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ECO251 QBA1
THIRD EXAM
NOVEMBER 21, 2000
Name _______key_________
Section Enrolled: MWF TR 10 11 12:30
Part I. Do all the Following (10+ Points) 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!
Assume that the following joint probability table corresponds to two variables x and y
x
0
1
2
0 .30 .10 .05 


y 2 .10 .20 .05 
4 .05 .05 .10 


a.
Are
P y 
.45
.35
.20
x and y independent? Why?(2)
c.
 x , the standard deviation of x . (2)
Compute  xy , the covariance of x and y , and interpret it. (3)
d.
Compute
e.
Find
b.
Compute
 xy , the correlation of x
and
y , and interpret it. (2)
f.
Px  y  4 (2)
x  y can take the following values: 0, 1, 2, 3, 4, 5 and 6. Write out the distribution of x  y
g.
by making a list with these values and their probabilities (2)
(i) Find the mean and standard deviation of x  y from the table you just made. (2)
(ii) Show that they are the same as the values of the mean and standard deviation of
x y
that you would have gotten using the method you used in the last part of Graded Assignment
2. (2)
Solution:
a) Check for independence: First you need to find Px  and P y  .
Look at the upper left hand
probability below. Its value is .20 and it represents Px  0   y  0 . If x and y were independent , we
would have Px  0   y  0  Px  0 P y  0  .45.45   .2025 . Since this is not true, we can say
that x and y are not independent..
x
0
y
2
4
Px 
xPx 
x 2 Px 
0
 .30

 .10
 .05

.45
1
.10
.20
.05
.35
2
.05 

.05 
.10 

.20
P y  yP y  y 2 P y 
.45 0.00
0.00
.35
0.70
1.40
.20
0.80
3.20
1.0
1.50
4.60
0.00 0.35 0.40
0.75
0.00 0.35 0.80
1.15
 Px   1 ,
  E x    xPx   0.75 ,
E x    x Px   1.15 ,  P  y   1 ,
  E  y    yP y   1.5 and
E y    y P y   4.6
To summarize
x
2
2
2
2
y
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E xy  

 .30 00  .10 10  .0520  0.0  0.0  0.0
xyPxy    .10 02  .20 12  .0522   0.0  0.4  0.2  1.6
 .0504  .0514  .10 24  0.0  0.2  0.8
 
 xy  Covxy  Exy   x  y  1.6  0.751.5  0.475 ,  x2  E x 2   x2  1.15  0.75 2  0.5875 and
 
 y2  E y 2   y2  4.6  1.52  2.35 . So that  xy 
 xy
 x y

0.475
 0.404255 .
0.5875 2.35
(  x  0.766485 ,  y  1.53297 )
b) Compute  x , the standard deviation of x . (2) Above, it says
 
 x2  E x 2   x2  1.15  0.75 2  0.5875 and  x  0.766485
c) Compute  xy , the covariance of x and y , and interpret it. (3) Above, it says
 xy  Covxy  Exy   x  y  1.6  0.751.5  0.475 Since this is positive x and y seem to move
together; if one rises, the other also tends to rise.
d) Compute  xy , the correlation of x and y , and interpret it. (2) Above, it says
 xy 
 xy

0.475
 0.404255 . If we square this correlation, we get .1634. Though the
0.5875 2.35
positive sign indicates that x and y seem to move together, .1634 is quite small on a zero to one scale, so
the correlation is rather weak.
 x y
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e) Find Px  y  4 (1) the original table is copied below, along with a second table with the sums of x
x
x
There are two ways to get 4: by adding x  0
0
1
2
0 1 2
0 .30 .10 .05 
0 0 1 2




y 2 .10 .20 .05 
y 2  2 3 4
4 .05 .05 .10 
4 4 5 6




and y  4 , or by adding x  2 and y  2 . The joint probability of the first pair is .05 and the joint
probability of the second pair is .05, so the total probability is Px  y  4  .05  .05  .10
f), g) x  y can take the following values: 0, 1, 2, 3, 4, 5 and 6. Write out the distribution of x  y by
making a list with these values and their probabilities (2) . (i) Find the mean and standard deviation of
x  y from the table you just made. (2) (ii) Show that they are the same as the values of the mean and
standard deviation of x  y that you would have gotten using the method you used in the last part of
Graded Assignment 2. (2) We put the probabilities together for each value of x  y in the same way we did
and y .
Px  y  4 . The table follows:
x  y Px  y 
0
.30

x  y Px  y  x  y 2 Px  y 
0
0
1
2
3
.10
.15
.20
.10
.30
.60
.10
.60
1.80
4
5
.10
.05
.40
.25
6
total
.10
1.00
.60
2.25
1.60
1.25
3.60


8.95
So Ex  y   2.25 and E x  y   8.95 . This means that Var x  y 

2
 E x  y   E x  y   8.95  2.25   3.8875 and  x  y  3.8875  1.9717 . If we compute
2
2
2
Ex  y  and Var x  y  from the results of b-d we get exactly the same results as we did above.
Ex  y   Ex  E y    x   y  0.75  1.5  2.25 and
Var x  y    x2   y2  2 xy  Var x   Var  y   2Covx, y   0.5875  2.35  20.475   3.8875
.
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Part II. Do the following problems. ( Do at least 40 points – but remember anything beyond 40 points is extra credit. ). Show your
work! Please indicate clearly what sections of a problem you are answering! And what columns you found table values in.. 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! You must do part a) of problem 3! (3 point penalty!)
1. Describe the meaning and give the probabilities for the problems below.
Example: For the Hypergeometric Distribution, P1  x  2 when N  18, n  5, and M  9 is the probability of between 1
and 2 successes when a sample of 5 is taken from a population of 18 where there are 9 successes in the population.
9! 9!
9! 9!
P1  x  2 
 18  1!8! 4!5!  2!7! 3!6!  .48529
18!
18!
C 518
C5
5!13!
5!13!
a. Binomial P4  x  10  , when p  .35 and n  15 (2)
b. Binomial P6  x  11 , when p  .85 and n  15 (2)
c. Poisson P6  x  11 , when m  8 (2)
d. Geometric P6  x  11 , when p  .35 (2)
C 29 C 39
C19 C 49
e. No meaning is needed for the following:
(i) Binomial P x  7 , when

(ii) Geometric
(iii) Binomial
(iv) Geometric

p  .35 and n  15 (1)
Px  7 , when p  .35 (1)
Px  7 , when p  .35 and n  15 (1)
Px  7 , when p  .35 (1)
Solution:
a. Binomial Distribution with p  .35 , n  15
The probability of between 4 and 10 successes in 15 tries when the probability of success on a
single try is .35 is P4  x  10   Px  10   Px  3  .99717  .17270  .82447
b. Binomial Distribution with p  .85 , n  15
The probability of between 6 and 11 successes in 15 tries when the probability of success on a
single try is .85 is P6  x  11 . It can't be done directly with tables that stop at p  .5 , so try to
do it with failures. (The probability of failure is 1 - .85 = .15.) 6 successes correspond to 9 failures
out of 15 tries. 11 successes correspond to 4 failures. So try 4 to 11 successes when
p  .45 and n  15 . P4  x  9  Px  9  Px  3  .99999  .82266  .17733
c. Poisson Distribution with parameter of 8 (parameter of 8 means 8  m     2 )
The probability of between 6 and 11 successes in a unit of space or time when the average number
of successes is 8 is P6  x  11  Px  11  Px  5  .88808  .19124  .69684
d. Geometric Distribution with p  .35
The probability that the first success will occur between try 6 and try 11 when the probability of
success on any one try is .35 is P6  x  11 . Remember that F c  Px  c  1  q c , because
success at try c or earlier implies that there cannot have been failures on the first c tries.
q  1  p  1  .35  .65 . P6  x  11  Px  11  Px  5  F 11  F 5  1  .6511  1  .655

 .65  .65
5
11

 .11603  .00875  .1073
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e. No meaning is needed for the following:
(i) Binomial Distribution with p  .35 , n  15 . Px  7  1  Px  6  1  .75484  .024516


(ii) Geometric Distribution with p  .35 Px  7  1  Px  6  1  1  .656  .656  .07542
(iii) Binomial Distribution with p  .35 , n  15 .
Px  7  Px  7  Px  6  .88677  .75484  .13193 or
P7  C 715 p 7 q 8  C 715 .35 7 .65 8  .1319
(iv) Geometric Distribution with p  .35

Px  7   q x 1 p  .65 6 .35   .02640 or


F 7  F 6  Px  7  Px  6  1  .657  1  .656  .950978 .924581 .02640
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2. If both parents are healthy carriers of sickle cell anemia, each of their children independently have a 25%
chance of being born with the disease.
In a family of six children,
a. What is the chance that at least one child is born with the disease?(2)
Solution: (Binomial n  6, p  .25 ) Px  1  1  Px  0  1  .17798  .82202
b. What is the chance that all the children are born with the disease?(1)
Solution: (Binomial n  6, p  .25 ) Px  6  Px  6  Px  5  1  .99976  .00024
c. What is the chance that at least half the children are born with the disease? (1)
Solution: (Binomial n  6, p  .25 ) Px  3  1  Px  2  1  .83057  .16943
d. What is the mean and standard deviation of the number born with the disease?(1.5)
Solution: (Binomial n  6, p  .25 ) ( q  1  p  .75 )   np  6.25   1.50 ,
 2  npq  1.50.75  1.125 and   1.125  1.0607
e. What is the chance that the last child born will be the first to be born with the disease?(1.5)
Solution: (Geometric p .25 ) P x  q x 1 p . P6  q 5 p  .75 5 .25   .05933
f. In a group of 100 children of healthy carriers of sickle cell anemia, what is the chance that at
least thirty are born with sickle cell anemia? (1)
Solution: (Binomial n  100, p  .25 ) Px  30   1  Px  29   1  .85046  .14954
g. Would it be appropriate to do part f) using the Poisson distribution? Why? Assume that your
answer is 'yes' and find an answer to f) from the Poisson table. (3)
n
n 100
 500 . In this case

 400  500 , so we shouldn't use it.
the test for using Poisson is
p
p .25
However the problem says to do it anyway, so use m  np  100 .25   25 . The Poisson table with
a parameter of 25 gives us Px  30   1  Px  29   1  .81790  .18210 which is off by about
20%.
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3 An airline reports the number of passengers it carries over 7 months and the amount spent on advertising
in the previous month. It collects data as follows:
Observation Advertising Passengers
($1000)
(1000s)
1
2
3
4
5
6
7
x
y
10
12
8
17
10
15
10
15
17
13
23
16
21
14
For your convenience the following values are given:
x  82 ,
y  119 ,
y 2  2105



a. You must do part a) of problem 3! (3 point penalty!) The sample standard deviation for
advertising (4).
b. The sample covariance between advertising and the passengers (3)
c. The sample correlation between advertising and the passengers (2)
d. Interpret the correlation . On the basis of this correlation, does the advertising seem effective?
(2)
The answer to question e) must be based on the results in questions a-d. Do not recompute
the answers after changing x and y !
e. If all the numbers reported in the advertising column were twice what you see there and all the
numbers in the passenger column were 2.5 times what you see there, what would the standard
deviation of advertising( x ), and the covariance and correlation computed above be? (2)
Solution:
a) From the computations below s x2  10.2381 so that s x  3.1997 .
b) From the computations below s xy  11.6667
c) From the computations below rxy  .9863
obs x
1 10
2 12
y
15
17
x2
100
144
y2
225
289
xy
150
204
3 8
4 17
5 10
13
23
16
64
289
100
169
529
256
104
391
160
6 15 21 225
7 10 14 100
total 82 119 1022
441 315
196 140
2105 1464
x
 x  82  11.71429
n
s x2 

7
x 2  nx 2
n 1

1022  711 .71429 2
6
 10 .2381
y
s 2y
 y  119  17.00000
n
y

7
2
 ny 2
n 1
 13 .6667
y 2  2105 and

2105  717 .00 2
6
 x  82 ,  x  1022 ,  y  119 , 
 xy  1464
 xy  nx y  1464  711.71429 17.00   69.9999  11.6667
s 
2
xy
rxy 
n 1
s xy
sx s y

6
11 .6667
6
so
 0.9728 .
10 .2381 13 .6667
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d) The correlation is positive and the square of the correlation is about .95, which on a zero to one scale is
very high. The two variables move in a very similar way. One interpretation of the data is that since
passengers are very closely associated with advertising expenditures from the previous month, advertising is
very effective.
e) From the outline Varax  b  a 2Varx  , so Var 2 x  0  22 Var x   22 10 .2381   40 .9524
(Standard deviation is 6.399). Also, Cov(ax  b, cy  d )  acCov( x, y) , so
Cov(2 x  0, 2.5 y  0)  22.5Cov( x, y)  22.511.6667   58.3333
and if w  ax  b and v  cy  d ,  wv  signac xy . signac  sign22.5   , so the correlation
between the two is 0.9863.
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4. a. A Computer company manufactures in two plants, one in Pennsylvania and the other in
California. Of its 54 employees, one-third are in the California plant. A random sample of 9
employees are asked to fill out a questionnaire about drug use.
(i) What is the probability that exactly one of the employees works at the California plant? (3)
(ii) What is the probability that at least one of the employees works at the California plant?
(2)
(iii) What is the expected value and standard deviation of the number of California employees
in the sample? (2)
C N M C M
1
Solution: (i) Px   n  x N x . N  54 , M  pN  54  18 and n  9 .
3
Cn
36!
36  35  34  33  32  31  30  29
18
18
28! 8!
8  7  6  5  4  3  2 1
P1 


54!
54  53  52  51  50  49  48  47  48
C 954
9  8  7  6  5  4  3  2 1
45! 9!
36  35  34  33  32  31  30  29 18  9

 .10242 Actually, if you know the trick just added to the
54  53  52  51  50  49  48  47  48
outline, you should do (ii) first.
36!
36  35  34  33  32  31  30  29  28
1
C 936 C 018 27! 9!
9  8  7  6  5  4  3  2 1
(ii) Px  1  1  P0  1 

 1
54!
54  53  52  51  50  49  48  47  46
C 954
9  8  7  6  5  4  3  2 1
45! 9!
C 836 C118
36 35 34 33 32 31 30 29 28 27
 1  .01770  .98230 (Particularly easy to calculate because
54 53 52 51 50 49 48 47 46 45
both numerator and denominator are divided by (9!). The appendix says
M  x 1  n  x 1 
Px  
 N  M  n  x  Px  1
x


 1
So P1 
18  1  1  9  1  1 
18 9
P0 
.01770  .10241


1
1 28
 54  19  9  1 
Note: C836  30260340 , C936  94143280 , and C954  5317936260 .
(iii) The outline says that if p 
2 
M 1
1
 ,   np 9  3 and
N 3
3
N n
54  9 2
npq 
3  1.622 . So   1.622  1.2738
N 1
54  1 3
b. Redo the three parts of the problem immediately above, assuming that the company has
many employees (over 1000), but that one-third are still in California. (5)
Solution: (i) (Binomial n  9, p 
1
) P1  C19 p1q8  9 13 2 3   .11706 (1)
8
3
(ii) Px  1  1  Px  0  1  C 09 p 0 q 9  1  2 3   1  .02601  .97399 (2)
9
(iii)   np 9
1
2
 3 and  2  npq  3  2 . So   2  1.4142 (2)
3
3
9
251y0031 11/28/00
c. An investor makes an average of 180 transactions a year.
(i) What is the mean and standard deviation of the number of transactions in a month? (1)
(ii) What is the probability of fewer than 20 transactions in a given month for this investor?
(2)
(iii) If I am designing a form on which to report monthly transactions, (each of which can be
reported in one line) and want to keep the probability that I will have to use more than one
page to at most 5%, how many lines for reporting transactions should be on the page? (2)
180
Solution: (i) (Poisson m 
 15 )   m  15 ,   m  15  3.8729
12
(ii) Px  20   Px  19   .87522
(iii) Looking at the table for a parameter of 15, we see that
Px  22   1  Px  22   1  .96726  .03274 and
Px  21  1  Px  21  1  .94689  .05311 . If we provide 22 lines we will have to use another
page less than 5% of the time.
10