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# Problem 2, Random Variable (15 points) The joint probability distribution of variables X and Y is shown in the table below. Amber and Bianca are real estate agents. Let X denote the number of houses...

Problem 2, Random Variable (15 points)
The joint probability distribution of variables X and Y is shown in the table below. Amber and Bianca are real estate agents. Let X denote the number of houses that Amber will sell in a month, and let Y denote the number of houses Bianca will sell in a month.

X
Y
1
2
3
1
0.30
0.18
0.12
2
0.15
0.09
0.06
3
0.05
0.03
0.02
a. Determine the marginal probability distribution of X. (3 points)
. Determine the marginal probability distribution of Y. (3 points)
c. Verify that E(X + Y) = E(X) + E(Y). (4 points)
d. Compute the covariance COV(X, Y). (5 points)
e.

Problem 3 Distributions (10 points, 5 points each)
A recent survey in Michigan revealed that 60% of the vehicles traveling on highways, where speed limits are posted at 70 miles per hour, were exceeding the limit. Suppose you randomly record the speeds of ten vehicles traveling on US 131 where the speed limit is 70 miles per hour. Let X denote the number of vehicles that were exceeding the limit.
a.    Identify the distribution of X
b.     Find the standard deviation of number of vehicles that are traveling on Michigan highways and exceeding the speed limit.
Problem 4, Distributions (15 points, 5 points each).
1. The average number of drop-in blood donors who visit a Red Cross clinic in a given week is 42.7 donors per week
a) Describe why a Poisson distribution might be appropriate to model the random number of drop-in blood donors in a given week.
) Assuming Poisson, what is the probability that 40 donors (exactly) visit the clinic?
c) Assuming Poisson, what is the probability that between 40 and 45 donors (inclusive) visit the clinic?

Problem 5, Normal Distributions (20 points, 5 points each).
If Z is a standard normal random variable, find the value z for which:
a.
the area (probability) between 0 and z is XXXXXXXXXXand z is positive
b.
the area (probability greater than) to the right of z is 0.7123
c.
the area (probability less than) to the left of z is 0.1736
d.
the area (probability) between z and z is 0.6630
Problem 6, Normal Distributions (10 points, 5 points each).
Scores of high school students on a national mathematics exam were normally distributed with a mean of 86 and a standard deviation of 4. (Total possible points = 100.)
(a) What is the probability that a randomly selected student will have a score between 80 and 90?
(b) What is the cutoff for the bottom 5% of the score?
Answered Same Day Aug 13, 2021

## Solution

Shreya answered on Aug 14 2021
Problem 2, Random Variable (15 points)
The joint probability distribution of variables X and Y is shown in the table below. Amber and Bianca are real estate agents. Let X denote the number of houses that Amber will sell in a month, and let Y denote the number of houses Bianca will sell in a month.

X
Y
1
2
3
1
0.30
0.18
0.12
2
0.15
0.09
0.06
3
0.05
0.03
0.02
a. Determine the marginal probability distribution of X. (3 points)
Answer : The marginal probability distribution of X is given by P(X=x) = sum of all probabilities of X=x
P(X=1) = 0.3+ 0.15+0.05 = 0.5
P(X=2) = 0.18+0.09+0.03 = 0.3
P(X=3) = 0.12+0.06+0.02 = 0.2
. Determine the marginal probability distribution of Y. (3 points)
Answer : The marginal probability distribution of Y is given by P(Y=y) = sum of all probabilities for (Y=y)
P(Y=1) = 0.3+0.18+0.12= 0.6
P(Y=2) = 0.15+0.09+0.06 = 0.3
P(Y=3) = 0.05+0.03+0.02 = 0.1
c. Verify that E(X + Y) = E(X) + E(Y). (4 points)
L.H.S: E(X+Y) = Σ (x+y) * P(X=x,Y=y)
E(X+Y) = (1+1)*0.3 + (1+2)*0.15 + (1+3)*0.05 + (2+1)*0.18 + (2+2)*0.09 + (2+3)*0.03 + (3+1)*0.12 + (3+2)*0.06 + (3+3)*0.02
E(X+Y) = 3.2
E(X) = Σ x* P(X=x)
E(X) = 1*0.5 + 2*0.3 + 3*0.2
E(X) = 0.5 + 0.6 + 0.6 = 1.7
E(Y) = Σ y* P(Y=y)
E(Y) = 1*0.6 + 2*0.3 + 3*0.1
E(Y) = 0.6 + 0.6 + 0.3 = 1.5
E(X) + E(Y) = 1.7 + 1.5 =3.2
This proves that E(X+Y) = E(X) + E(Y) = 3.2
d. Compute the covarianceCOV(X, Y). (5 points)
ANSWER: E(XY) = Σ (xy) * P(X=x,Y=y)
E(XY) = (1*1)*0.3 + (1*2)*0.15 + (1*3)*0.05 + (2*1)*0.18 + (2*2)*0.09 + (2*3)*0.03 + (3*1)*0.12 + (3*2)*0.06 + (3*3)*0.02
E(XY) = 2.55
COV(X, Y) = E(XY) – ( E(X)* E(Y))
COV(X, Y) = 2.55 – (1.7* 1.5) =2.55 – 2.55 = 0
COV(X, Y) = 0
Problem 3 Distributions (10 points, 5 points each)
A recent survey in Michigan revealed that 60% of the vehicles traveling on highways, where speed limits are posted at...
SOLUTION.PDF