) = p(0,0) + p(1,0) + p(2,0) + p(3,0) + p(4,0) + p(5,0) = .5
p(0,0) = .025
Since, .025 = (.05)(.5), X and Y are independent. But, we must remember the definition
of independence states that all (x,y) need to meet this proposition.
It is now easily verified that for every (x,y), p(x,y) = px(x) § py(y), so X and Y are
independent.
Correct Answer: Option (A)