251y0641 5/10/06
ECO 251 QBA1
FINAL EXAM, Version 1
MAY 8, 2006
Name KEY
Class ________________
Part I. Do all the Following (14 Points) Make Diagrams! Show your work! Illegible and poorly
presented sections will be penalized. Exam is normed on 75 points. There are actually 123+ possible
points. If you haven’t done it lately, take a fast look at ECO 251 - Things That You Should Never Do on a
Statistics Exam (or Anywhere Else).
x ~ N 13, 5.6
28  13 
 0  13
z
 P 2.32  z  2.68 
1. P0  x  28   P 
5.6 
 5.6
 P2.32  z  0  P0  z  2.68   ..4898  .4963  .9861
For x ~ N 13, 5.6 make a Normal curve centered at 13 and shade the area from 0 to 28; for z make a
Normal curve centered at zero and shade the area from -2.32 to 2.68. Since the diagrams show an area on
both sides of the mean, you add.
12  13 

 Pz  0.18 
2. F 12 .00  (Cumulative Probability) Px  12   P  z 
5.6 

 Pz  0  P0.18  z  0 =.5-.0714=.4286
For x ~ N 13, 5.6 make a Normal curve centered at 13 and shade the area below 12; for z make a
Normal curve centered at zero and shade the area below -0.18. Since the diagrams show an area below the
mean that does not touch the mean, you subtract.
28  13 

 Pz  2.68   Pz  0  P0  z  2.68   .5  .4963  .0037
3. Px  28   P  z 
5.6 

For x ~ N 13, 5.6 make a Normal curve centered at 13 and shade the area above 28; for z make a
Normal curve centered at zero and shade the area above 2.68. Since the diagrams show an area above the
mean that does not touch the mean, you subtract
32  13 
 28  13
z
 P2.68  z  3.39   P0  z  3.39   P0  z  2.68 
4. P28  x  32   P 
5.6 
 5.6
.4997  .4963  .0034
For x ~ N 13, 5.6 make a Normal curve centered at 13 and shade the area above 28; for z make a
Normal curve centered at zero and shade the area between 3.39 and 2.68. Since the diagrams show an
area above the mean that does not touch the mean, you subtract
3  13 
  3  13
z
 P 2.68  z  1.79   P2.68  z  0  P1.79  z  0
5. P3  x  3  P 
5.6 
 5.6
 .4963  .4663  .0300
For x ~ N 13, 5.6 make a Normal curve centered at 13 and shade the area between -3 and 3, both of which
are below 13; for z make a Normal curve centered at zero and shade the area between -2.68 and -1.79.
Since the diagrams show an area below the mean that does not touch the mean, you subtract.
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251y0641 5/10/06
x ~ N 13, 5.6
6. x.23 (Find z .23 first) .
Solution: Make a diagram. The diagram for z will show an area with a probability of 100 - 23% =
77%. below z .23 . The area below z .23 is split by a vertical line at zero into two areas. The lower one has
a probability of 50% and the upper one a probability of 77% - 50% = 27%. The upper tail of the
distribution above z .23 has a probability of 23%, so that the entire area above 0 adds to 27% + 23% =
50%. From the diagram, we want one point z .23 so that Pz  z 23   .77 or P0  z  z 29   .2700 . If
we try to find this point on the Normal table, the closest we can come is P0  z  0.74   .2704 , but
P0  z  0.73   .2673 not as close, but is acceptable. So z .23  0.74
Since x ~ N 13, 5.6 , the diagram for x would show 77% probability split in two regions on
either side of 13 with probabilities of 50% below 13 and 27% above 13 and below x.23 , and with 23%
above x.23 . z .23  0.74 , so the value of x can then be written x.23    z.23
 13  0.745.6  13  4.144  17.144 .
To check this:
17 .144  13 

Px  17.144   P  z 
  Pz  0.74   Pz  0  P0  z  0.74 
5.6


 .5000  .2704  .2296  .23.
7. A symmetrical region around the mean with a probability of 23%. [14, 14]
Solution: Make a diagram. The diagram for z will show a central area with a probability of 23%. It is
split in two by a vertical line at zero into two areas with probabilities of 11.5%. The tails of the
distribution each have a probability of 50% - 11.5% = 38.5%. From the diagram, we want two points z .385
and z .615 so that Pz.615  z  z.385   .2300 . The upper point, z .385 will have
P0  z  z .385  
23 %
 .1150 , and by symmetry z .615   z .385 . From the interior of the Normal table the
2
closest we can come to .1150 is P0  z  0.29   .1141 , which is slightly too low. The next best point
would be 0.30 since P0  z  0.30   .1179 . We can say z .385  0.29 , and our 23% symmetrical interval
for z is -0.29 to 0.29.
Since x ~ N 13, 5.6 , the diagram for x (if we bother) will show 23% probability split in two
11.5% regions on either side of 13, with 38.5% above x.385 and 38.5% below x.855 . The interval for x
can then be written x    z .385  13  0.295.6  13  1.624 or 11.376 to 14.624.
To check this:
14 .624  13 
11 .376  13
z
P11 .376  x  14 .624   P 
  P 0.29  z  0.29 
5.6
5.6


 2P0  z  0.29   2.1141   .2281  23%
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II. (10 points+, 2 point penalty for not trying part b.) Show your work! Mark individual sections clearly.
Wynn, Anthony and Avronovic give us the following data for a sample of eight professional golfers. Ea or
x  is earnings in thousands of dollars and SA or  y  is average score.
Row
1
2
3
4
5
6
7
8
Ea x 
71.6
55.8
147.4
117.4
112.3
82.7
22.8
58.6
SA  y 
71.50
72.75
71.34
71.27
70.95
71.65
72.49
71.46
In order to speed things up, I have computed the sum of the first seven observations.
7

7
x  610.000,
i 1

7
y  501.950,
i 1

7
x 2  63720.1,
i 1

7
y 2  35996.0 and
i 1
 xy  43607.4.
i 1
Calculate the following:
a. The sums that you will need to calculate whatever parts you do. (1 point if you don’t quit at b)
Make sure that I can tell how you did these sums.
b. The sample standard deviation s y of average score. (2)
c. The sample covariance s xy between x and y . (2)
d. The sample correlation rxy between x and y . (2)
e. Given the size and sign of the correlation, what conclusion might you draw on the relation
between x and y ? (1) Can you guess why the correlation isn’t stronger?
f. Assume that the earnings of the golfers were 15% lower ( w  .85 x ). Find w (the sample mean
of earnings), s w2 , s wy and rwy . Use only the values you computed in a-d and rules for functions of
x and y to get your results. If you state the results without explaining why, or change x1 and x 2
and recompute the results, you will receive no credit. (4).
g. Do an 80% confidence interval for the population average score of professional golfers. (2)
[14, 28]
Solution: Here is the table that you should have generated.
Row
Ea x  SA  y 
x2
xy
y2
1
2
3
4
5
6
7
8
71.6 71.50
55.8 72.75
147.4 71.34
117.4 71.27
112.3 70.95
82.7 71.65
22.8 72.49
58.6 71.46
668.6 573.41
5126.6 5112.25
3113.6 5292.56
21726.8 5089.40
13782.8 5079.41
12611.3 5033.90
6839.3 5133.72
519.8 5254.80
3434.0 5106.53
67154.1 41102.57
5119.4
4059.5
10515.5
8367.1
7967.7
5925.5
1652.8
4187.8
47794.9
Of course, you we only responsible for the last two lines.
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 x  668.6,  y  573.41,  x
 xy  47794.9 and n  8.
2
y
 67154.1,
2
 41102.57,
b. The sample standard deviation s y of average score comes from the variance.
 y  573 .41  71.67625 and s   y
y
2
y
n
8
2.69919

 0.3456
7
 ny 2
2

n 1
41102 .6  871 .67625 2

7
s y  .6210
 x  668 .6  83.575
x
67154 .1  883 .575 2
 1610 .84
n
8
n 1
7
s x  40.1352 Your answers will differ due to rounding error.
c. The sample covariance s xy between x and y is
x
s xy 
s x2 
2
 nx 2

 xy  nxy  47794 .9  883.575 71.67625   18.2584 .
n 1
7
d. The sample correlation rxy between x and y is
rxy 
s xy
sx s y
 18 .2584

 0.7326
1610 .84 0.3456
e. The best way to look at the strength of a correlation is to compute rxy2  .536 and to look at it on
a zero to one scale. It’s not particularly weak or strong. The negative sign is expected because
golfers win with low scores. However, correlations track linear relationships and there is no reason
to expect that the relationship between scores to be linear. We would, in fact expect much higher
earnings for low scores than could be explained by a straight line, because winners get large prizes.
f. Assume that the earnings of the golfers were 15% lower ( w  .85 x ). Find w (the sample mean
of earnings), s w2 , s wy and rwy . Use only the values you computed in a-d and rules for functions of
x and y to get your results. If you state the results without explaining why, or change x1 and x 2
and recompute the results, you will receive no credit. (4).
Using formulas from 251var2, when w  ax  .85 x and v  cy  1y . Recall x  83 .575 ,
s x2  1610.84 , s xy  18.2584 and rxy  0.7326 .
i) Just as E ax  aEx  , w  .85 x  .8583.575   71.039
ii) Varax  a 2Varx  .852 1610.84  1163 .83
iii) Covw, v   acCovx, y   .85 118.2584   15.5196
iv) Corr w, v   Corr ax, cy   SignacCorr x, y   .7326
g. Do an 80% confidence interval for the population average score of professional golfers. (2)
Recall that y  71.67625 s 2y  0.3456 and remember that when  is unknown   x  tn1 s x ,
2
sx
where s x 
n
, or s x 
sx
n
N n
when the sample is more than 5% of the population.
N 1
7
  1  .80  .20 . t n1  t.10
 1.415 and s x 
2
sx
n

0.3456
 0.0432  0.2078
8
 y  y  t n1 s y  71.676  1.4150.2078  71.676  0.294 or 71.381 to 71.970.
2
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III. Do at least 5 of the following 6 sections (at least 12 each) (or do items adding to at least 48 points Anything extra you do helps, and grades wrap around) . Show your work! Please indicate clearly what
sections of the problem you are answering! If you are following a rule like E ax  aEx  please state it! If
you are using a formula, state it! If you answer a 'yes' or 'no' question, explain why! If you are using the
Poisson or Binomial table, state things like n , p or the mean. Avoid crossing out answers that you think
are inappropriate - you might get partial credit. Choose the problems that you do carefully – most of us are
unlikely to be able to do more than half of the entire possible credit in this section!) This is not an opinion
questionnaire. Answers without reasons or supporting calculations or table references will not be
accepted (except in multiple choice questions)!!!! Answers that are hard to follow will be penalized.
Note that some sections extend over more than one page.
A. Answer the following 6 multiple choice questions. (These should be 2 each, but to discourage guessing,
how about 2.5 each for right answers and 0.5 penalty for wrong answers.)
1.
The t distribution should be used when the parent (underlying) population
a) *Is Normal, the population standard deviation is unknown and we are testing a mean.
b) Is Normal, the population standard deviation is known and we are testing a mean.
c) Is Normal, the mean of the population is unknown and we are testing a mean.
d) Is binomial and we are testing for a proportion.
e) The t distribution should be used in all of these cases.
2.
(Was 5) It is desired to estimate the average total compensation of CEOs in the Service industry.
Data were randomly collected from 18 CEOs and the 97% confidence interval was calculated to be
($2,181,260, $5,836,180). Which of the following interpretations is correct?
a) 97% of the sampled total compensation values fell between $2,181,260 and $5,836,180.
b) We are 97% confident that the mean of the sampled CEOs falls in the interval $2,181,260
to $5,836,180.
c) In the population of Service industry CEOs, 97% of them will have total compensations
that fall in the interval $2,181,260 to $5,836,180.
d) *We are 97% confident that the average total compensation of all CEOs in the Service
industry falls in the interval $2,181,260 to $5,836,180.
3.
(Was 9) In the construction of confidence intervals, if all other quantities are unchanged, an
increase in the sample size will lead to a __________ interval.
a) *narrower
b) wider
c) less significant
d) biased
4.
(Was 13) For air travelers, one of the biggest complaints is of the waiting time between when the
airplane taxis away from the terminal until the flight takes off. This waiting time is known to have
a skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes.
Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean
waiting time between when the airplane taxis away from the terminal until the flight takes off for
these 100 flights.
a) Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes.
b) Distribution is skewed-right with mean = 10 minutes and standard error = 8 minutes.
c) *Distribution is approximately normal with mean = 10 minutes and standard error = 0.8
minutes.
d) Distribution is approximately normal with mean = 10 minutes and standard error = 8
minutes.
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5.
(Was 16) Why is the Central Limit Theorem so important to the study of sampling distributions?
a) It allows us to disregard the size of the sample selected when the population is not normal.
b) It allows us to disregard the shape of the sampling distribution when the size of the population is
large.
c) It allows us to disregard the size of the population we are sampling from.
d) *It allows us to disregard the shape of the (parent) population when n is large.
6.
(Was 21) What type of probability distribution will the consulting firm most likely employ to
analyze the insurance claims in the following problem?
An insurance company has called a consulting firm to determine if the company has an
unusually high number of false insurance claims. It is known that the industry proportion for
false claims is 3%. The consulting firm has decided to randomly and independently sample
100 of the company’s insurance claims. They believe the number of these 100 that are false
will yield the information the company desires.
a) *binomial distribution.
b) geometric distribution
c) continuous uniform distribution
d) Poisson distribution.
e) hypergeometric distribution.
f) none of the above.
[15]
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B. A random sample of 36 Kleenex users taken at a college found that the average sneezer used 52.10
tissues in the course of a cold. The researcher previously believed that the average was 59. Do a 95%
confidence interval for the population mean and answer if the population mean is significantly different
from 59.00 for parts a-d. Note that each number here is stated to the nearest hundredth. Please maintain at
least that level of significance throughout the problem. Mark your individual questions clearly!!!
a. Assume that the population standard deviation is known to be 20.95. (4)
b. Assume, more realistically, that 20.95 was a sample standard deviation. (4)
c. Now assume that the sample of 36 was drawn from a population of 401. (4)
d. Assume that the population standard deviation is known to be 20.95 but that we want a
confidence level of 96% (You cannot use the t-table for this part.).(4)
e. Assume that the researcher was right and that the average number of Kleenex used by a sneezer
has a population mean of 59 and a population standard deviation of 20.95. Assume that the
researcher has 2250 tissues on hand for the 36 subjects. What is the probability that the researcher
runs out of tissues? (4) [35]
Solution: From the introduction to problem O1. We are using the following formulas.
When  is known   x  z  x , where  x 
2
x
n
, or  x 
x
n
N n
when the sample is
N 1
more than 5% of the population.
s
s
N n
When  is unknown   x  tn1 s x , where s x  x , or s x  x
when the sample is
2
n
n N 1
more than 5% of the population. For a)-c),   1  .99  .01 .
a) We find a 95% confidence interval for the mean number of tissues assuming that
x  52 .10 ,   20 .95, n  36 and N is large. Because we are given  we should use

20 .95
 3.4917 z  z.025  1.960
z . Because of the large population  x  x 
2
n
36
  x  z  x  52.10  1.9603.4917  52.10  6.8437 so
2
P45 .26    58 .94   .95 . Note that 59.00 is not on the confidence interval, so that
our sample mean is significantly different from 59.
b) We find a 95% confidence interval for the mean number of tissues assuming that
x  52 .10 , s  20 .95, n  36 and N is large. Because we are given s we should use
s
20 .95
 3.4917 t n1  t .35
t . Because of the large population s x  x 
025  2.030
2
n
36
  x  t n1 s  52.10  2.0303.4917  52.10  7.08 so

2
x
P45 .02    59 .18   .95 . Note that 59.00 is on the confidence interval, so that our
sample mean is not significantly different from 59.
c)
We find a 95% confidence interval for the mean processing time assuming that
x  52 .10 , s  20 .95, n  36 and N  401 . Because we are given s we should use t .
Because of the small population use a finite population correction.
s
N  n 20 .95 401  36
365
sx  x

 3.49167
 3.48167 .95525   3.3259 .
401  1
400
n N 1
36
t n1  t 35  2.030   x  t n1 s  52.10  2.0303.3259  52.10  6.75 so

2
.025
P45 .35    58 .85   .95

2
x
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d) We find a 96% confidence interval for the mean number of tissues assuming that
x  52 .10 ,   20 .95, n  36 and N is large. Because we are given  we should use

20 .95
 3.4917 To find the value of z  that we need,
z . We still have  x  x 
2
n
36
note that   1  .96  .04 , so that z   z.02 . Make a diagram. Draw a horizontal
2
line at zero. There is 50% of the distribution below zero and 2% over z  , which
2
implies that there is 48% between z  and zero. If we use table 17, we see that
2
P0  z  2.05   .4798 and P0  z  2.06   .4803 . So both 2.05 and 2.06 are
acceptable values of z ; 2.054 seems about right to me. .
  x  z 2  x  52.10  2.0543.4917  52.10  7.17 or
P44 .93    59 .27   .95 . Note that 59.00 is on the confidence interval, so that our
sample mean is not significantly different from 59.
e)
If we assume that the researcher was right and that the average number of
Kleenex used by a sneezer has a population mean of 59 and a population standard
deviation of 20.95, the sample mean for a sample of 36 has a Normal distribution

20 .95
 3.4917 . 2250 tissues
with a mean of 59 and a standard error of  x  x 
n
36
is 62.5 tissues for each of the 36 subjects.
62 .5  59 

Px  62 .5  P  z 
 Pz  1.00   .5  .3413  .8413 .
3.4917 

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C. Assume that P A  .25 and PB   .70 . (Making joint probability tables will help. – do not round
excessively, you should retain at least 3 figures to the right of the decimal point.)
1. Find (i) P A  B  , (ii) P A  B (iii) P B A if


 
a. A and B are independent. (3)
b. A and B are mutually exclusive. (3)
c. P A B  .20 (3)
 
B B
sum


A
.25


Solution: In each case start with
and use the addition rule, which states
A 
.75



sum .70 .30 1.00
 
P A  B   P A  PB   P A  B  , and the multiplication rule, which states P A  B  P A B PB,
P A  B 
P A  B 
or P B A 
.
P B 
P A
a) If A and B are independent, (i) P A  B   P A PB   .25.75   .175 . If we simple fill in the
P A  B  PB AP A , P A B  
 
B B
sum
.175 .075  .25
blanks so that the numbers add up, we get
.


A .525 .225  .75
sum .700 .300 1.00
A


 
(i) P A  B   .25  .70  .175  .775 . (ii) We can read P A  B  .525 from the table. (iii) P B A
 PB   .70.
b) If A and B are mutually exclusive, by definition P A  B   0. If we simple fill in the blanks so
B
 0
that the numbers add up, we get

A .70
sum .70
A


P A  B  .70 , (iii) PB A 
 
B
sum
.25  .25
. (i) P A  B   .25  .70  0  .95 , (ii)

.05  .75
.30 1.00
P A  B 
0

0.
P A
P A
 
c) If P A B  .20 , P A  B  P A B PB  .20  .70  .14 . If we simple fill in the blanks so that the
numbers add up, we get
B B
sum
A .14 .11 .25
. (i) P A  B   .25  .70  .14  .81 , (ii) P A  B  .56 ,


A .56 .19  .75
sum .70 .30 1.00


P A  B  .14

 0.56 .
P A
.25
2. If P A  .40 and PB   .70 , show that A and B cannot be mutually exclusive. (3)
Solution: If A and B are mutually exclusive, by definition P A  B   0.
(iii) PB A 
P A  B   .40  .70  0  1.10 . Since a probability cannot be above 1, this is impossible.
9
251y0641 5/10/06
3. At a Pennsylvania state college, 60% of freshmen come from Eastern Pennsylvania, 30% from
Western Pennsylvania, and 10% from out of state. They are given a math test during freshman week.
70% of the students from Eastern Pennsylvania pass it as do 60% from from Western PA and 90%
from out of state. Tara Bulsnob comes from Eastern PA and has passed the math test and will only be
friends with someone else from Eastern PA who has passed the math test..
a. If she picks someone at random, what is the chance that that person has passed the math
test? (2)
b. If she picks someone who has passed the math test at random, what is the chance that
Tara will be willing to be friends with that person? (3)
[52]
Solution: Define the following events: EP  , Being from Eastern PA; WP  , Being from Western PA;
OS  , Being from out of state and PM  , Passing the math test.
We have been given PEP   .60 , PWP   .30 , POS   .10 , PPM EP  .70 , PPM WP  .60


and P PM OS  .90 .
Assume that we have 100 students. Since we know PEP   .60 , PWP   .30 and POS   .10 , we
EP WP OS total
can divide the 100 students as follows.


PM
PM
total


. Since P PM EP  .70 , 70% of

60 30 10

100
60 is 42. Similarly P PM WP  .60 and P PM OS  .90 give us numbers of people who passed in
each group.
PM
PM
EP WP
42 18
OS
9
total
. If we add these and fill in the remainding numbers, we get
total
60 30 10
100
EP WP OS total
PM
42 18 9
69
. So a) If she picks someone at random, the chance that that person has
PM
18 12 1
31
total 60 30 10
100
69
 .69 . b) If she picks someone who has passed the math test at
passed the math test is PPM  
100
42
 .6086 .
random, the chance that Tara will be willing to be friends with that person is PEP PM  
69
10
251y0641 5/10/06
D. If x is binomial, and n  11, find solutions to 1-5 (If you substitute another distribution for the
binomial, or any other distribution, justify it!):
1) P3  x  6 when p  .35 . (2) For any discrete distribution for integer values of x ,
P3  x  6  F 6  F 2 . For the binomial distribution,
F 6  F 2  Px  6  Px  2  .94986  .20013  .7497
2)
3)
P5  x  8 when p  .65 . (2) P5  x  8 can be expressed as the probability of between 11 –
8 = 3 failures and 11 – 5 = 6 failures, when the probability of failure is 1 - .65 = .35.
P3  x  6  F 6  F 2  Px  6  Px  2  .94986  .20013  .7497
P2  x  7 when p  .465 . (3 or 3.5) If n  11, the expected number of successes is
  np  11.465   5.115 and the expected number of failures is 11 – 5.115 = 5.885. Since both of
these are (barely) above, we can use the Normal distribution.
  npq  5.115 .535   2.736525  1.6524
.
If we use the Normal distribution with a continuity correction we have
7.5  5.885 
1.5  5.885
P1.5  x  7.5  P 
z
1
.
6524
1.6524   P2.65  z  0.98 

 P2.65  z  0  P0  z  0.98   .4960  .3365  .8325
Of course, if we do not use the continuity correction, we get
7  5.885 
 2  5.885
P2  x  7   P 
z
 P 2.35  z  0.67 
1.6524 
 1.6524
 P2.35  z  0  P0  z  0.67   .4906  .2486  .7392
4)
5)
P1  x  3 when p  155 . (2) If n  11 , n p  1155  550 .
So, since this is above 500, we can use the Poisson distribution with a parameter of
 1 
np  11   0.2 . P1  x  3  Px  3  Px  0  .99994 - .81873 = .1812
 55 
Px  1 when p  .17 . (2)
 1  .12878  .8712 .
Px  1  1  P0  1  C 011 p 0 q11  1  q11  1  .83 11
6) If x follows the continuous uniform distribution with c  11 and d  17 , find P3  x  14  . (2)
Make a diagram. Represent the distribution by a rectangle by a rectangle with a base from 11 to 17
and a height of 1c d   11711  16 . Shade the area in the rectangle below 14. (There is no area
between 3 and 11.) This is the area with base 14 – 11 = 3. The area is 3 1 6   0.5000 .
7) If x follows the Poisson distribution with a parameter of 1.1 find P3  x  15  (2)
P3  x  15   Px  15   Px  2  1  .90042  .0996
8) If x follows the Poisson distribution with a parameter of 37 find P15  x  45  (2 or 2.5)
If we use the Normal distribution with a continuity correction we have   15  3.8730 .
45 .5  15 
14 .5  15
P14 .5  x  45 .5  P 
z
 P 0.13  z  7.88  
3.8730 
 3.8730
 P0.13  z  0  P0  z  7.88   .0517  .5000  .5517
45  15 
 15  15
z
 P0  z  7.75   .5000 se
Of course, if we do not u P15  x  45   P 
3
.
8730
3
.8730 

the continuity correction, we get.
11
251y0641 5/10/06
9) If x follows the Hypergeometric distribution with N  380 , M  133 and n  11, find
P2  x  7 (2) Since the population size is more than 20 times the sample size, we can use the
M
 .35 .
binomial distribution p 
N
[71]
P3  x  6  F 6  F 2  Px  6  Px  2  .94986  .20013  .7497
10) (Extra Credit) Find P15  x  45  for an exponential distribution with c  137 , and explain the
relation of this problem to 8) above. (3)
1
1
1
F x  1  ecx , when the mean time to a success is . So if  37 , c  . The average number
c
37
c
of occurances in a given unit of time is 37 and would be used in a Poisson problem. If an average
1
. So
of 37 events occur in one minute, the average wait for the first event to occur is c 
37


Px1  x  x 2   F x 2   F x1   1  e cx2  1  e cx1  e cx1  e cx2 . Note that if
x1  15, cx1 
15
45
 .4054 and if x 2  45, cx 2 
 1.2162
37
37
So P15  x  45  e 0.4054  e 1.2162  .66671 .29635  .3704
12
251y0641 5/10/06
E. Assume that we are considering buying two stocks. For the first of these two stocks  x  1.35 and
 x  3.5 . For the second stock  y  2.00 and  y  8.0 . Mark individual questions and parts of
questions clearly. (Note: y appears as v in some equations due to a bug in Word formatting: it prints correctly)
1. Assume that the amounts above are in dollars per year and that you buy one share of each stock.
Show the mean, standard deviation and coefficient of variation of your return x  y  if a)  xy  .8 , b)
 xy  .0 and c)  xy  .8 . (8.5)
2. Assume that some time in the future the first stock doubles in value, so that its return is now
w  2 x and the second stock rises by 50% so that its return is now v  1.5 y . Find  v ,  wv ,  w v and
[83.5]
 wv if  xy  .8 (4)
3. Again  x  1.35 ,  x  3.5 ,  y  2.00 and  y  8.0 . Assume that you have one dollar to
invest, so that you will buy P1 shares of stock 1 and P2 shares of stock 2, where P1  P2  1 . Then
R  P1 x  P2 y is your return and  x  1.35 ,  x  3.5 ,  y  2.00 and  y  8.0 . Show how  xy affects
the minimum risk point, by computing coefficients of variation for R  P1 x  P2 y for P1 equal to 0, .25,
.50, .75 and 1 (You have already done this for zero and one) and for both  xy  .8 and  xy  .8 . This will
give you a table with 10 values, some of which are duplicates. Comment on this and from what you learned
on the last exam and try to identify ‘no fly’ zones, if there are any. (10+)
Solution: From the solution to grass1, we had Ex  y   Ex  E y    x   y and
Var x  y    x2   y2  2 xy  Varx   Var y   2Covx, y  . If we check 251varmin, we find
 xy   xy  x y and the coefficient of variation case C 
1) For all sections the mean is Ex  y    x y
R
.
E R 
  x   y  1.35  2.00  3.35 . Recall that  x  3.5 and
 y  8.0
Recall that  x  3.5 and  y  8.0
 x2 y  3.5 2  8.0 2  2 xy  12 .25  64  2 xy  76 .25  2 xy
1a)  xy  .8  xy   xy  x y  .83.58.0  22.4  x2 y  76 .25  222 .4  31 .45
 x y  31.45  5.6080 C 
5.6080
 1.674
3.35
1b)  xy  0  xy   xy  x y  03.58.0  0  x2 y  76 .25  x  y  76.25  8.7321
C
8.7321
 2.607
3.35
1c)  xy  .8  xy   xy  x y  .83.58.0  22.4  x2 y  76 .25  222 .4   121 .05
 x y  121.05  11.0023 C 
11 .0023
 3.284
3.35
2) Using formulas from 251var2, when w  ax  2 x and v  cy  1.5 y . Recall  xy  22.4 and
 xy  .8 .
1a) Varcx  c 2Varx   1.5 2 8.02 , so  v  1.58  12
1b) Covw, v   acCovx, y   21.522.4  67.2
1c) Varax  cy   a 2Var x   2acCovx, y   c 2Var  y   2 2 3.52  221.5 22 .4  1.5 2 8.02
 49  134 .4  144  58.60 .  wv  58 .60  7.6551
1d) Corrw, v   Corrax, cy   SignacCorrx, y   .8.
13
251y0641 5/10/06
3) If we check 251varmin, we find the following. “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  is the variance of
the return. Thus if P1 and P2 are both .50 , we can say VarR =.25VarR1 +.25VarR2 +.50CovR1 R2  .
…… Since variance is a measure of risk, minimizing variance minimizes risk, though actually, the best
measure of risk is probably the coefficient of variation, the standard deviation divided by the mean, in this
case C 
R
E R 
.”  R  Var R 
Recall that  x  1.35 ,  y  2.00 ,  x  3.5  y  8.0 . So that  x2  12.25 and  y2  64
3a)  xy  .8 and  xy  22.4 .
VarR  P12VarR1   P22VarR2   2P1 P2CovR1 , R2 
E R 
Row
1
2
3
4
5
1.0
1.35
12.25
0.75 .75(1.35)+ .25(2)= 1.5125 .5625(12.25)+ .0625(64)-.375(22.4)= 2.4906
0.5 .5(1.35)+ .5(2)= 1.675
.25(12.25)+ .25(64)-.50(22.4)= 7.8625
0.25 .25(1.35)+ .75(2)= 1.7375 .0625(12.25)+ .5625(64)-.375(22.4)= 28.3656
0
2.00
64
Row
1
2
3
4
5
1.0
0.75
0.5
0.25
0
P1
P1
E R 
1.35
1.5125
1.675
1.7375
2.00
R
CR
3.5
1.578
2.804
5.326
8.0
2.59
1.04
1.67
3.07
4.00
3a)  xy  .8 and  xy  22.4 .
VarR  P12VarR1   P22VarR2   2P1 P2CovR1 , R2 
E R 
Row
1
2
3
4
5
1.0
1.35
12.25
0.75 .75(1.35)+ .25(2)= 1.5125 .5625(12.25)+ .0625(64)+.375(22.4)= 19.2906
0.5 .5(1.35)+ .5(2)= 1.675
.25(12.25)+ .25(64)+.50(22.4)= 30.2625
0.25 .25(1.35)+ .75(2)= 1.7375 .0625(12.25)+ .5625(64)+.375(22.4)= 45.1656
0
2.00
64
Row
1
2
3
4
5
1.0
0.75
0.5
0.25
0
P1
P1
E R 
1.35
1.5125
1.675
1.7375
2.00
R
CR
3.5
4.392
5.501
6.720
8.0
2.59
2.90
3.28
3.87
4.00
With the negative covariance, putting 75% in stock 1 lowers risk, creating a ‘no fly’ zone between 100%
and 75%, where a lower return gives higher risk. Even the 50-50 split gives better results than all in stock 1.
But now we are in a ‘normal’ region where accepting higher risk yields higher return. With the positive
covariance, there seems to be no ‘no fly’ zone, accepting increasing risk always yields higher return.
14
251y0641 5/10/06
F. (BLK 38) Assume that the amount of gasoline purchased per car at a large service station has an
approximately Normal distribution with a mean of $15 and a standard deviation of $4. Find the following.
Mark individual questions clearly!
1. The probability that a given car will purchase between $14 and $16 worth of gas. (2)
2. The probability that a random sample of 16 cars will have a sample mean between $14 and $16
(3)
3. The probability that all 16 cars in the sample will purchase an amount between $14 and $16. (2)
4. The probability that at least one of the 16 cars will purchase an amount between $14 and $16.(2)
5. The 95th percentile of the distribution of the sample mean for the 16 cars. (1.5)
6. The 10th percentile of the distribution of the sample mean for 16 cars. (1.5)
[95.5]
4
Solution: Note that x ~ N (15,4) . This means that for a sample of 16, we will have x ~ N (15,
) or
16
x ~ N (15,1)
16  15 
14  15
z
 P 0.25  z  0.25 
1. P14  x  16   P 
4
4 

 P0.25  z  0  P0  z  0.25   .0987  .0987  .1974
For x ~ N (15,4) make a Normal curve centered at 15 and shade the area from 14 to 16; for z make a
Normal curve centered at zero and shade the area from -0.25 to 0.25. Since the diagrams show an area on
both sides of the mean, you add.
16  15 
14  15
z
 P 1.00  z  1.00 
2. P14  x  16   P 
1
1 

 P1.00  z  0  P0  z  1.00   .3413  .3413  .6826
For x ~ N (15,1) make a Normal curve centered at 15 and shade the area from 14 to 16; for z make a
Normal curve centered at zero and shade the area from -0.25 to 0.25. Since the diagrams show an area on
both sides of the mean, you add.
3. This is a binomial problem. Let y be the number of cars that purchse an amount between $14 and $16.
Then, assuming independence, y will have a binomial distribution with n  16 , p  .1974 and
16 16 0
p q  p16  .1974 16  5.31  10 12
q  1  .1974  .8026. Px  C xn p x q n x . So P16   C16
4. This is also a binomial problem. y still has a binomial distribution with n  16 , p  .1974 and
q  1  .1974  .8026. Px  C xn p x q n x . So
Px  1  1  P0  1  C 016 p 0 q16  1  q16  1  .8026 16  1  .02965  .9704
4. x ~ N (15,1) . The 95th percentile is x.05 and has 95% of the data below it. According to the t table,
z .05  1.645 . If we employ the usual transformation x    z , we conclude that
x.05  15  1.645 1  16.645 .
5. x ~ N (15,1) . The 10th percentile is x.90 and has 10% of the data below it. Because the mean of x is 15,
x.90 is below 15. Because the standardized Normal distribution is summetrical about zero,
z .90   z10 . According to the t table, z .10  1.282 . This means that z .90  1.282 . If we employ the usual
transformation x    z , we conclude that x.90  15  1.282 1  13.718 .
15
251y0641 5/10/06
ECO 251 QBA1
Name
FINAL EXAM, Version 1
Class _________________
MAY 8, 2006
Student Number _____________
Supplementary Correlation Problem
This problem is an edited version of a problem due to Ben-Horim and Levy. (14 points)
A textile firm operates in two companies, the United States x  and Japan  y  . Its anticipated profits for
2007 are approximated by the following joint distribution. Please turn in any scratch paper that you use.
x
2
y
4
12
Px 
2
 .24

 .08
 0

.32
2
.06
.25
.08
.39
6
0

.11
.18 

.29
P y 
.30
.44
.26
1.00
a. Find the expected net profit for both countries and the countries combined. (1.5)
b. Find the (population) standard deviation of profits for one or both countries (2)
c. Find the (population) covariance and correlation between profits in both countries (3)
d. Comment on the strength and the sign of the results and explain from general economic knowledge why
you would be very surprised to get a different sign. (1.5)
e. What is the standard deviation of the firm’s total profit? (2)
f. To verify that the firm’s mean and standard deviation are the values that you presumably got above by
using formulas that you learned in class, note that, though there are 9 joint probabilities, x  y can only
take 8 values. Fill in the following table and compute Ex  y  and  x y using it. If there is a discrepancy,
find your error. (4 if they agree)
x y
Px  y 
x  y Px  y 
 x  y  2 P x  y 
-4
0
2
4
6
10
14
18
16
251y0641 5/10/06
Solution: To compute the means and variances we do the following tableau.
x
2
 .24

 .08
 0

.32
2
y
4
12
Px 
xPx 
x 2 Px 
2
.06
6
0

.11
.18 

.29
.25
.08
.39
 0.64  0.78 
1.74 
P y  yP y 
.30  0.60
y 2 P y 
1.20
.44
1.76
7.04
.26
3.12
37 .44
1.00
4.28
45 .68
1.88
1.28  1.56  10 .44  13 .28
 
 
This tells us Ex   1.88 , E x 2  13.28 , E  y   4.28 , E y 2  45.68 . We also need to do
0 
.24  2 2  .06 2 2  06 2    0.96   0.24 



E xy    .08 24 
.2524  .1164     0.64 
2.00  2.64    19 .6 .
 0 212   .08212   .18612  
0
1.92  12 .96 
a. The expected net profit for both countries and the countries combined is Ex   1.88 , E y   4.288 and
Ex  y   Ex  E y    x   y  1.88  4.28  6.16 .
b. The (population) standard deviation of profits for both countries is found by computing the square root
of the variance.
 
 x Px  
 Var y   E y     y P y   
 x2  Varx  E x 2   x2 
2
 y2
2
2
2
y
 13.28  1.882  9.74560 ,  x  9.7456  3.12179 ,
2
x
2
y
 45.68  4.282  27.3616 and  y  27.3616  5.23083.
c. The (population) covariance and correlation between profits in both countries is found from E xy  .
 xy  Covxy  Exy   x  y  19.6  1.884.28  11.5536 .
So that  xy 
 xy
 x y
11 .5536

 .707526 .
9.7456 27 .3616
2
 .5005 This can be termed
d. The strength of the correlation is best measured on a zero to one scale by  xy
neither weak or strong. The sign of the results is positive, indicating that the profits in the two countries tend
to move together as we would expect in an increasingly internationalized economy where world-wide
prosperity and depression are major factors in profits of transnational firms.
e. The standard deviation of the firm’s total profit is found from the variance of the total? (2)
Var x  y    x2   y2  2 xy  Var x   Var  y   2Covx, y   9.74560  27.3616  211.5536   60.2144
 x  y  60.2144  7.7597 .
Row
x  y Px  y 
1 -4
2
0
3
2
4
4
5
6
6 10
7 14
8 18
Total
0.24
0.06
0.08
0.00
0.25
0.11
0.08
0.18
1.00
x  y Px  y  x  y 2 Px  y 
-0.96
0.00
0.16
0.00
1.50
1.10
1.12
3.24
6.16
3.84
0.00
0.32
0.00
9.00
11.00
15.68
58.32
98.16
So Ex  y   6.16 and Varx  y   98.16  6.16 2  60.2144
17
251y0641 5/10/06
Extra Credit Problem Version 1
(Stevens, modified) A sample of 350 parents of children 19 to 24 months is taken. They were asked the
following question:
Did your child consume a hot dog today? 110 said yes.
According to a USA Today Article the national percentage for hot dogs was 25%
1) Using the information above, check to see if the percent consuming the food was significantly different
from the percent given in the USA Today Article by creating a 95% confidence interval for the proportion.
(3.5)
2) (Extra extra credit!) Assume the above results came from a daycare system where the population of
parents of children attending is only 1400. Do part 1) again with this additional information. (3.5)
x
Solution: 1) Let p  and q  1  p be the sample quantities. Remember that a proportion is always
n
between zero and one. For the interval write p  p  z s p , where s p 
2
x  110 , so p 
pq
. We have n  350 and
n
x 110

 .3143 and q  1  .3143  .6857 . Our estimate of the standard error of the
n 350
.3143 .6857 
 0.0006157  .02481 . If 1    .95 ,
350
 1.960 .
proportion is s p 

2
 .05 2  .025 . According to
the t table, z .025
Thus our confidence interval is p  .3143  1.960 .02481   .3143  .0486 . So we can say
P.2657  p  .3629   .95. Since .25 is not on this interval, we can say that the proportion found in this
survey was significantly different from .25. Just to give myself some credibility, I ran this on Minitab and
got the following.
Version 1
MTB > POne 350 110;
SUBC>
Test 0.25;
SUBC>
UseZ.
Test and CI for One Proportion
Test of p = 0.25 vs p not = 0.25
Sample
1
X
110
N
350
Sample p
0.314286
95% CI
(0.265651, 0.362921)
Z-Value
2.78
P-Value
0.005 .0248
2) For the interval we still write p  p  z s p , but we have N  1400 , so that our sample is more than one
2
twentieth of the population. Then we must use the finite population correction and s p 
have n  350 and x  110 , so p 
N n
N 1
pq
. We
n
x 110

 .3143 and q  1  .3143  .6857 . Our estimate of the standard
n 350
error of the proportion is
1400  350 .3143 .6857 
sp 
 0.700467 0.0006157  000431  .020768 . Thus our confidence
1400  1
350
interval is p  .3143  1.960 .020768   .3143  .0407 . So we can say P.2735  p  .3550   .95. Since .25
is not on this interval, we can still say that the proportion found in this survey was significantly different
from .25
18