251y0332 12/01/03
ECO251 QBA1
THIRD HOUR EXAM
Nov 24, 2003
Name: ____KEY_____________
Social Security Number: _____________________
Class Time (Circle) 10am 11am
Part I: (30+ points) Do all the following: All questions are 2 points each except as marked. Exam is normed
on 50 points including take-home. (Showing your work can give partial credit on some problems!)
1.
In its standardized form, the normal distribution
a) *has a mean of 0 and a standard deviation of 1.
b) has a mean of 1 and a variance of 0.
c) has an area equal to 0.5.
d) cannot be used to approximate discrete probability distributions.
2.
For some value of z , the probability that a standardized normal variable is below z is 0.2090.
The value of z is
a) *– 0.81.
b) – 0.31.
c) 0.31.
d) 1.96.
Solution: Your diagram of the normal distribution should show a Normal curve with a mean
at zero. 50% of the area is below zero, so a number with 20.9% below it must be below zero
too. If the number has 20.9% below it, it must have 79.1% above it. The probability that a
value of z is above zero is 50%, so the probability between the number we want and zero is
50% - 20.9% = 20.1%. According to the Normal table, P0  z  0.81  .2910 . This means
that P0.81  z  0  .2910 too. So our value is -0.81.
3.
If we know that the length of time it takes a college student to find a parking spot in the library
parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1
minute, find the probability that a randomly selected college student will find a parking spot in the
library parking lot in less than 3 minutes. Make a diagram!
a) 0.3551
b) *0.3085
c) 0.2674
d) 0.1915
Solution: x ~ N 3.5, 1
3  3.5 

Px  3  P  z 
 Pz  0.50 
1 

 Pz  0  P 0.50  z  0  .5  .1915  .3085
Your diagram for x should show a Normal curve with a mean at 3.5. 3 lies to the left of 3.5.
Shade the area below 3. You must take the area below 3.5 and subtract the area between 3 and
3.5. If you make this diagram, put 3.5 in the middle, not zero!
Your diagram for z should show a Normal curve with a mean at zero. -0.5 lies to the left of
zero. Shade the area below -0.5. You must take the area below 0 and subtract the area between 0.5 and zero. Most people seem to be happier if they make the z diagram instead of the x
diagram.
251y0332 11/18/03
4.
If we know that the length of time it takes a college student to find a parking spot in the library
parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1
minute, find the probability that a randomly selected college student will take between 2 and 4.5
minutes to find a parking spot in the library parking lot. Make a diagram!
a) 0.0919
b) 0.2255
c) 0.4938
d) *0.7745
Solution: x ~ N 3.5, 1
4.5  3.5 
 2  3.5
P2  x  4.5  P 
z
  P 1.50  z  1.00 
1
1


 P 1.50  z  0  P0  z  1.00   .4332  .3413  .7745
Your diagram for x should show a Normal curve with a mean at 3.5. 2 lies to the left of 3.5, and
4.5 to the right of 3.5. Shade the area between 2 and 4.5. You must take the area between 2
and 3.5 and add the area between 3.5 and 4.5.
Your diagram for z should show a Normal curve with a mean at zero. -1.5 lies to the left of
zero, and 1 lies to the right of zero. Shade the area between -1.5 and 1. You must take the area
between -1.5 and 0 and add the area between zero and 1.00.
Not only is this almost identical to Grass2. The probabilities for x are still the same
as on the last exam! Why are so many of you still trying to compute sample means and variances
when there is no sample here?
TABLE 5-4
The following table contains the joint probability distribution for
x and y .
Show your work neatly in questions 6-9.
x
y
0
1
2
0
.10
.20
.25
0
0
0
6
0
0
.20
.15
.05
.05
Total
.10
.20
.45
.15
.05
.05
3
4
5
5. Referring to Table 5-4, the probability that x is at least 1 is ________.
ANSWER: 0.90 = .20 + .45 + .15 + .05 + .05 = 1 - .10
Work for questions 6-9 follows.
x
y
0
6
Px 
xPx 
x 2 Px 
0
1
2
3
.10 .20 .25
0

0 .20
.15
 0
.10 .20 .45 .15
0 .20 .90 .45
0
4
0
.05
.05
.20
5
P y 
0
.55

.05 
.45
.05  1.00
.25  2.00
0.20 1.80 1.35 0.80 1.25
yP y  y 2 P y 
0
0
2.70
16 .20
2.70
16 .20
 5.40
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251y0332 11/18/03
 Px  1 ,   E x    xPx  2.00 , E x    x Px  5.40
 P y   1 ,   E y    yP y   2.70 and E y    y P y   16.20
2
To summarize
2
x
2
2
y
E xy  
 xyPxy
 030
 040
 050
.10 00  .20 10  .2520

 016  .20 26  .1536  .0546  .0556
 006
0
0
0
 0
 0 0

  7.8
 0  0  2.4  2.7  1.2  1.5
 xy  Covxy  Exy   x  y  7.8  2.002.70  2.400 ,
 
 
 x2  E x 2   x2  5.40  2.00 2  1.40 and  y2  E y 2   y2  16.20  2.702  8.91 . So that
 xy 
 xy
 x y

2.400
1.40 8.91

2.400 2 
1.40 8.91
.46176  .6795 .
(  x  1.40  1.1832,  y  2.9850 ) Note that 1   xy  1 always!
6. Referring to Table 5-4, the mean or expected value of x is ________.
ANSWER: 2.0
7. Referring to Table 5-4, the standard deviation of x is ________.
ANSWER: 1.18322 because  x2  E x2   x2  5.40  2.00 2  1.40 and 1.18322 is the square root.
 
8. Referring to Table 5-4, the covariance of x and y is ________. (4)
ANSWER: 2.40
9.
Referring to Table 5-4, the correlation between x and y is ________.
ANSWER: .6795
10. If x has a binomial distribution with n = 4 and p = 0.3, then Px  1 = ________ .
ANSWER: Px  1  1  Px  1  1  .6517  .3483
11. Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5
students is to be selected. The probability that at least 1 prefers brand C is ________.
ANSWER: Binomial Px  1  1  Px  0  1  .01024 = 0.9898
To find P0 Note that zero successes correspond to 5 failures when n  5 . The probability of failure
is .4 and P5  Px  5  Px  4  1  .98976  .01024 or use the Binomial formula.
0
5
5
Px  C xn p x q n x , so P0  C 05 .6 .4  .4  .01024
12. A professor receives, on average, 24.7 e-mails from students the day before the midterm exam. To
compute the probability of receiving at least 10 e-mails on such a day, he will use what type of
probability distribution?
a) binomial distribution.
b) *Poisson distribution.
c) hypergeometric distribution.
d) none of the above.
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251y0332 11/18/03
13. A multiple-choice test has 30 questions. There are 4 choices for each question. A student who
has not studied for the test decides to answer all questions randomly. What type of probability
distribution can be used to figure out his chance of getting at least 20 questions right?
a) *binomial distribution.
b) Poisson distribution.
c) hypergeometric distribution.
d) none of the above.
14. The number of 911 calls in Butte, Montana, has a Poisson distribution with a mean of 10 calls a
day. The probability of between six and nine 911 calls in a day is ____________.
Solution: P6  x  9  Px  9  Px  5  .45793  .06709  .3908
15. A debate team of 4 members for a high school will be chosen randomly from a potential group of
15 students. Two thirds of the 15 students have no prior competition experience while the others
have some degree of experience. What is the probability that exactly half of the members chosen
for the team have some prior competition experience?
Solution: Hypergeometric distribution. N  15, n  4, M  Np  13 15   5,
P2 
C 25 C 210
C 415
5! 10!
5  4 10  9
3! 2! 8! 2!
5 10  9

 2 1 2 1 
 .3297
15!
15 14 13 12 15  7 13
11! 4!
4  3  2 1
16. The on-line access computer service industry is growing at an extraordinary rate. Current
estimates suggest that only 20% of the home-based computers have access to on-line services.
This number is expected to grow quickly over the next 5 years. Suppose people with home-based
computers were randomly and independently sampled. Find the probability that the first person
that has access to on-line services is between the 5th and the tenth person sampled.
Solution: Geometric distribution
F c  Px  c  1  q c , because success at try c or earlier implies that there cannot have been
failures on the first c tries. q  1  p  1  .2  .8 . P5  x  10   Px  10   Px  4


 1  .810  1  .8 4  .8 4  .810  .40960 .10737  .3022
17. (Extra Credit)A catalog company that receives the majority of its orders by telephone conducted a
study to determine how long customers were willing to wait on hold before ordering a product.
The length of time was found to be a random variable best approximated by an exponential
distribution with a mean equal to 2.8 minutes. What proportion of callers is put on hold longer
than 2.8 minutes?
a) 0.60810
b) *0.367879
c) 0.50
d) 0.632121
Solution: The mean is 1  2.8 minutes. F x  1  ecx . Since 1  2.8 , c  1
 .3571 calls
c
c
2.8
 1 
per minute. If x  2.8, cx  
2.8  1 and Px  2.8  1  Px  2.8  1  F 2.8
 2.8 
 1  (1  e 1 )  e 1  .367879
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251y0332 11/18/03
ECO251 QBA1
THIRD EXAM
Nov 24, 2003
TAKE HOME SECTION
Name: _________________________
Social Security Number: _________________________
Throughout this exam show your work! Please indicate clearly what sections of the problem you are
answering and what formulas you are using.
Part II. Do all the Following (20 Points) Show your work!
1. The Lees give the following rates of return for two securities.
Time Period
Security A
Security B
1
.10
-.10
2
-.05
.05
3
.15
.00
4
.05
-.10
5
.00
.10
a) Take the second to last digit in your Social Security Number and use it to replace the zero in the -.10 in
security B returns. For example, Seymour Butz’s SS number is 123456789 and he will change -.10 to -.18.
Find the sample mean and standard deviation for Security B. For Security A, the sample mean is 0.05 and
the sample standard deviation is .07906. Use a coefficient of variation to show which seems riskier.(3)
b) Find the sample covariance and correlation between the returns on Security A and Security B. Comment
on the meaning of these results. (3)
c) If we invest 60% of our money in security A and 40% of our money in Security B, what is the mean and
standard deviation of the return? How does the riskiness of this portfolio compare with that of the
component stocks. (2)
d) Assume that the return of security B was 20% higher. So that you could write v  1.2 y , where y is the
original return on security B and v is the new return. Find (the sample mean and) standard deviation of the
new security B and the sample covariance and correlation with security A. Do not multiply the numbers
you were given for security B by 1.2 and recompute your answers – you will get no credit! You can get
your answers from the results you already have in a-c). (3)
Solution: This is a sample!
x
x2
Row
Note that there was no reason to do
xy
y
y2
1
0.10
2 -0.05
3
0.15
4
0.05
5
0.00
sum 0.25
0.0100
0.0025
0.0225
0.0025
0.0000
0.0375
-0.10
0.05
0.00
-0.10
0.10
-0.05
0.0100
0.0025
0.0000
0.0100
0.0100
0.0325
 x  0.25,  x  0.0375 ,  y  0.05,  y
 x  0.25  0.05 and y   y   0.05  0.01 .
Then x 
a) So n  5,
s x2 
s 2y
x
2
n
2
 nx
5
y

2
 ny 2
n 1
2
 0.0325 , and
 xy
 0.0175 .
5

0.0375  50.05 2
 0.00625
4
s x  0.00625  0.0790569 . Done for you!

0.0325  5 0.012
 0.00800
4
s y  0.00800  0.0894427 .
2
n 1
n
the x and x 2 columns.
-0.0100
-0.0025
0.0000
-0.0050
0.0000
-0.0175
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251y0332 11/18/03
.0790569
0.0894427
 1.581 and for y it is
 8.944 . Though it is
0.05
 0.01
not completely obvious how to compare negative and positive coefficients of variation, it should be clear
that stock y is riskier.
The coefficient of variation for x is
b) s xy 
rxy 
 xy  nxy   0.0175  50.05 0.01  0.00375
s xy
sx s y
n 1

5 1
 0.00375
and
  0.28125  .530000 . The correlation and covariance are negative,
0.00625 0.00800
indicating a tendency of y to fall when x rises. rxy2  .28125 is not very large on a zero to one scale,
indicating that the relationship is relatively weak. Note that 1  rxy  1 always!
c) From the outline, if R  P1 R1  P2 R2 and P1  P2  1 , then E R   P1 E R1   P2 E R2  and
VarR  P12VarR1   P22VarR2   2P1 P2CovR1 , R2  . Let R1  x, R2  y, P1  .6 and P2  .4 .
We know x  0.05 and y  0.01. s x2  0.00625, s 2y  0.00800 and s xy  0.00375. Since the
formulas can be used for sample data as well as expected values, R  .60.05   .4 0.01  0.03  0.004
 0.026 . .
s R2  Var R   .62 0.00625   .42 0.00800   2.6.4 0.00375   0.00225  .00128  0.00180
= 0.00173, so s R  0.04159 . The coefficient of variation is
.04159
 1.596 , which is comparable to that
0.026
of the first security.
d) The outline says that Cov(ax  b, cy  d )  acCov( x, y)
and if w  ax  b and v  cy  d ,  wv  signac xy , E v   cE y  and Varx  c 2Var y  .
a  1, b  0, c  1.2 and d  0.
Since y  0.01, v  1.2 y   0.012 .
Since s 2y  0.00800 , s v2  1.22 0.00800   0.01152 and s v  0.01152  0.107331 .
Since s xy  0.00375, s xv  11.2 0.00375  0.0045
Since rxy  .530000. , rxv  sign11.2rxy   .530000  .530000 .
Your results: In order to check results I wrote a Minitab routine. This program, cov5, was written for the
student version of Minitab, since our regular version was not working and it is not as clear about what it is
printing as newer versions. First it computes the x , y , x 2 , xy and y 2 . It prints out their sums in the
same order, each indicated by the word ‘SUM’ and then prints the actual columns of observations. It then
produces a small table with x and y on both sides that contains the variance of x, the covariance of x with y
and the variance of y. It then computes a correlation between x and y. To find these results go to
251y0332T and look for the number that you used to modify the ANOVA on the page heading. Note that
the y column there represents a consensus of what I meant in my instructions – I accepted your
interpretation, whatever it was, so that your solution may differ from the posted one without penalty.
6
251y0332 11/18/03
2. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that the
jorcillator only works as long as both components work.
The probability of the phillinx failing is given by a continuous uniform distribution with a range between
c  1.5 and d  3.5, so , if x represents the life of the phillinx, the probability of the phillinx failing in the
first year is P0  x  1 , the probability of it failing in the second year is P1  x  2 and the probability
of it lasting beyond the 2nd year is Px  2 .
The probability of the flubberall failing is given by a Normal distribution with a mean of 2 and a standard deviation of 0.5, so , if
y represents the life of the phillinx, the probability of the phillinx failing in the first year is P 0  y  1 , etc.


Failure of components is assumed to be independent, so that, if the probability of the phillinx failing in the first year is .1 and the
probability of the flubberall failing in the first year is .7, the probability of both components failing in the first year is (.1) (.7) =.07
(This is not necessarily the probability that the jorcillator will fail in the first year!)
a) What is the probability the Phillinx will fail in the first year? The second year? After the second year? (1.5)
b) What is the probability the Flubberall will fail in the first year? The second year? After the second year? (1.5)
c) What is the probability that the jorcillator will fail in the first year? (2)
d) What is the probability that the jorcillator will fail in the second year? (2)
e) What is the probability that the jorcillator will last beyond 2 years? (2)
Solution: a) The probability of the phillinx failing is given by a continuous uniform distribution with a
range between c  1.5 and d  3.5, so , if x represents the life of the phillinx, the probability of the phillinx
failing in the first year is P0  x  1 , the probability of it failing in the second year is P1  x  2 and the
probability of it lasting beyond the 2nd year is Px  2 . What is the probability the Phillinx will fail in the
first year? The second year? After the second year?
1
1
1

  0.5 . Make a diagram of a box with height 0.5 between 1.5
The density is
d  c 3. 5  1. 5 2
and 3.5. Name the following events A1 : Phillinx fails in first year; A2 : Phillinx fails in second year and
A3 : Phillinx fails after second year. Wake up! Some of you have been caught on this 3 times. Make a
diagram with a box of height 0.5 between 1.5 and 3.5 and nothing outside of that range.
(i)
Event A1 : Surprise! There is no area between zero and 1. P A1   P0  x  1  0
(ii)
Event A2 : Shade the part of the box below 2. It has area P A2  
(iii)
2  1.5
 .25
3 . 5  1 .5
Event A3 : Shade the part of the box between 2 and 3.5. It has area P A3   Px  2
P1  x  2 
3 .5  2
 .75
3 . 5  1 .5
Note that these probabilities add to one.
b) The probability of the flubberall failing is given by a Normal distribution with a mean of 2 and a
standard deviation of 0.5, so , if y represents the life of the phillinx, the probability of the phillinx failing in

the first year is P0  y  1 , etc. What is the probability the Flubberall will fail in the first year? The
second year? After the second year? y ~ N 2, 0.5 Name the following events B1 : Flubberall fails in first
year; B 2 : Flubberall fails in second year and B3 : Flubberall fails after second year.
(i)
(ii)
1 2 
0  2
z
 P 4.00  z  2.00 
Event B1 : PB1   P0  y  1  P 
0
.
5
0.5 

 P4.00  z  0  P2.00  z  0  .5000  .4772  .0228
2  2
1  2
z
 P 2.00  z  0  .4772
Event B 2 : PB2   P1  y  2  P 
0.5 
 0.5
2  2

 Pz  0   .5 I warned you!
Event B3 : PB3   Px  2  P  z 
0.5 

Note that these probabilities add to one.
(iii)
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251y0332 11/18/03
Now, note that whichever component fails first downs the jorcillator. The following nine joint events can
be enumerated.
Joint Event
Probability
Jorcillator fails
In Year 1
0.0228   0
A1  B1
In Year 1
0.4772   0
A1  B2
A1  B3
A2  B1
A2  B2
A2  B3
A3  B1
A3  B2
A3  B3
0.5  0
.25.0228   .0057
.25 .4772   .1193
.25 .5  .1250
.75 .0228   .0171
.75 .4772   .3579
.75 .5  .3750
In Year 1
In Year 1
In Year 2
In Year 2
In Year 1
In Year 2
After Year 2
If we add together the probabilities of possible joint events, we find the following.
c) Probability of failing in first year = .0057 + .0171 = .0228
e) Probability of it lasting beyond second year = .3750
d) Probability if it failing in second year = 1 - .0228 - .3750 = .1193 + .1250 + .3579 = .6022
Most of you were not thinking! Thinking is what probability is all about. If you told me that the
probability of the jorcillator failing in the first year was 0.0228   0 , you are saying that it cannot fail in
the first year. But you know that there is a probability of .0228 of the Phillinx failing in the first year, and
the rules say it needs the Phillinx. This is an easier version of PROBLEM H4, which I told you that you
would see again. One way of looking at it is (i) that if both components are needed the probability of failing
in first year = P A1  B1   P A1   PB1   P A1  B1   0  .0228  0  .0228 , the probability of it lasting
beyond second year = P A3  B3   P A3  PB3   .75 .5 = .3750 because to last beyond the second
yeast both components must last beyond the second year. Then the probability if it failing in second year =
1  P A1  B1   P A3  B3   1  .228  .3750 because it has to fail sometime so the probabilities have to
add to one. But (Are you ready to try a little more thinking?) (ii) if the jorcillator will work with only one of
the two components the probability of failing in first year = P A1  B1  , the probability of it lasting beyond
second year = P A3  B3   P A3   PB3   P A3  B3  because the jorcillator will last beyond the
second year if either or both components last beyond the second year. Then the probability if it failing in
second year = 1  P A1  B1   P A3  B3  because it has to fail sometime so the probabilities have to add
to one.
3. (Extra Credit) A new phillinx has an average life of 1 year and its failure time follows the exponential
distribution.
a) What is the probability that the phillinx will fail in the first month? (2)
b) What is the probability the Phillinx will fail in the first year? The second year? After the second
year? (1.5)
c) If we assume that the Flubberall has the same distribution as in problem 2, what is the
probability of the jorcillator failing in the first year? The second year? After the second year? (3)
Solution: The mean is 1  1 year. F x  1  ecx . Since 1  1 , c  1  1 .
c
c
1
1
1
1
 
a) If x  , cx  1  
and the probability that the phillinx will fail in the first month is
12
 12  12
1
1
1

F    P x    1  e  1 2  1  .9200  .0800
12 
 12 

8
251y0332 12/01/03
b) Name the following events B1 : Flubberall fails in first year; B 2 : Flubberall fails in second year and
B3 : Flubberall fails after second year.
(i)
If x  1, cx  11  1 , and the probability that the Flubberall fails in the first year is
(ii)
PB1   P y  1  F 1  1  e 1  1  .3679  .6321
If x  2, cx  12  2 , and the probability that the Flubberall fails in the second year is
PB2   P1  y  2  F 2  F 1


 1  e 2  1  e 1  e 1  e 2  .3679  .1353  .2326
(iii)
The probability that the Flubberall fails after the second year ( not just in the third year
but in the fourth, fifth etc.)is
PB3   Px  2  1  Px  2  1  F 2  1  1  e 2  e 2  .1353
c) Following the results of the last problem, for the first year the probability is .6321, for beyond the second
year the probability is .75 .1353   .1015 and for the second year the probability is 1 - .6321
- .1353 = .2326.


9