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ECO251 QBA1
THIRD EXAM
Due June 27, 2008
Name: _____________________
Exam is normed on 50 points, but any points above 50 wrap around.
Part I. (15 points – 1.5 each) Do all the following (2 points each unless noted otherwise). Make
Diagrams! Show your work! In particular you must briefly explain how you got the answer to the values
of z at the bottom of this page.
A. z ~ N (0,1)
1. P 3.70  z  0.73 
2. F 1.23
3. P 1.23  z  2.50 
 
4. The second percentile z .98
5. A symmetrical interval about the mean with 76% probability.
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B. x ~ N (1.2, 5)
1. P3.70  x  0.73
2. Px  1.23 
3. P 123
.  x  2.50
 
4. The second percentile x. 98 .
5. A symmetrical interval about the mean with 76% probability.
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Part II. Do all the Following (35+ Points) Show your work! Neatness counts! Answers of ‘zero’ or
‘one’ especially are unacceptable without an explanation. Do not use one distribution to approximate
another without justifying the replacement! Points on an individual question are in parentheses and a
running total is in brackets.
1. (15+ points)
9 4
3
0 .10
2 .05
0
y
4
0
0
5
0 .25
x
 3 1
0 .05
0
0
.10
0
.05 .05
0
.10
0
.05
.20
Use the following events: X 9 , which is the event x  9 , X 4 x  4 , X 3 x  3 , X 1 x  1 ,
X 0 x  0 , Y 3 y  3 , Y 2 y  2 , Y 4 y  4 and Y 5 y  5 .
Find the following.
a) Are x and y independent? Why? (1)


b) Identify the probabilities below by symbols like P X 2Y 2 , P X 2  Y 2 or P X 2  Y 2 and give their
values. (4)
i) The conditional probability of Y4 given X3.
ii) The joint probability of X3 and Y4
iii) The probability of X3
iv) The probability of Y4 or (not X3).
c) Demonstrate your understanding of Bayes’ rule by finding the conditional probability of X3 given Y4
using, with another probability, two probabilities from b). (2)
d) Find P y  4 x  0 . (2) [9]


e) Find the variance of y . (1)
f) Find the covariance of x and y . (2)
g) Find the correlation between x and y . (1)
h) Find the variance of x  y . (1)
g) Find the variance of 3x  y (2)
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2. (10+ points) The education department of the State of Confusion is worried about the size of private
college endowments in a period of rising costs. It takes a sample of 9 schools and gets the following
numbers (in millions of dollars). Assume that this is a sample taken from the Normal distribution.
60
47
235
900
27
3909
1001
20
833
a) Find the sample standard deviation s of the endowments. (2)
b) Find a 99% confidence interval for the population mean using the mean and the sample
standard deviation that you found in a) (2)
c) Repeat b) under the assumption that your sample of 9 was taken in a state in which there are
only 20 colleges. (2)
d) Assume that, instead of the sample standard deviation you found in a) that the population
standard deviation is known to be 1000. Forget about there only being 20 colleges in the
population. Find a 95% confidence interval for the population mean (1)
e) What you are doing is a simple type of hypothesis testing. On the base of the interval that you
found in d) can you say that the mean endowment is significantly different from 1500? Why? (1)
f) Let’s continue with hypothesis testing. There are actually 3 ways to do a hypothesis test. One is
a confidence interval, which you have already done. You have also found a sample mean for the
endowment that is quite a bit below 1500. Let’s say that the education department believes that the
population of college endowments is Normally distributed with a mean of 1500 and a population
standard deviation of 1000. The sample mean will also have a Normal distribution with a mean of
1500. What will be the standard error (the standard deviation of the sample mean)? (1)
g) The critical value method (for the sample mean) is to use the information in f) to construct a
symmetrical interval around 1500 with a probability of 95%. If your sample mean does not fall in
that interval we can conclude that there is less than a 5% chance that the hypothesis of a mean of
1500 is true. Construct the interval and come to a conclusion. (2)
h) The p-value is the probability of getting the value of the sample mean that you actually got or
something more extreme. If the p-value is below 5% we can reject the hypothesis that the mean is
1500. Use the values of the sample mean that you found in a) and the distribution that you found
in f). Find the probability that the sample mean is as low as or lower   than the value that you
actually got. To make this probability into a p-value, double it. What is your conclusion? (Because
you have to compute a z-score to get this, I call this a test-ratio method. (2)
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3. (20+ points) Identify the distribution that you are using in each problem. Make it very clear what values
of n, p, m or other parameters you are using. If I have to guess what part of what table you used, you will be
penalized. Look at the solved problems for Section L, the solution to Grass3 and ‘Great Distributions’
(especially the hints on the 3rd page) before you start.
a) On a Wednesday morning trucks arrive at a weighing station at an average rate of 6 per hour. (No! these
problems do not have the same answer.)
i) What is the probability that exactly eight arrive in a one hour period? (1)
ii) What is the probability that more than 8 arrive in a one hour period? (1)
iii) What is the probability that more than 16 arrive in a two-hour period? (1)
iv) What is the probability of more than 64 in an 8-hour day? (1)
v) How many trucks must I be able to weigh in an 8-hour day to be 99% sure that I don’t have to
turn any away? (Hint: find the 99th percentile of the distribution in iv) (1) [5]
b) Let us assume that we believe that 10% or fewer of our population are unemployed. To test this
proposition we take a sample and find that x 0 are unemployed. We then compute the probability of x 0 or
more being unemployed in a sample of that size. If that probability is below 5%, we can doubt that the
actual (population) proportion is 10% of fewer.
i) So, if we take a sample of 10 and the probability of an individual being unemployed is 10% and
we find that three are unemployed, what is the probability of 3 or more being unemployed? Would
we doubt the 10% proportion? (2)
ii) So, what would we conclude if 5 out of a sample of 25 were unemployed? (I want calculations
– not opinions.) (1)
iii) Out of 100, what is the lowest number that could be unemployed that would make us doubt the
10% population proportion? Why? (2)
iv) Out of a sample of 250, you find that 34 are unemployed. To figure out the probability of 34 or
more being unemployed can you use the Poisson distribution? Why? What is the probability? (2)
v) Out of a sample of 250, you find that 34 are unemployed. To figure out the probability of 34 or
more being unemployed can you use the Normal distribution? Why? What is the probability? On
the basis of iv) and v) can you doubt the hypothesis? (2) [14]
c) Assume that 2% of the cattle in the US have mad cow disease.
i) A slaughterer takes a random sample of 10 cattle from the entire US herd, tests it and concludes
that, since none of the cattle in the sample have mad cow disease, there is no problem. What, in
fact, is the chance that, if 2% of the population is infected, at least one out of the sample is
infected? (2)
ii) Assume that you are inspecting a large number of cows, what is the chance that the first
infected cow will occur in the first 100 cows you inspect? In the first 1000? (2)
iii) Now assume that the slaughterer buys 100 cattle, that 2% of the 100 are infected and that the
slaughterer takes a sample of 10. Is the chance that at least one of the cattle is infected higher or
lower than in i)? Show your work!!! (2) [20]
iv) (Extra credit) If, in fact 2% of the US herd is infected, how large would a sample have to be for
there to be a 95% chance that at least one cow in the sample is infected? (2)
d) The amount of time it takes to inspect a vehicle has a mean of 23 minutes with a standard deviation of 4
minutes. An inspector is scheduled to inspect 100 vehicles in 40 hours (2400 minutes). What is the chance
that the inspector will finish the inspections in the allotted time? (3)
e) (Extra credit) According to Ken Black, a manufacturer has determined that in a 20-minute interval 1.38
defects occur in a manufacturing process. This means that the average time between defective items is
20
 14 .49 minutes. An inspector observes the line for 15 minutes. What is the chance that the inspector
1.38
will observe a defective item? (2)
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4. (10+ points)
a) A bank knows that 90% of its borrowers have been employed at their current jobs for 3 years or more.
5% of the borrowers that have been at their current jobs for 3 years or more are behind in their mortgage
payments. However, 15% of the borrowers who have not been at their jobs for 3 years or more are behind
in their payments. Let B be the event that a borrower is behind in mortgage payments. Let E be the event
that the borrower has been at the job for 3 years or more. What proportion of the people who are behind in
mortgage payments have been at their jobs for 3 years or more? You should start by identifying the
probability that I have requested in terms of B and E. Then identify 90% and 5% the same way. (4)
b) As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that our new
jorcillator only works as long as both components work (so that it fails in the first period if either
component fails).
The probability of the phillinx failing is given by a continuous uniform distribution with a lower bound c 
of 11 and an upper bound d  of 31. For example, if the life of the phillinx is represented by x1 , the
chance of the phillinx failing in the first ten years (Period 1) is P0  x1  10  and the probability of it
failing in the second ten years (Period 2) is (NOTE) P10  x1  20  . The flubberall x 2  also follows the
continuous uniform distribution with c  1 and d  41 .
In order to maintain my sanity, use the following events. Period 1 is the first ten years, period 2 is the
second ten years and period three is happily ever after.
Failure of the phillinx in period 1, 2, 3 are events A1, A2 and A3
Failure of the flubberall in period 1, 2, 3 are events B1, B 2 and B 3
Failure of the jorcillator in period 1, 2, 3 are events C1, C 2 and C 3 .
i) What is the probability that the phillinx will fail in period 1? Period 2? Period 3? (1.5)
ii) What are the mean and the standard deviation of the failure time for the phillinx? (1)
iii) What is the probability that the Flubberall will fail in period 1? Period 2? Period 3? (1.5)
iv) What is the probability that the jorcillator will fail in the first period? (1)
v) What is the probability that the jorcillator will fail in the second period 2? (1)
vi) What is the probability that the jorcillator will fail in the third period? (1)
If you haven’t figured it out already, one of the easiest ways to do this is to make a joint probability
table. Put the A events across the top. Put the B events down the side. Figure out what the
probability of the joint events must be if they are independent. Now make a similar table. This time,
instead of probabilities, fill in the period in which the jorcillator fails.