251y0321 4/18/03
ECO251 QBA1
THIRD HOUR EXAM
April 21, 2003
Name: ______KEY______________
Social Security Number: _____________________
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. Engineers designed a monorail so that the amount of time that people have to wait at airport concourse B
has a continuous uniform distribution ranging from 0 to 5 minutes. The mean waiting time for some one
who just arrived at concourse B is:
a)
*2.5 minutes
b)
3 minutes
c)
4 minutes
d)
5 minutes.
2. A recent study on teen-age pregnancy indicates that 5% of the female population will get pregnant
during their teenage years and that teenage pregnancies occur independently of one another. Suppose that
ten female teen-agers are selected at random. Find the approximate probability that at least two get
pregnant some time in their teenage years.
a)
.914
b)
.988
c)
.012
d)
*.086
Explanation: This is a binomial problem with n  10 and p  .05. Px  2  1  Px  1  1  .91386
 .08614
3. Assume that x is a random variable with a mean of 200 and a standard deviation of 2. Assume that
u  2 x  3 . The standard deviation of u is:
a)
-1
b)
*4
c)
7
d)
-4
Explanation: If u  ax  b, Varu   a 2Varx . So here a  2 and Varu    22 22  16
So the standard deviation is
16  4.
4. Let x and y represent the quantity of two goods that you sell and assume that the correlation between
their sales is  xy  .6 . Assume that the first good sells for $0.90, so that the amount you make on it is
u  .90 x and the second good sells for $0.80, so that the amount that you make on it is v  .80 y . What is
the correlation between u and v ? .6
Explanation: If u  ax  b and v  cy  d , Corr u, v   SignacCorrx, y  Since a  .90 and c  .80 ,
( b  d  0 ) Signac   and Corr u, v   Signac.6  .6.
251y0321 4/18/03
5. What type of probability distribution will most likely be used to analyze the number of chocolate chip
parts per cookie in the following problem?
The quality control manager of Marilyn’s Cookies is inspecting a batch of chocolate chip cookies.
When the production process is in control, the average number of chocolate chip parts per cookie
is 6.0. The manager is interested in analyzing the probability that any particular cookie being
inspected has fewer than 5.0 chip parts.
a) binomial distribution.
b) *Poisson distribution.
c) normal distribution.
d) Hypergeometric distribution.
Explanation: Probability of the number of successes when you know the mean.
6. The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6
outages per year. The probability that there will be between 1 and 3 power outages in a year is _.14872.
Explanation: P1  x  3  Px  3  Px  0  .15120  .00248  .14872 . I went through this several
times in class and it’s in the outline. Is the problem simply that you didn’t believe me?
7. A debate team of 4 members for a high school will be chosen randomly from a potential group of 15
students. Ten of the 15 students have no prior competition experience while the others have some degree
of experience. What is the probability that at most 1 of the members chosen for the team have some prior
competition experience?
Solution: Hypergeometric N  15, M  5 have experience and n  4.
Px  1  P0  P1 Px  

P0 
C 05 C 410
C 415
P1 
C15 C 310
C 415
C xM C nNxM
C nN
Crn 
n!
n  r !r!

10  9  8  7
4  3  21
10  9  8  7



 .15385
15 14 13 12 15 14 13 12
 15! 


4  3  21
 11! 4! 
 5!  10! 
10  9  8



5
 4!1!  7! 3! 
3  2 1  5 10  9  8  4  .43956


15

14
13 12 15 14 13 12
 15! 


4  3  2 1
 11! 4! 
1 10! 
 6! 4! 
Px  1  P0  P1  .15385  43956  .59341
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251y0321 4/18/03
TABLE 13-2
A candy bar manufacturer is interested in trying to estimate how
sales are influenced by the price of their product. To do this, the
company randomly chooses 6 small cities and offers the candy bar at
different prices.(Note that there are no probabilities here – please
don’t invent them!)
City
Price($) Sales
Row price sales
River Falls 1.30
100
Hudson
1.60
90
x
y
xy
x2
y2
Ellsworth
1.80
90
1
1.3
100
1.69
10000
130.0
2
1.6
90
2.56
8100
144.0
Prescott
2.00
40
3
1.8
90
3.24
8100
162.0
Rock Elm
2.40
38
4
2.0
40
4.00
1600
80.0
Stillwater
2.90
32
5
2.4
38
5.76
1444
91.2
Note: For price
 x  12,  x
6
2.9
12.0
 25 .66 , x  2 and
2
32
390
8.41
25.66
1024
30268
92.8
700.0
s x2  0.3320.
8. Compute a sample variance for sales (3) Answer: 983.6 – see below.
9. Compute a sample correlation between sales and price. (3) Answer: -.8854 – see below.
Computations:
x 2  25 .66 , x  2 and s x2  0.3320.
x  12 ,
Given


390
 65
So  y  390 ,  y  30268 ,  xy  700 and y 
6
 x  nx  25.66  62  25.66  24  0.3320 . s  0.3320  .5762 
s 
2
2
2
x
you.
s 2y 
s xy 
rxy 
2
2
n 1
y
2
5
 ny 2
n 1

x
5
30268  665 2 30268  25350

 983 .6.
5
5
s
y
This was done for

 983.60  31`.3624
 xy  nxy  700  6265  700  780  16.00
n 1
s xy
sx s y

5
 16 .00
0.3220 983 .6
5

 16 2

0.3320 983 .6
.78394  .8854
10. Suppose z has a standard normal distribution with a mean of 0 and standard deviation of 1. (Make a
diagram) The probability that z is less than -2.20 is .0139 .
Explanation: Make a diagram of a Normal curve with zero in the middle. Shade the area below -2.20.
Pz  2.2  Pz  0  P2.20  z  0  .5  .4861  .0139 .
11. Suppose z has a standard normal distribution with a mean of 0 and standard deviation of 1. (Make a
diagram) The probability that z is less than 2.20 is .9861.
Explanation: Make a diagram of a Normal curve with zero in the middle. Shade the area below 2.20 .
This will be both the area below zero and the area between zero and 2.20.
Pz  2.2  Pz  0  P2.20  z  0  .5  .4861  .9861 .
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251y0321 4/18/03
12. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams
and  = 25 grams. What is the probability that a randomly selected vitamin will contain between 82 and
100 grams of pyridoxine? (Make a diagram) _________
Explanation: Make a diagram of a Normal curve with 110 or zero in the middle. If you did the diagram
with 110 in the middle, shade the area between 82 and 100. x ~ N 110 ,25  .
100  110 
 82  110
P82  x  100   P 
z
  P 1.12  z  0.40 
25
 25

 P1.12  z  0  P0.40  z  0  .3686  .1554  .2132
If you did the diagram with zero in the middle, shade the area between -1.12 and -0.40.
13. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with  = 110 grams
and  = 25 grams. What is the probability that a randomly selected vitamin will contain at least 100 grams
of pyridoxine?
Explanation: Make a diagram of a Normal curve with 110 or zero in the middle. If you did the diagram
100  110 

with 110 in the middle, shade the area above 100. Px  100   P  z 
  Pz  0.40 
25


 P1.04  z  0  Pz  0  .1554  .5  .6554
If you did the diagram with zero in the middle, shade the entire area above -0.40.
14. Using the probability that you found in the problem 13, what is the probability that exactly two out of
three randomly selected vitamins will contain at least 100 grams of pyridoxine?
This is a Binomial problem. Px   C xn p x q n x , so if n  3 and p  .6554,
P2  C 23 p 2 q 1 
3!
.6554 2 1  .6554   3.42955 .3446   .4441
1!2!
15. Using the probability that you found in the problem 13, what is the probability that if you check
vitamins in sequence, the third one you check will be the first to contain at least 100 grams of pyridoxine?
Solution: This is a Hypergeometric problem. Px   q x 1 p So P3  .3446 2 .6554   .0778
16. Scientists in the Amazon are trying to find a cure for a deadly disease that is attacking the rain forests
there. One of the variables that the scientists have been measuring involves the diameter of the trunk of the
trees that have been affected by the disease. Scientists have calculated that the average diameter of the
diseased trees is 42 centimeters. They also know that approximately 95% of the diameters fall between 32
and 52 centimeters and almost all of the diseased trees have diameters between 27 and 57 centimeters.
When modeling the diameters of diseased trees, which distribution should the scientists use?
a) Binomial distribution
b) Poisson distribution
c) *Normal distribution
d) Exponential distribution
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251y0321 4/18/03
ECO251 QBA1
THIRD EXAM
April 21, 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. (15 points) Do problem 5.11 in the text. As usual there are sharp differences between the text and the
Instructor’s Solutions Manual as to what is in the problem.. The table that should be in the book should
read:
RE TURNS
Probability
Stock X
Stock Y
0.1
-100
50
0.3
0
100
0.3
80
-20
0.3
150
100
Before you start, fill in the table below with the probabilities that you see above and a lot of zeros (1). Use
it to compute the expected values, variances and covariance. You may want to re-read section 5-2 before
you do the rest of the problem. In addition to what the text requests, compute m.) the correlation between
x and y . The neatness of your presentation and the way you explain the recommendation count.
x
150
80
0  100
100
y
50
 20
Solution:
x
y
100
80
0
0
.3
 100
0
50
0
0
0
.1
.1
5
250
 20
0
.3
0
0
.3
6
120
xPx 
.3
.3
.3
.1
1.0
59
6370
45
 24
0
 10
 59
 0  1000
 9670
Px 
x 2 Px  6750
 1920
 Px  1 ,   E x    xPx   59 E x    x
 E  y    yP y   59 and E y    y P y   6370
To summarize
y
P y  yP y  y 2 P y 
.6
60
6000
150
.3
2
x
2
2
Px   9670
 P y   1 ,
2
5
251y0321 4/18/03
.30100 
0 100 100 
 .3150 100   080 100 

E xy  
xyPxy     0150 50   080 50 
 0050   .1 100 50 
 0150  20  .380  20   00 20   0 100  20 
0 0
 0
4500

  0
 0  0  500   3520
  0  480  0
 0

 xy  Covxy  Exy   x  y  3520  5959  39 ,
 
 
 x2  E x 2   x2  9670  59 2  6189 and  y2  E y 2   y2  6370  592  2889 . So that
 xy 
 xy
 x y
39


6189 2889
(  x  78 .6702 ,  y  53.7494 )
39 2
 .00008507  .009223 .
6189 2889 
So the answers for the questions in the text are
a. E x   59
(1)
b. E y   59
(1)
c.  x  78 .6702 (1)
d.  y  53.7494 (1)
e.  xy  39
(1)
f. From the Instructor’s Solutions Manual -- Stock Y gives the investor a lower standard deviation
while yielding the same expected return as investing in stock X, so the investor should select
stock Y. (1)
g. From the posted solution to Exercise 5.10. -- We find that E ( R)  wE ( x)  1  wE ( y) and
Var ( R)  w 2Var ( x)  1  w2 Var ( y)  2w1  wCov( x, y) or
 R2  w 2 x2  1  w2  y2  2w1  w xy .
So if w=.1 E ( R)  .1E ( x)  .9E ( y)  .159   .959   59
(1)
 R2  .12 6189  .92 2889  2.1.939  61 .89  2340 .09  7.02  2409
and  R  2409  49.0816 .
Obviously all the expected returns are the same. My results for the proportions suggested are below.
Needless to say, they are not the same as those in the Instructor’s Solutions Manual but the
conclusion is the same.
g.
h.
i.
j.
k.
w
1  w
0.1
0.3
0.5
0.7
0.9
0.9
0.7
0.5
0.3
0.1
w 2 6189 + 1  w2 2889 2w1  w39   R2
61.89
557.01
1547.25
3032.61
5013.09
2340.09
1415.61
722.25
260.01
28.89
7.02
16.38
19.50
16.38
7.02
2409
1989
2289
3309
5049
R
49.0816
44.5982
47.8435
57.5239
71.0563
(1)
(1)
(1)
(1)
(1)
l. From the Instructor’s Solutions Manual -- Based on the results of (g)-(k), you should
recommend a portfolio with 30% stock X and 70% stock Y because it has the same
expected return as other portfolios ($59) but has the smallest portfolio risk ($44.41). (1)
m.  xy  .009223
(1)
6
251y0321 4/18/03
2. Use the second to last digit of your Social Security number divided by 10 to figure out the fraction of the
items produced by a supplier that are defective. If the number you get is above .5, subtract .5 from it. If the
second to last number is a 0, use .1. For example, my Social Security Number is 265398248, so I used .4,
Seymour Butz’s Social Security number is 123456789, so he used .8 - .5 = .3. Let us call this number, p * .
a. Assume, that the entire amount of the product made by the supplier is a population of 100 units and that
you buy the whole batch. (If you use my value of p * , there are 40 defective items in the box) Take a
sample of 10 items and give me the probability that at least one is defective. (2)
b. Assume that the batch you buy is much larger, well over 200 and that you still take a sample of 10. What
is the chance that at least one is defective? (You should not need to use the number 200 or any larger
number in your calculations.) (1)
c. Assume that you have bought at least a million units, but that p * still represents the proportion of the
product that is defective. How large a sample do you need to be able to use the Poisson distribution to
figure out the probability of at least one defective item in the sample? Whatever the number you think is
correct, add 10 to it and assume this is your sample size. Find the probability of at least one defective item
in the sample. It is extremely likely that you will not be able to use a table and you may get some incredibly
tiny numbers in your calculations, which after explaining them, you may want to replaced with zero. (2)
Solution for all 5 versions:
a. Hypergeometric n  10, N  100 , M  p * 100  are defective.
Px  1  1  P0 Px  
i)
1  P0   1 
C nN
Crn 
n!
n  r !r!
Px  1  1  P0
p*  .1, M  10
90
C 010 C10
100
C10
C xM C nNxM
 1
 90! 

 80!10! 
1
90  89  88  87  86  85  84  83  82  81
10  9  8  7  6  5  4  3  21
 1
100  99  98  97  96  95  94  93  92  91
10  9  8  7  6  5  4  3  21
 100! 


 90!10! 
90  89  88  87  86  85  84  83  82  81
 1
 1  .33048  .6695
100  99  98  97  96  95  94  93  92  91
ii)
Px  1  1  P0
p*  .2, M  20
1  P0   1 
80
C 020 C10
 1
 80! 

 70!10! 
1
80  79  78  77  76  75  74  73  72  71
10  9  8  7  6  5  4  3  21
 1
100  99  98  97  96  95  94  93  92  91
10  9  8  7  6  5  4  3  21
 100! 


 90!10! 
80  79  78  77  76  75  74  73  72  71
 1
 1  .09512  .9049
100  99  98  97  96  95  94  93  92  91
iii)
100
C10
Px  1  1  P0
p*  .3, M  30
1  P0   1 
70
C 030 C10
100
C10
 1
 70! 

 60!10! 
1
70  69  68  67  66  65  64  63  62  61
10  9  8  7  6  5  4  3  21
 1
100  99  98  97  96  95  94  93  92  91
10  9  8  7  6  5  4  3  21
 100! 


 90!10! 
70  69  68  67  66  65  64  63  62  61
 1
 1  .02292  .9771
100  99  98  97  96  95  94  93  92  91
7
251y0321 4/18/03
iv) p*  .4, M  40
1  P0   1 
60
C 040 C10
100
C10
Px  1  1  P0
 1
 60! 

 50!10! 
1
60  59  58  57  56  55  54  53  52  51
10  9  8  7  6  5  4  3  21
 1
100  99  98  97  96  95  94  93  92  91
10  9  8  7  6  5  4  3  21
 100! 


 90!10! 
60  59  58  57  56  55  54  53  52  51
 1
 1  .0004355  .99564
100  99  98  97  96  95  94  93  92  91
Px  1  1  P0
v) p*  .5.M  50
1  P0   1 
50
C 050 C10
100
C10
 1
 50! 

 40!10! 
1
50  49  48  47  46  45  44  43  42  41
10  9  8  7  6  5  4  3  21
 1
100  99  98  97  96  95  94  93  92  91
10  9  8  7  6  5  4  3  21
 100! 


 90! 10! 
50  49  48  47  46  45  44  43  42  41
 1
 1  .000013187  .999987
100  99  98  97  96  95  94  93  92  91
b) Binomial n  10
i) p*  .1
Px  1  1  P0  1  .34868  .65132
ii)
p*  .2
Px  1  1  P0  1  .10737  .89263
iii)
p*  .3
Px  1  1  P0  1  .02825  .97175
iv) p*  .4
Px  1  1  P0  1  .00605  .99395
v) p*  .5
Px  1  1  P0  1  .00098  .99902
n
 500 and that will happen if n  500 p . The problem says to
p
use n  500 p  10 . Note that we use the Poisson distribution with m  np. The Poisson formula is
c) We can use the Poisson distribution if
Px  
e m m x
, so that, since m 0  1 and 0!  1 , P0  e m .
x!
i) p*  .1
n  500 p  10  60 , m  np  60.1  6
Px  1  1  P0  1  .00248  .9975 From table.
ii)
p*  .2
n  500 p  10  110 , m  np  110 .2  22
Px  1  1  P0  1  e 22  1  .0000000003 1
iii)
p*  .3
n  500 p  10  160 , m  np  160 .3  48


Px  1  1  P0  1  e 48  1 1.4 10 21  1
iv) p*  .4
n  500 p  10  210 , m  np  210 .4  84
Px  1  1  P0  1  e 84  1  3.310 37  1
8
251y0321 4/18/03
v) p*  .5
n  500 p  10  260 , m  np  260 .5  130
Px  1  1  P0  1  e 130  1  3.4810 57  1
3. (Extra credit) According to Bowerman and O’Connell, a plant manager says that the breakdowns of his
jorcillator follow an exponential distribution with an average of 1 breakdown every 250 hours. (Be very
careful of definitions when you calculate c. )
a. Using my formula for the cumulative distribution, the formula for the curve of the exponential
distribution is f x   ce cx . Mark off your y axis from 0 to .0005 and your x axis from 1 to 1000. Find the
value of the function for x  0. This will be the highest point on the curve. Compute height of the curve
for several points between zero and 1000 and draw the curve. (2)
b. Find the probability that the time between 2 successive breakdowns is between 100 and 300
hours. Compute this using the formula for the cumulative distribution and show it as an area under your
curve. (2)
c. Find the probability that the time we have to wait for a breakdown is less than or equal to 5
hours. In a hypothesis test of a statement like we have an average of only one breakdown in 250 hours, if
the researcher believes that the actual interval is smaller, the researcher finds the probability of getting a
number as small as or smaller than what actually happened. Assume that you next breakdown actually
occurs in 5 hours.. On the basis of the probability that you found, do you believe that the manager is
truthful? (2)
1
Solution: From the outline F x   1  e cx , when the mean time to a success is . If there is one
c
1
1
 250 means c 
 .004 .
breakdown every 250 hours, the mean time to a breakdown is 250 hours.
c
250
1
Note that this is only for x  0 . There is no probability below zero.
  
c
a. f x   ce cx , so f 0  .004e .0040  .004 , f 100  .004e .004100  .004e .4  .00268,
f 1000  .004e .0041000  .004e 4  .00007 . There is no point in letting your y axis go higher than .005
and you should get a smooth curve that merges with the x axis as x gets larger.
b. This is just the area under the curve from 100 to 300. P100  x  300   F 300   F 100  
 1  e .004300  1  e .004100  e .4  e 1.2  .67032 .30119  .3691 .


.0045

c. F 5  1  e
 1 e
 1  .98020  .0198 . The probability of such a short wait is well below 5%
and should make us suspicious that the average time to a breakdown is less than 250 hours.
.02
9