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ECO251 QBA1
THIRD EXAM
NOVEMBER 24, 1998
Name __________________
Section Enrolled: MWF TR 10 11 12:30
Part I. Do all the Following (10+ Points) Penalty for not doing question 1. Show your work!
You are following a stock and wish to compare its return to the Dow-Jones index. The return for 9
periods on the stock ( y ) is compared below to the return on “buying the Dow”( x ).
Period
x (Dow)
1
2
3
4
5
6
7
8
9
Total
y (Stock)
12
6
2
4
4
5
6
-6
3
36
12
15
-4
1
2
-1
-8
-2
3
18
y2
144
225
16
1
4
1
64
4
9
468
xy
144
90
-8
4
8
-5
-48
12
9
206
Note that x  4, s x  4.7170
Compute the following:
1.
The sample variance for the stock (4).
2.
The sample covariance between the stock and the Dow (2)
3.
The sample correlation between the stock and the Dow (2)
4.
Interpret the correlation (1)
Answers to questions 5) and 6) must be based on the results in questions 1-4. Do not
recompute the answers after changing y !
5.
If all the numbers in the y column were higher by 3 (i.e. 15, 18, -1, 4 etc.), what would
the variance, covariance and correlation computed above be? (1.5)
6.
If all the numbers in the y column were twice as high (i.e. 24, 30, -8, 2 etc.), what would
the variance, covariance and correlation computed above be? (1.5)
Solution:
1)
 y  18  2 ,
y
2)

n
9
xy  196 , s xy 
sx s y

y

2
 ny 2
n 1

468  922
 54
8
( s y  54  7.34847 )
 xy  nx y  206  942  16.75 .
n 1
9 1
16 .75
 0.4832
4.7170 54
4) The positive sign of rxy , the sample correlation, indicates that x and y tend to move together. If we
3)
rxy 
s xy
s 2y
square rxy , we get approximately .23, which on a zero to one scale indicates a relatively weak
relationship.
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5) We are leaving x alone, but replacing y by y  3 . From the syllabus supplement and the outline if
v  cy  d  y  3 so that c  1 and d  3
a)  v2  Varv  c2Var y   c2 y2  12 y2  54
b) If w  ax  b and v  cy  d ,  wv  Cov(ax  b, cy  d )  acCov( x, y) and a  1 and b  0
 wv  acCov( x, y)  1116.75   16.75 .
c)
Corrw, v  rwv  signacrxy  sign11.4471  sign 1.4832  .4832
6) We are leaving x alone, but replacing y by 2 y . From the syllabus supplement and the outline if
v  cy  d  2 y so that c  2 and d  0
a)  v2  Varv  c2Var y   c2 y2  22 y2  454  216
b) If w  ax  b and v  cy  d ,  wv  Cov(ax  b, cy  d )  acCov( x, y) and a  1 and b  0
 wv  acCov( x, y)  1216.75   33.50 .
c) Corrw, v  rwv  signacrxy  sign12.4471  sign 2.4832  .4832
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Part II. Do the following problems ( do at least 40 points ). Show your work! Note: You only need 40
points out of the 72 below to get an A+ -. Do the parts that look easiest!
1.
The following table represents the joint probability of x and y .
Event
x
2
y 3
4
Px 
0
P y  yP y  y 2 P y 
1
2
.25 .10
.05 .25
.15 .10
.05
.05
?
.40 .80
.35 1.05
.25 1.00
1.60
3.15
4.00
.10
1.00 2.85
8.75
.45
.45
 
So E  y    y  2.85 and E y 2  8.75
0+ .45 + .20 = 0.65  Ex  x
xPx 
 
x P x 
0+ .45 + .40 = 0.85  E x 2
2
a.
b.
Fill in the missing number. (1)
Are x and y independent? Why?(2)
c.
Compute  xy , the covariance of x and y , and interpret it. (3)
d.
Compute  xy , the correlation of x and y , and interpret it. (3)
e.
f.
Find the probability that x  y is less than 6. (1)
(i) Find the distribution of x  y . (2)
(ii) Using only the results of a)-d), find the mean and variance of x  y .(3)
Solution: a)
b)
c)
Since the table must total 1.00, the missing number is 0.
x and y are independent if Px, y   Px   P y  . In the lower right corner 0  10 .25  ,
so x and y are not independent .
E  x, y  
 xyPx, y 
 xy  Covx, y   Exy   x  y
 1.85  0.65 2.85   0.0025
Negative, so x and y move in opposite
directions.
= 0(2)(.25) + 1(2)(.10) + 2(2)(.05)
+ 0(3)(.05) + 1(3)(.25) + 2(3)(.05)
+ 0(4)(.15) + 1(4)(.10) + 2(4)(0) = 1.85
d)
 xy 
 xy
 .0025 
 
 Ey  

.0025 2

0.4275 0.6275 
0.0000232  0.0048
0.4275 0.6275
We measure the strength of a correlation by squaring it. If we square -.0048, we get .000023.
On a zero to one scale, this is tiny, so correlation is very weak.
 x y

Var x    x2  E x 2   x2  0.85  0.65 2  0.4275
Var y    y2
e)
2
2
y
 x  0.4275  0.6538
 8.75  2.85  0.6275  y  0.6275  0.79215
2
See below. Since the sums of x and y are all below 6, Px  y   6  1
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sums
probabilit ies
x
x
0 1 2
0 1 2
f-i)
2  2 3 4
2 .25 .10 .05 




y 3 3 4 5
y 3 .05 .25 .05 
4 4 5 6
4 .15 .10 0 




The tables above show the sums of x and y and their probabilities . If we add the
probabilities we find the distribution below.
P x  y 
x y
3
P2  0 
P0  3  1  2 
4
P0  4  1  3  2  2 
2
5
6
P1  4  2  3 
P2  4 
x  y Px  y 
 x  y 2 P x  y 
.25 =
.25
.50
1.00
.05 +.10 =
.15
.45
1.35
.15 +.25+.05 =
.45
1.80
7.20
.10 +.05 =
.15
.75
3.75
0=
.00
.00
0.00
1.00
3.50
13.30
Only the x  y and Px  y  columns are needed. From this, Ex  y   3.50 and


Varx  y   E x  y 2   x  y
Ex  y  and Var x  y  .
2
 13 .30  3.50 2  3.05 . But this is not how I asked you to compute
f-ii)  x y  Ex  y    x   y  Ex  E y   0.65  2.85  3.50
Varx  y   Varx   Var y   2Covx, y   0.4275  0.6275  2(0.0025 )  1.05
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2.
I sell both coal and oil. Although the prices of oil and coal fluctuate, I maintain a constant markup of
$16.00 per ton for coal and $0.06 per gallon on oil. My fixed costs for coal storage are $600 a month.
My fixed costs for oil storage are $300/month.
My mean sales of coal are 3000 tons, with a standard deviation of 50 tons.
My mean sales of oil are 4500 gallons with a standard deviation of 300 gallons.
The correlation between sales of coal and oil is –0.3.
a. What are the mean and standard deviation of profits on coal? (3)
b. What are the mean and standard deviation of profits on oil? (1)
c. What is the covariance between sales of coal and oil? (1.5)
d. What is the covariance between profits on coal and oil? (1.5)
e. What is the mean and standard deviation of total profits (oil profits plus coal profits)? (4)
Solution: If coal is x and oil is y ,  x  3000,  x  50,  y  4500,  y  300 and  xy  0.3 .
a)
Let profits on coal be w . From the syllabus supplement and the outline if w  ax  b  16 x  600
so that a  16 and b  600 , then:
E w  E ax  b  aEx   b or E16 x  600   16 Ex   600  163000   600  47400
 w2  Varw  a 2Varx   a 2 x2  16 2  x2  16 2 50 2  640000 ,  w  640000  800
b) Let profits on oil be v . From the syllabus supplement and the outline if v  cy  d  .06 y  300 so
that c  .06 and d  300 , then:
E v   E cy  d   cE y   d or E.06 y  300   .06 E y   300  .064500   300  30 .
 v2  Varv  c 2Var y   c 2 y2  .062  y2  .062 3002  324,  v  324  18
c)
Cov( x, y)   xy   xy  x y  0.350300  4500
d) Cov( x, y)   wv  Cov(ax  b, cy  d )  acCov( x, y)  16 .06 4500   4320
e) From the previous page substituting w for x and v for y ,
 x y  Ew  v   w   v  Ew  Ev  47400 30  47370
Varw  v   Varw  Varv   2Covw, v   640000  324  2(4500 )  631324
 wv  631324  794 .559
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3.
Assume that my father gives me a present of $1000 every time my wife has a boy baby and $2000
every time she has a girl baby. (Assume that the chance of a girl baby is 50%)
a.
What is the chance that out of 10 births exactly 3 will be girls? (2)
b.
What is the chance that out of 10 births more than half will be girls? (2)
c.
What is the mean and variance for the amount of money that I receive on any one birth?
(2)
d.
What is the mean and variance for the number of girls born in 10 births? (1.5)
e.
What is the chance that the first girl will be born on the 10 th birth ? (1.5)
If a variable has the binomial distribution with p  .03 and n=10
(i)
Find the probability that you have at least one success (2)
(ii)
Is it appropriate to use the Poisson distribution in this problem? Why? If it is
appropriate, use the Poisson distribution to find the answer to (i). (2)
f.
Solution:
Binomial n  10, p  .5 . (Because we are asking for the probability of x successes in n tries, and the
probability of success on any one try is given and constant.)
a)
b)
c)
P3  Px  3  Px  2  .17188  .05469  .11719 , also C319 .53 .57 .
Px  5  Px  6  1  Px  5  1  .62305  .37695
y
P y  yP y 
y 2 P y 
So  y  1500 and  y2  250000 15002  250000
.5 500
.5 1000
1.0 1500
500000
2000000
2500000
 y  500
1000
2000
d)   np  10.5  5.0  2  npq  10.5.5  5.0.5  2.50
e) Geometric p  .5 (Because we are asking for the probability of the first success on try x .)
P10   q 9 p  .59 .5  .0009766
f) (i) Binomial n  10,
(ii) Since
0
10
10
p  .03 . Px  1  1  P0  1  C 010  p  q   1  .97   1  .73742  26258
n 10

 333 .33
p .03
is less than 500, the Poisson distribution is not appropriate.
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4 . . Make sure that you state clearly what distribution you are using in each section of this problem.
If I find that my paint jobs have an average of 0.10 blisters per square foot and a refrigerator has 14 square
feet of painted surface,
a. What is the chance that there will be exactly one blister on a refrigerator? (1)
b. What is the chance that there will be at least one blister on a refrigerator? (2)
c. What is the standard deviation of the number of blisters on a refrigerator? (1)
d. Using your results from a), what is the chance that, in a delivery of 10 refrigerators, all will
have at least one blister? (2)
e. In d) what is the chance that more than 8 will have at least on blister? (2)
f. If a store sells an average of 72 refrigerators in a 16 hour day and it takes 2 hours to get a
delivery, what is the fewest number of refrigerators it should have in stock before it reorders if
it wishes to limit the chance of running out to not more than 1% ? Explain. (3)
Solution:
Poisson with parameter (mean) of 0.1(14) = 1.4. (Because we are asking for the probability of x successes
when the average number successes in a unit of space or time is given and constant.)
a) P1  Px  1  .345236
b) Poisson Px  0  Px  1  1  P0  1  .246597  .753403
c)
d)
Since, for the Poisson distribution, the mean and variance are identical,   m  1.4  1.1832
Binomial n  10, p  .753403. (Because we are asking for the probability of x successes in n
tries, and the probability of success on any one try is given and constant.) Px  C xn p x q n x so
e)
10
.753403 10 1  .753403 0  .753403 10  .058921
P10   C10
Binomial n  10, p  .753403
Since P9  C910 .753403 9 1  .753403 1  10.753403 9 .241597   10.018894597   .1889460 ,
Px  8  P9  P10   .1889460  .058921  .247867
f)
72
9.
8
From the Poisson (9) table, the first value of k with Px  k   .99 is 17 with Px  17   .99468 .
So the reorder point is 15.
Since there are 8 two-hour periods in a 16 hour day, this is a Poisson problem with m 
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5.
Assume that in any batch of soft drink cans 5% are defective. Make sure that you state clearly
what distribution you are using in each section of this problem.
a. If I take 4 cans from a package of 20 cans what is the chance that exactly 2 are defective? (2)
b. If I take 4 cans from a package of 20 cans, what is the chance that at least one is defective? (2)
c. What are the mean and variance of the number of defective cans in a sample of 4 taken from a
package of 20? (2)
d. If I take 4 cans from a package of 2000 cans, what is the chance that exactly 2 are defective? (2)
e. If I take 4 cans from a package of 2000 cans, what is the chance that at least one is defective? (2)
f. What are the mean and variance of the number of defective cans in a sample of 4 taken from a
package of 2000? (2)
g. If I am taking cans from a package of 2000, what is the chance that I will find my first defective
can in the first 100 cans (2) ?
Solution: Hypergeometric with N  20, M  pN  .05 20   1 and n  4 (Because we are asking for the
probability of x successes when the population is of limited size and the number of successes in the
population is also limited.)
a) Since M  1, P2  0
b) Since Px  
C xM C nNxM
C nN
,
Px  0  1  P0  1 
C 01 C 419
C 420
19!
 19 18 17 16 


15!4!  1   4  3  2 1   1  19  18  17  16  1  16  1  .80  .20
 1
20!
20 19 18 17
20
 20 19 18 17 


16!4!
 4  3  2 1 
1
 N n
 20  4 
 16 
npq  
4.05 .95    0.1900  .1600
 N 1 
 20  1 
 19 
d) Though hypergeometric is not wrong here, because of the large population we are better off using
the binomial. So the distribution is either Binomial with n  4, p  .05 or Hypergeometric with
c)
  np  4.05   0.20
2 
N  2000 , M  pN  .05 2000   100 and n  4 .
P2  Px  2  Px  1  .99952  .98598  .01354 or
100  99 1900 1899

100  99 1900 1899  4  3
2 1
2 1
P2 


 .01344
2000
2000

1999

1998

1997
2000
1999 1998 1997  2 1
C4
4  3  2 1
Px  0  1  P0  1  .81451  .18549 or
C 2100C 21900
e)
Px  0  1  P0  1 
C 0100C 41900
C 42000
2
 1
1900 1899 1898 1897
 1  .8143775  .18562
2000 1999 1998 1997
f)   np  4.05   0.20
  npq  4.05.95  0.1900
g) Geometric p  .05 (Because we are asking for the probability of the first success on try x .)
Remember F x  1  q x where q  1  p .
Px  100   F 100   1  q100  1  .95 100  1  .0059  .9941
8
251y9872 11/27/98
6.
A jorcillator has two components, a phillinx and a flubberall. As long as both are working the
jorcillator will not be junked. If either fails the jorcillator must be junked.
a. If a phillinx has a lifespan approximated by a uniform distribution between 4 and 14 years,
find:
(i)
The mean and standard deviation of its life. (2)
(ii)
The probability that it will last between 0 and 5 years (2)
(iii)
The probability that it will last between 5 and 10 years (1)
(iv)
The probability that it will last more than 10 years (1)
b. If a flubberall has a lifespan approximated by a uniform distribution between 5 and 20 years,
do (i) a(ii)-a(iv) above for the flubberall (2) and then (assuming that the lifespan of the two
components is independent) find:
(ii)
The probability that the jorcillator will last between 0 and 5 years. (2)
(iii)
The probability that the jorcillator will last between 5 and 10 years. (3)
(iv)
The probability that the jorcillator will last more than 10 years. (2)
Solution:
a) For the Phillinx: c  4, d  14
c  d 4  14

9
2
2
d  c2  14  42  8.3333
2 
12
12

(i)
  8.3333  2.8868
For the logic of the rest of this
problem, see problem H4.
1
1
P A  P0  x  5  5  4 
10 10
=.100
1
5
PB   P5  x  10   10  5 
10 10
=.500
1
4
PC   Px  10   14  10  
10 10
=.400
(ii)
(iii)
(iv)
Phillinx (Draw a diagram for the uniform
distribution)
Flubberall (Draw a diagram for the uniform
distribution)
b) For the Flubberall: c  5, d  20
(i) PD   P0  y  5  0
PE   P5  y  10   10  5
1 1
 =.3333
15 3
1 2
PF   P y  10   20  10   =.6667
15 3
b-ii) A joint probability table follows on the left with the unions of joint events that make up the events that
were specified summed to the right. Note that the probabilities sum to one.
D
A
0
B
0
C
0
0
E
1
30
5
30
4
30
1
3
F
2
30
10
30
8
30
2
3
1
10
5
10
4
10
1
Event
Fails in
0-5
Fails in
5-10
Fails in
10+
Sum
Component Joint Events
P A  D   A  E    A  F 
 B  D  C  D 
PB  E   B  F   C  E  
PC  F  
Probability
1
2
3
0

00 
 .10000
30 30
30
5 10 4



30 30 30
8

30
19
 .63333
30
8
 .26667
30
1.0000
9