251x0531 4/18/05
(Preliminary!!!!)
ECO251 QBA1
THIRD EXAM
Apr 25, 2005
Name: _____________________
Student Number: _____________________
Class Time (Circle) 1pm 2pm
Part I: 16 points.
z follows the standardized Normal distribution z ~ N 0,1 .
Find the following. Make diagrams! Add a vertical line at zero to these diagrams.
1. P2.63  z  2.63  P2.63  z  0  P0  z  2.63  2.4957   .9914
Normal Curve with Mean 0 and Standard Deviation 1
The Area Between -2.63 and 2.63 is 0.9915
0.4
Density
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Dat a A xis
1
2
3
4
2. P3.11  z  2.63  P3.11  z  0  P2.63  z  0  .4991  .4957  .0034
Normal Curve with Mean 0 and Standard Deviation 1
The Area Between -3.11 and -2.63 is 0.0033
0.4
Density
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Data A xis
1
2
3
4
3. F 3.11  Pz  3.11  Pz  0  P0  z  3.11  .5  .4991  .9991
Normal Curve with Mean 0 and Standard Deviation 1
The Area to the Left of 3.11 is 0.9991
0.4
Density
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Data A xis
1
2
3
4
4. z .115 Solution: Make a diagram. z .115 is defined as a point with 11.5% above it and thus 100% 11.5% =88.5% below it, so it is the 88.5 percentile. The diagram for z will show an area with a
probability of 88.5% below z .115 . It is split by a vertical line at zero into two areas. The lower one has a
251y0531 4/22/05
(Preliminary!!!!)
probability of 50% and the upper one a probability of 88.5% - 50% =38.5%. The upper tail of the
distribution above z .115 has a probability of 11.5%, so that the entire area above 0 adds to 50%. From the
diagram, we want one point z .115 so that Pz  z.115   .8850 or P0  z  z.115   .3850 . The closest we
can come is P0  z  1.20   .3849 . We can say z.115  1.20 .
Normal Curve with Mean 0 and Standard Deviation 1
The Area to the Right of 1.2 is 0.1151
0.4
Density
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Data A xis
1
2
3
4
x follows the Normal distribution x ~ N 2.63, 2 . This means   2.63 and   2.00 .
Find the following. Make diagrams!
x
Remember z 
. As usual, people made diagrams of x with zero in the middle. Make up your

mind! If you are diagramming x , put the mean in the middle; if you are diagramming z put zero
in the middle.
2.63  2.63 
  2.63  2.63
z
5. P2.63  x  2.63  P 
  P 2.63  z  0  .4957
2
2


0.20
0.4
0.15
0.3
Density
Density
Norm al Curv e with Mean 2.63 and Standard Dev iation
Norm
2
al Curv e with Mean 0 and Standard Dev iation 1
The Are a Between -2.63 and 2.63 is 0.4957
The Are a Between -2.63 and 0 is 0.4957
0.10
0.05
0.00
0.2
0.1
-5
0
5
0.0
10
-5.0
-2.5
Da ta A x is
0.0
Da ta A x is
2.5
5.0
 2.63  2.63 
  3.11  2.63
z
6. P3.11  x  2.63  P 
  P 2.87  z  2.63 
2
2


 P2.87  z  0  P2.63  z  0  .4979  .4957  .0022
0.20
0.4
0.15
0.3
Density
Density
Norm al Curv e with Mean 2.63 and Standard Dev iation
Normal
2
Curv e with Mean 0 and Standard Dev iation 1
The Area Between -3.11 and -2.63 is 0.0022
The Area Between -2.87 and -2.63 is 0.0022
0.10
0.05
0.00
0.2
0.1
-5
0
5
Da ta A x is
10
0.0
-5.0
-2.5
0.0
Da ta A x is
2.5
5.0
2
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(Preliminary!!!!)
3.11  2.63 

7. F 3.11  Px  3.11  P  z 
  Pz  0.24   Pz  0  P0  z  0.24 
2


 .5  .0948  .5948
0.20
0.4
0.15
0.3
Density
Density
Norm al Curv e with Mean 2.63 and Standard Dev iation
Norm
2
al Curv e with Mean 0 and Standard Dev iation 1
The A re a to the Le ft of 3.11 is 0.5948
The Area to the Left of 0.24 is 0.5948
0.10
0.05
0.00
0.2
0.1
-5
0
5
0.0
10
-5.0
-2.5
Da ta A x is
8. x.115 In problem I4, we found z.115  1.20 . The opposite of z 
  z.115  2.63  1.202  5.03. Check:
x

0.0
Da ta A x is
2.5
5.0
is x    z , so x.115
5.03  2.63 

Px  5.03   P  z 
  Pz  1.20   Pz  0  P0  z  1.20   .5  .3849  .1151
2


0.20
0.4
0.15
0.3
Density
Density
Norm al Curv e with Mean 2.63 and Standard Dev iation
Norm
2
al Curv e with Mean 0 and Standard Dev iation 1
The Area to the Right of 5.03 is 0.1151
The Area to the Right of 1.2 is 0.1151
0.10
0.05
0.00
0.2
0.1
-5
0
5
Da ta A x is
10
0.0
-5.0
-2.5
0.0
Da ta A x is
2.5
5.0
3
251y0531 4/22/05
(Preliminary!!!!)
Part II: (15+ points) Do all the following: All questions are 2 points each except as marked. Exam is
normed on 50 points including take-home. Most questions come from Wonnacott and Wonnacott (1990)
(Showing your work can give partial credit on some problems! In open-ended questions it is
expected. Please indicate clearly what sections of the problem you are answering and what formulas
you are using. Neatness counts!) Remember that you may not be able to finish this section, so ration your
time on each problem. [Numbers in brackets are a cumulative total]
1.
If y tends to decrease as x increases, what can we say about the population correlation  ?
a)   1 .
b)   1 .
c)   0 .
d) *   0 .
2.
The riskiness of a portfolio made up of two investments
a) will be higher when the covariance is zero.
b) will be higher when the covariance is negative.
c) *will be higher when the covariance is positive.
d) does not depend on the covariance.
[2]
Explanation: Remember from the portfolio material, 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 P2 CovR1 , R2 
is the variance of the return. Here P 1 is the portion of $1 invested in stock 1, P 2 is the
portion of $1 invested in stock 2, R1 is the return on stock 1 and R 2 is the return on stock 2.
The variance is a measure of riskiness, and the product of P 1 and P 2 must be positive. So
the higher CovR1 , R2 , the higher VarR  .
3.
Should read 70!!!!Fifteen items are randomly selected from a pilot production run of N items to
check their quality. x is the number of defective items in the sample. The distribution of x can be
considered approximately binomial if N is
a) 15
b) 152
c) 400
d) *1500
e) All of the above
f) None of the above.
Explanation: We can consider the population infinite if N  20n. n  70 and
20 n  2070   1400 . The only number above 1400 is 1500.
4.
There are 24 Million people living in California. Approximately 8 million are in Los Angeles. The
Census Bureau takes a random sample of 200 people from the state. Let x be the number of
people in the sample who live in Los Angeles, then x is most conveniently treated as having the
a) Binomial distribution with n  24 million and p  8 24 .
b) Poisson distribution.
c) *Binomial distribution with n  200 and p  8 24 .
d) Hypergeometric distribution
e) none of the above.
4
251y0531 4/22/05
5.
(Preliminary!!!!)
Let x represent the number of times a fair die (6 sides) comes up with a 1 on top when the die is
cast 300 times. What is a) E x  , b) Var (x), c) E x 2 Show your work! (5)
[11]
Solution: Since getting a one on top can be describes as a success and the probability of a 1 is a
constant 1 6 , we have a binomial distribution with n  300 . We know   np  300  1 6   50 ,
 2  npq  505 6   41.6667 . Since  2  E x 2   2 ,
E x 2   2   2  41.6667  50 2  2583 .3333
 
 
 
6.
7.
8.
Assume that a fair 6-sided die is cast a large number of times. What is the chance that the first
time it comes up with a 1 on top is in the first 10 throws? Show your work! (2)
Solution: Since we are talking about first success, we need the geometric distribution.
10
F x  1  q x . So Px  10   F 10   1  5 6   1  .16151  .83849 .
A student calculates that P9  x  35   F 35   F 9 . The student could be working with the
a) Binomial Distribution
b) Poisson Distribution.
c) *Continuous Uniform Distribution
d) Geometric distribution
e) All of the above
f) None of the above.
[15]
Explanation: For any discrete distribution P9  x  35   F 35   F 8 .
If we are taking a sample of 3 from a population that is one-third defective, what is the probability
that at least one item in our sample is defective if a) N  30 ? and b) N is very large? Show your
work! (4)
Solution: In both sections of the problem, n  3 , p 
1
3
and Px  1  1  P0 .
a) Let x be the number of defective items in the sample, which we will call successes. x  0 ,
N  30 and M  13 30   10 is the limited number of available successes, which pushes us to
using a Hypergeometric distribution where Px  
C nNxM C xM
C nN
. So P0 
C nNxM C xM
C nN

C 320 C 010
C 330
 20 19 18 


20 19 18
 3  2 1 


=.2808. Finally Px  1  1  P0  1  .2808  .7172
 30  29  28  30  29  28


 3  2 1 
b) If N is very large (more than 20 times n ), we can treat the population as infinite.
q  1  p  2 3 and we can use the binomial term, Px  C xn p x q n x . So
P0  C 03 p 0 q 3  q 3  2 3   .2963 . Finally, Px  1  1  P0  1  .2963  .7037 .
3
9.
(Extra credit) The time it takes to complete an exam has an exponential distribution with a mean
of 40 minutes. Luce Cash is in a hurry to finish so she can go shopping. What is the probability
that she will finish the exam in 30 minutes or less? What is the probability that she will still be
working when the instructor collects the exams after 55 minutes?
1
Solution:    40 , so c  1 40  .025 . F x 1  e cx when x  0 and the mean time to a
c
1
success is . Both are zero if x  0 . a) So Px  30  F 30  1  e .02530  1  e .75  1  .4724
c
 .5276 . b) Px  55  1  F 50  e .02550  e 1.25  .28650
5
251y0531 4/22/05
(Preliminary!!!!)
ECO251 QBA1
THIRD EXAM
Apr 25, 2005
TAKE HOME SECTION
Name: _________________________
Student 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 III. Do all the Following (19+ Points) Show your work!
1. Before you start, personalize the data below as follows. Take the third to last digit of your student
number and add it to the last digit of .06. Take half the third to last digit of your social security number and
subtract it from each of the .09s.
A photographic processor is trying to determine the distribution of the times that the business can promise
to customers. x is the time in days it takes the package to arrive and y is the time in days it takes the
package to return to the customer. It takes 2 days to process the film, so that the total time is t  x  y  2
days.
x
y
2
.10
.15
.05
1
2
3
3
.15
.26
.09
4
.05
.09
.06
For example Robin dePoore’s student number is 999799 so she changes the .06 to .06 + .07 = .13 and the
two .09s to .09 - .035 = .055. This should not change the total sum of the numbers in the joint probability
table. Find.
a) The mean and standard deviation of both x and y . (2)
b) The covariance and correlation between x and y . (3)
c) Using only the results in a) and b), find the mean and variance of t (2).
d) Find the probability that the total time t  x  y  2 is no more than 5 days (2).
e) Assume that because of an anthrax scare, x , the time it takes the package to arrive, doubles. Using only the numbers that
you found in a) and b) and appropriate formulas, find the new covariance and correlation and the mean and variance of t
f)
(2).
Go back to the original values of x and
y and assume that the marginal probabilities of x and y are correct, what
y were independent? (1)
Recall what the covariance of x and y would be if they were independent. What would the variance of t  x  y  2
would the joint probability table look like if x and
g)
be now? (1)
[13]
Solution: a) The mean and standard deviation of both x and y . (2)
x
1
2
y
3
4
Px 
xPx 
x Px 
2
P y 
.300
3
yP y  y 2 P y 
0.600 1.200
 .10

.260
 .15
 .05
.055

0.300  0.465
.050 

.055 
.130 

 0.235
0.300
 0.930
 0.705
 1.935
0.300
 1.860
 2.115
 4.275
.465
1.395
4.185
.235
0.940
3.760
 1.000
2.935
9.145
 Px  1 ,   E x    xPx   1.935 E x    x
 E  y    yP y   2.935 and E y    y P y   9.145 .
To summarize
y
2
.150
2
x
2
 
2
Px   4.275
 P y   1 ,
2
 
 x2  E x 2   x2  4.275  1.935 2  0.530775 and  y2  E y 2   y2  9.145  2.9352  0.530775
6
251y0531 4/22/05
(Preliminary!!!!)
b) The covariance and correlation between
x and y . (3)
 .10 12  .150 22  .050 32  0.20  0.60  0.300 
E xy  
xyPxy     .1513  .260 23  .055 33   0.45  1.56  0.495   5.805
 .0514  .055 24  .130 34  0.20  0.44  1.560 
 xy  Covx, y   Exy   x  y  5.805 1.9352.935  0.125775 ,

. So that
 xy  Corr x, y  
 xy
0.125775

0.125775 2 
0.728543 .728543 
.0.05616 /  .2370 .
0.530775 .530775
(  x  0.728543 ,  y  0.78543 ) The correlation and covariance are positive, indicating a tendency of
 x y

2
 .05615 hardly exists on a zero to one scale, indicating that the
y to rise when x rises.  xy
relationship is barely there. Note that 1   xy  1 always!
c) Using only the results in a) and b), find the mean and variance of t (2).
Recall that t  x  y  2 . Since means add, E t   E x   E  y   2  1.935  2.935  2  6.870
Recall that Varx  y   Varx   2Cov( x, y)  Var( y)  0.530775  2.125775   0.530775  1.3131 and
that adding a constant does not affect the variance. So Vart   1.3131
d) Find the probability that the total time t  x  y  2 is no more than 5 days (2).
Recall the original joint probability table.
x
y
1
2
3
2
.10
.15
.05
3
.15
.26
.09
4
.05
.09
.06
2
1
2
3
5
6
7
3
6
7
8
4
7
8
9
Now repeat the table with x  y  2 in place of the probabilities.
x
y
Thus Pt  5  Px  2, y  1  .10.
e) Assume that because of an anthrax scare, x , the time it takes the package to arrive, doubles. Using
only the numbers that you found in a) and b) and appropriate formulas, find the new covariance
and correlation and the mean and variance of t (2).
 y  E y   2.935 ,  x  Ex   1.935 ,  xy  Corrx, y   .2370
 xy  Covx, y   0.125775,  x2  0.530775 and  y2  0.530775 .
Remember that if w  ax  b , and v  cy  d , where a, b, c, and d are constants that we had the
following.
 w  Ew  Eax  b  aEx   b  w2  Varw  a 2Varx  a 2 x2
 v  E v   E cy  d   cE y   d
 v2  Var v   c 2Var  y   c 2 y2
Covw, v   wv  acCovx, y   ac xy and  wv  signac xy
Let w  2 x and v  y , So that a  2 , b  0, c  1 and d  0. Then
 w  Ew  Eax  b  2Ex   21.935   3.87  w2  Varw  a 2Varx  a 2 x2  4.530775  2.1231
 v  Ev   Ecy  d   cE y   d  1.935
 v2  Var v   c 2Var  y   c 2 y2  0.530775
Covw, v   wv  acCovx, y   ac xy  2.125775  .25155 and  wv  signac xy  .2370
t  x  y  2 becomes t  w  v  2 Et   Ew  Ev   2  3.87  2.935  2  8.805
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251y0531 4/22/05
(Preliminary!!!!)
Recall that Varw  v   Varw  2Cov(w, v)  Var (v)  2.1231  2.125775   0.530775  2.9054
f) Go back to the original values of x and y and assume that the marginal probabilities of x
and y are correct, what would the joint probability table look like if x and y were independent?
(1)
Each number inside the table would be the product of the marginal probabilities. For example .0900 is
the product of .3000 and .3000
x
2
y
3
4
Px 
xPx 
1.0000
2.0000
 .0900
.139500

.216225
 .1395
 .0705
.109275

.3000  .465000
0.300
 0.930
3 .0000
P y 
.070500 
.300

.109275 
.465
.055225 
.235

 .235000
 1.000
 0.705
yP y  y 2 P y 
0.600 1.200
1.395
4.185
0.940
3.760
2.935
9.145
 1.935
x Px 
0.300
 1.860
 2.115
 4.275
Recall what the covariance of x and y would be if they were independent. What would the variance
of t  x  y  2 be now? (1) The covariance and correlation would be zero.
So Varx  y  2  Varx   2Cov( x, y)  Var ( y)  0.530775  20  0.530775  1.06155
2
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251y0531 4/22/05
(Preliminary!!!!)
2. Personalize the data below by adding the third to last digit of your student number to 10.
x
y
c)
Find the covariance and correlation between
w  2 x  3 and v  7 y  9. Use only
23
8
a)
b)
16
13
20
9
21
10
Using the computational formula, find the sample
variance of x . (2)
Find the covariance and correlation between x
and y .(2)
what you found in b) and appropriate formulas.
If w  2 x  3 and v  7 x  9, find the
covariance and correlation between w and v .
Use only what you found in b) and appropriate
formulas. (Hint: figure out or find from problems
what Cov x, x is.
d)


Solution: a) Using the computational formula, find the sample variance of x . (2)
If we do the necessary tables, we find the following.
Row
x
x2
xy
y
y2
1
2
3
4
23
16
20
21
80
529
256
400
441
1626
8
13
9
10
40
64
169
81
100
414
184
208
180
210
782
 x  80,  x  1626 ,  y  40,  y  414 , and  xy
 x  80  20.000 and y   y  40  10.000 .
Then x 
So n  4,
s x2 
2
x
n
2
4
 nx
n 1
2
2

n
4
1626  420 .000 2
 8.6667 , s 2y 
3
y
2
 ny 2
n 1

 782 .
414  410 .000 2
 4.6667 .
3
( s x  8.6667  2.9439 and s y  4.6667  2.16025 ).
b) Find the covariance and correlation between x and y .(2)
s xy  Covx, y  
rxy 
s xy
sx s y
 xy  nxy  782  420.000 3.500   6.0000
n 1
 Corr x, y  
4 1
 6.000
and
 0.9435 .
8.667 2.16025
Find the covariance and correlation between w  2 x  3 and v  7 y  9. Use only what you
found in b) and appropriate formulas.
Remember that if w  ax  b , and v  cy  d , where a, b, c, and d are constants that we had the
following.
 w  Ew  Eax  b  aEx   b  w2  Varw  a 2Varx  a 2 x2
c)
 v  E v   E cy  d   cE y   d
 v2  Var v   c 2Var  y   c 2 y2
Covw, v   wv  acCovx, y   ac xy and  wv  signac xy
s xy  6.0000 and rxy 
s xy
sx s y
 Corr x, y   0.9435 .
If w  2 x  3 and v  7 y  9. v  y , So that a  2 , b  3, c  7 and d  9. Then
Covw, v   wv  acCovx, y   ac xy  276  84 and
 wv  signac xy  sign 2 7Corrx, y   Corrx, y   .9435
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251y0531 4/22/05
(Preliminary!!!!)
d) If w  2 x  3 and v  7 x  9, find the covariance and correlation between w and v . Use only
what you found in b) and appropriate formulas. (Hint: figure out or find from problems what
Covx, x  is.)
 xx  nxx   x
2
 nx 2
s xx
s2
 x2  1
n 1
n 1
sx sx sx
If w  2 x  3 and v  7 x  9. v  y , So that a  2 , b  3, c  7 and d  9. Then
Covw, v    wv  acCovx, x   ac xx  278.6667   121 .3338 and
 wv  signac xx  sign 2 7Corrx, x   Corrx, x   1.
s xx  Covx, x  
 s x2
rxx  Corr x, x  
10