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251y0322 10/27/03
ECO251 QBA1
SECOND HOUR EXAM
March 21, 2003
Name: ____KEY_____________
Social Security Number: _____________________
Part I: (48 points) Do all the following: All questions are 2 points each except as marked. Exam is normed
on 50 points including take-home. Please re-read, ‘Things that You should never do on an Exam or
Anywhere Else, ‘ and especially recall that a probability cannot be above 1!
The following joint probability table shows the relation between two sets of events. Let event A be that the
individual is below 22 ( A is over 21), and the event B 0 be that the individual had no traffic violations in
the last 18 months, the event B1 be that the individual has one traffic violation in the last 18 months and the
event B 2 be that the individual has 2 traffic violations over the last 18 months. No individuals in this pool
of drivers has more than 2 violations.
B0
B1
B2
A
.22 .12 .06
A
.41 .18 .01
B0
A
Note, to do the problems below, at least total the rows and columns - A
1.
B1
B2
.22 .12 .06
.40
.41 .18 .01 .60
.63 .30 .07 1.00
The probability that someone who is over 21 has no traffic violations is (To 2 decimal places):
a) .63
b) .60.
c) .41
P B0  A .41
d) *.68 You have been asked for P B0 A 

 .683
.60
PA
e) None of the above.

   
2.
The probability that someone picked at random is over 21 and has no traffic violations is (To 2
decimal places):
a) .63
b) .60.
c) *.41 Joint probabilities are what the table shows!
d) .68
e) None of the above.
3.
The probability that someone chosen at random is either under 22 or has 2 violations is:
a) .40
b) .06
c) .46
d) .47
e) *.41 P A  B2   P A  PB2   P A  B2   .40  .07  .06  .41
f) None of the above
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B0
4.
B1
B2
A
.22 .12 .06
.40
A
.41 .18 .01 .60
.63 .30 .07 1.00
Which two events are independent?
a) A and A
Note that ‘mutually exclusive’ and ‘independent are almost opposites.
b) A and B 2
c)
* A and B1 .
 


P A  .60 and P A B1 


  
The definition of independence is P A B1  P A . In this case
  


P A  B1
.18

 .60 . But a better way to do this is to note
PB1 
.30
that P A  B1  P A PB1   .60.30  .18 .
d) A and B 0
e)
f)
5.
A and B 2
None of these.
Which two events are mutually exclusive?
a) * A and A
Complements are always mutually exclusive. None of the other pairs
have a joint probability of zero.
b) A and B 2
A and B1 .
d) A and B 0
e) A and B 2
f) None of these.
c)
In questions 6 and 7 you need to know what PB0  , PB1  and PB2  are to do the problems. Show your
work.
Event Px  xPx  x 2 P x 
x  0 .63
0
0
Solution: We can use the following table. x  1 .30 0.30
0.30
x  2 .07 0.14
0.28
Total 1.00 0.44
0.58
6. What is the probability that a person picked at random has at least one violation?
Solution: Px  1  P1  P2  .30  .07  .37 ‘For the 200th time, ‘at least one’ and ‘exactly
one’ are rarely the same thing.
7.
What is the mean and the standard deviation of the number of violations our drivers have? (6) 18
Solution: From the work above  
  .3864  0.6216
 xPx  0.44 . 
2
 
 E x 2   2  .58  .44 2  .3864
This can’t be x and s . These are a sample mean and variance, but
there is no sample.
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8. Which of the following statements must be true if P A  .6 , PB   .4 and P A  B   0
(i) A and B are mutually exclusive.
Explanation: P A  B   0
(ii) A and B are independent.
Explanation: P A  B   P A PB 
(iii) A and B are collectively exhaustive. Explanation: P A  B   1
 
(iv) P A B  0
a) All are true but (i)
b) *All are true but (ii)
c) All are true but (iii)
d) All are true but (iv)
e) All are true
f) None are true.
Explanation:
P A B   P  A  B 
P B 
9. According to a survey of American households, the probability that the residents own 2 cars if
annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over
$25,000 and 70% had 2 cars. The probability that the residents of a household own 2 cars and have an
income less than or equal to $25,000 a year is (3):
a) 0.12.
b) 0.18.
c) *0.22.
d) 0.48.
Solution: The easiest way I know is to start with the given facts. Let ‘ two ’ be the event that a
household owns two cars, and ‘ 25  ’ be the event that a family has an income over $25000.



The problem says P two 25   .80 , P25    .60 , Ptwo  .70 and asks for P two  25 
Using the second two facts we get
two
two
25  25 
__
__
__
.60
.70
__
__

. But 80% of the 60% of families
1.00


that own two cars have incomes over 25000, that is Ptwo  25   P two 25  P25 
 .80 .60   .48 . So now we have
two
two

25  25 
.48
.22
.12
.60

__
__
.70
1.00
25  25 
.48
__
two
two
and, finally,
__
.60
two
two
__
__
.70
1.00
25  25 
.48
.22
.12
.60
If we fill in more, we get
.18
.40
.70
.30
1.00
. From the table we read
P two  25   .22
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10. According to a survey of American households, the probability that the residents own 2 cars if
annual household income is over $25,000 is 80%. Of the households surveyed, 60% had incomes over
$25,000 and 70% had 2 cars. The probability that annual household income is over $25,000 if the
residents of a household own 2 cars is: (4)
a) 0.42.
b) 0.48.
c) 0.50.
P25  two .48
d) 0.69. If we look at the table above P 25  two 

 .6857 . Or we
Ptwo
.70
Ptwo 25  P25   .80 .60 

 .6857
could use Bayes’ rule to say P25  two 
Ptwo
.70


11. If you toss a coin 6 times, what is the chance that you get at least one head? (Show your work) (4)
Solution: The probability of at least one head is 1  PTTTTTT   1  .56  1  .015625  .984375 .
12. If you have five pennies and four dimes in your pocket and you pick three coins, what is the
chance that you get exactly 2 dimes? (Show your work) (4)
n!
Solution: Remember that C rn 
The number of ways you can get one penny from 5 is
n  r ! r!
C15  5 . The number of ways you can get two dimes from four is C 24 
of ways that you can get 3 coins from 9 is C 39 
4! 4  3

 6 . The number
2!2! 2 1
9! 9  8  7

 84 . our solution, which is an example
6!3! 3  2 1

56
 0.3571 . We could also say that your
84
chance of getting a dime on the first and second try is
4 3 5 60 10


, your chance of getting a
9 8 7 504 84
of the Hypergeometric distribution is P2 
C15 C 24
C 39
4 5 3 60 10


and your chance of getting a dime on the second
9 8 7 504 84
5 4 3 60 10
10


 .3571
and third try is
. Thus, your chance of getting exactly $0.20 is 3
9 8 7 504 84
84
dime on the first and third try is
13. Show, by using the appropriate formula, that the number of ways that you can pick 52 items from
55, if order is not important, is the same as the number of ways you can pick 3 items from 55.(4)
55!
55!
55!
55!
55
C 52



and C 355 
. These are obviously the same.
55  3!3! 52!3!
55  52 !52! 3!52!
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TABLE 4-3
A survey is taken among customers of a fast-food restaurant to determine preference for hamburger or
chicken. Of 200 respondents selected, 75 were children and 125 were adults. 120 preferred hamburger
and 80 preferred chicken. 55 of the children preferred hamburger.
Let ‘ H ' be hamburger, ‘ C ’ be chicken, ‘ A ’ be Adult and ‘ A ’ be child. The numbers given are as
H C
H C
A
__ __
125
A
65 60
125
follows.
. Fill in the table and get
or, if you divide by
A
55 __
75
A
55 20
75
120
80 200
120
80 200
H C
A .325 .300
.625
200,
A .275 .100
.375
.600
.400 1.000
14. Referring to Table 4-3, the probability that a randomly selected individual is a child and prefers
20
 .100
chicken is __________. Solution: P A  C 
200


15. Referring to Table 4-3, the probability that a randomly selected individual is a child or prefers
75 120 55


hamburger is __________. Solution: P A  H  P A  PH   P A  H 
200 200 200
140

 .70 or P A  H  P A  PH   P A  H  .375  .600  .275  .700
200


  
  




16. Referring to Table 4-3, assume we know that a person has ordered chicken. The probability that
60
P A  C  200
60


 .75 or
this individual is an adult is __________. Solution: P A C  
80
PC 
80
200
P A  C  .300
P AC 

 .75
.400
PC 
 
20. Assume that z is a standardized random variable. What is the mean and standard deviation of
6 z  3 ? Solution: Remember that Ez   0 and Varz   1. The outline says E ax  b  aEx   b
and Varax  b  a 2Varx , Substitute -6 for a , z for x and 3 for b and we have
E 6 z  3  6Ez   3  60  3  3 and Var  6 z  3   62 Var z   36 1  36 . The standard
deviation is thus
36  6.
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ECO251 QBA1
SECOND EXAM
March 21, 2003
TAKE HOME SECTION
Name: ____KEY ________________
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 (9 Points) Show your work!
1. My Social Security Number is 265398248.
Take the following set of numbers:
10
23
17
16
32
35
44
45
33
21
and use your Social Security Number to provide the first digit of the first nine numbers, so that, for me, the
numbers would become
210, 623, 517, 316, 932, 835, 244, 445, 833, 21. Compute a sample standard deviation for the resulting
numbers. (3- 2point penalty for not doing)
Solution: Please recall the following.
x
 x , (Computational Formula) s   x
2
n
So doing it both ways, we compute
Row
1
2
3
4
5
6
7
8
9
10
x2
x
210
623
517
316
932
835
244
445
833
21
4976
44100
388129
267289
99856
868624
697225
59536
198025
693889
441
3317114
xx
-287.6
125.4
19.4
-181.6
434.4
337.4
-253.6
-52.6
335.4
476.6
0.0
2
 nx 2
n 1
 x  x 2
82714
15725
376
32979
188703
113839
64313
2767
112493
227148
841056
(Definitional formula) s 2 
So
 x  4976 ,
x
2
 3717114
 x  x 
2
n 1
 x  x   0 (A check
on the validity of the mean.),
 x  x 
2
 841056 (with a slight
rounding error) and n  10 . This means
x
s
2
 x  4976  497 .6 .
n
x

10
2
 nx 2

3317114  10 497 .62
9
n 1
841056 .4

 93450 .71111
9
s  93450.7111  305.697
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2. At a local university 30% of women, 40% of men and 80% of monsters are business majors. 60% of
students are women, 30% of students are men and 10% of students are monsters. (6)
a. What percent of students are business majors?
b. What percent of business majors are monsters?
c. List the following events W - a person is a woman, Ma - a person is a man, Mo - a person is a
monster, B a person is a business major. Find P B Mo , P Mo B , PB  , PMo  and show that your

 

solution in b. satisfies Bayes’ rule.
Solution: According to the problem statement P B W  .30 , P B Ma  .40 and P B Mo  .80 . Also






PW   .60 , PMa   .30 and PMo  .10 . If 100 students walk into a room, 60 will be women, 30
W Ma Mo Total
B
men and 10 monsters. We can show this as
Of the 60 women, 30% or
B
Total 60 30 10
100
18 will be business majors. Of the 30 men, 40% or 12 will be business majors. Of the 10 monsters,
W Ma Mo Total
B 18 12
8
38
80% or 8 will be business majors. We now have
B
Total 60 30 10
100
W Ma Mo Total
B 18 12
8
38
We really ought to finish our table.
or we might present the joint
B 42 18 2
62
Total 60 30 10
100
W Ma Mo Total
B .18 .12 .08
.38
probability table
. This serves as a check that our probabilities
B .42 .18 .02
.62
Total .60 .30 .10
1.00
add to 1.
a) Thus 38 out of 100 or 38% are business majors. PB   .38
b) 8 out of 38 business majors are monsters.
c) From the work above: PB Mo  


8
 .2105 or 21.05%. P Mo B  .2105
38


8
 .80 , P Mo B  .2105 , PB   .38 and PMo  .10 .
10
PB AP A
P Mo B PB 
Since Bayes’ rule says PA B  
we can either write P B Mo  
and say
P B 
PMo 
P B Mo PMo 
.2105 .38 
.80 .10 
or we can try P Mo B  
as .2105 
. Both of these are true
P B 
.10
.38
except for rounding error.
.80 
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