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```							251y0321 3/20/03                            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. (There are 57 possible points.)

1.   If two events are mutually exclusive, what is the probability that both occur at the same time?
a) *0.
b) 0.50.
c) 1.00.
d) Cannot be determined from the information given.

2.   If two events are collectively exhaustive, what is the probability that one or the other occurs?
a) 0.
b) 0.50.
c) *1.00.
d) Cannot be determined from the information given.

3.   If two events are independent, what is the probability that they both occur?
a) 0.
b) 0.50.
c) 1.00. Explanation: P A  B  P A PB , but we don’t know either.
d) *Cannot be determined from the information given.

4.   A business venture can result in the following outcomes (with their corresponding chance of
occurring in parentheses): Highly Successful (10%), Successful (25%), Break Even (25%),
Disappointing (20%), and Highly Disappointing (?). If these are the only outcomes possible for the
business venture, what is the chance that the business venture will be considered Highly
Disappointing?
a) 10%
b) 15%
c) *20%
d) 25%

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251y0321 3/20/03

Show your work in problems 7-9.

TABLE 5-4
The following table contains the probability distribution for X = the
number of traffic accidents reported in a day in Corvallis, Oregon.
X     0     1     2     3     4     5
P(X)     0.10     0.20     0.45      0.15    0.05       0.05

5. Referring to Table 5-4, the probability of at
least 1 accident is ___.90_____. Explanation: Px  1  1  P1

6.   Referring to Table 5-4, the mean or
expected
value of the number of accidents is
__2.00__.                                               x       Px      xPx      x 2 Px 
0       .10       0.00       0.00
7.  Referring to Table 5-4, the standard                1       .20       0.20       0.20
deviation                                           2       .45       0.90       1.80
of the number of accidents is _1.1832_. (4)             3       .15       0.45       1.35
Ex         xPx   2.00 ,                        4       .05       0.20       0.80
  
E x2       x 2 P x   5.40 ,                         5       .05       0.25       1.25
 
 2  E x 2   2  5.40  2 2  1.40 ,                 Sum    1.00       2.00       5.40
  1.40  1.1832

8.    We sell a product with a markup of \$7 per unit and have fixed costs of \$500 monthly. Thus, if we
sell 100 units, our profits will be \$7(100) – 500 =\$200. If our expected sales are 250 units and the
standard deviation is 20, what is the mean and standard deviation of our profits? (4) Solution: Use
the solution of problem J2. The formula says that if y  ax  b,  y  a x  b
and  y  a 2 x , so, if y  7 x  500, a  7 and b  500 . Then
2         2

 y  7  x  500  7250  500  1250 , and  y  a 2 x  72 202  19600 . Thus
2        2

 y  19600  7 20  720  140. .
2     2

2
251y0321 3/20/03

TABLE 4-2
An alcohol awareness task force at a Big-Ten university sampled 200
students after the midterm to ask them whether they went bar hopping
the weekend before the midterm or spent the weekend studying, and
whether they did well or poorly on the midterm. The following result
was obtained.
Did Well on Midterm Did Poorly on Midterm
Studying for Exam              80                  20
Went Bar Hopping               30                  70
If we finish the table we get:
Did Well    Did Poorly Total
on Midterm on Midterm
Studying             80         20         100
for Exam
Went Bar             30         70         100
Hopping
Total              110          90         200

9.   Referring to Table 4-2, what is the probability that a randomly selected student who went bar
hopping will do well on the midterm?
a. *30/100 Explanation: This is a conditional probability. Out of the 100 who went bar
hopping, only 30 did well on the exam.
b. 30/110
c. 30/200
d. (100/200)*(110/200)

10. Referring to Table 4-2, the events "Did Well on Midterm" and "Studying for Exam" are
a) *statistically dependent. Explanation: To be independent the probability that someone
did well on the midterm, which is 110 out of 200, has to be the same as the conditional
probability that someone did well on the midterm, given that he/she studied for the exam,
which is 80 out of 100. These fractions are not equal.
b) mutually exclusive.
c) collective exhaustive.
d) None of the above.

11. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A  B) = 0.1. Then P(B|A) =
.25 .     Explanation: Note that P A  B  PB  A  .1 and, by the multiplication rule,
PB  A .1
PB A              .25 . Note that you can’t use Bayes’ Rule here because you don’t know
P A  .4
PA B .

12. Suppose A and B are events where P(A) = 0.4, P(B) = 0.5, and P(A  B) = 0.1. Then P(A  B ) =
.8 .     Explanation: Note that by the addition rule,
P A  B  P A  PB  P A  B  .4  .5  .1  .8 .

52
13. Evaluate C 4       270725 .
n!                    52!      52  51 50  49
Explanation: C rn                 . So C 4 
52
                   270725
n  r ! r!            48! 4!     4  3  2 1

3
251y0321 3/20/03

14. If I am playing a card game with hands of 4 cards. The deck contains 52 cards of which 13 are
hearts. What is the chance that I get 2 hearts? Do not finish this problem – write out the answer
 13!  39! 

 11!2!  37!2!  13!39!48!4!
       
                                      13!39!
using factorial notation. (4)                                  
 52!           52!11!2!37!2! 11!37!2!2 270725
       
 48!4! 
       
n!
Explanation: C rn                    . There are 13 hearts and 39 non-hearts in the deck. We need 2 of
n  r ! r!
13!                                   39!
each. There are C 2 
13
to pick the hearts and C 2 
39
ways to pick the non-hearts
11! 2!                              37! 2!
52!           52  51 50  49
out of a total of C 4 
52
                     270725 ways to pick a hand of 4. The answer
48! 4!           4  3  2 1
 13!  39! 

 11!2!  37!2!  13!39!48!4!
       
13 12  39  38  4  3  2 1
13 39
C2 C2                    
is P2                                                                                 .2135
C452
 52!           52!11!2!37!2! 2 1  2 1  52  51 50  49

 48!4! 

       

TABLE 4-4
Suppose that patrons of a restaurant were asked whether they
preferred beer or whether they preferred wine. 70% said that they
preferred beer. 60% of the patrons were male. 80% of the males
preferred beer.

15. Referring to Table 4-4, the probability a                             B W
randomly selected patron is a female is                          M          60
.40.                                                                               . We know that 80% of
F         40

16. Referring to Table 4-4, the probability a                            70 30 100
randomly selected patron is a female                            the 60 males or 48 of the men prefer beer.
who prefers beer is .22.                                                                      B W
M 48  60
17. Referring to Table 4-4, suppose a                               Our table now reads                      and
randomly selected patron prefers wine.                                                   F        40

Then the probability the patron is a                                                         70 30 100
male is .40. (4)                                                we can fill in the rest to give
B W
Explanation: Assume that 100 people walk                             M 48 12 60
into the restaurant. Let M be ‘male,’ F be                                               .
‘female,’ B be ‘beer’ and W be ‘wine.’                               F 22 18 40
         
Given:                                                                     70 30 100
PB  .70, PM   .60, PB M   .80 .                           So the probability that a randomly selected
Since PM   .60, PF   1  .60  .40.
patron is a female who prefers beer is 22 out
of 100 or PF  B  .22 , and the
So 40 are women and 60 are men. 70 out of
probability that a wine-drinker is male is 12
out of 30 or PM W   12 30  .40 .
100 prefer Beer, so 30 prefer wine. If we
make this into a table, we get

4
251y0321 3/20/03
PW M PM 
Using Bayes’ Rule, PM W                        . But, since a male must prefer either beer or wine
PW 
PW M   1  PB M   1  .80  .20. Similarly PW   1  PB  1  .70  .30.


So P M W    .20.60  .40 .
.30

TABLE 4-7
The next state lottery will have the following payoffs possible with
their associated probabilities.
Payoff Probability
\$2.00   0.0500
\$25.00   0.0100
\$100.00   0.0050
\$500.00   0.0010
\$5,000.00   0.0005
\$10,000.00   0.0001

18. Referring to Table 4-7, the probability that you win any money is .0666.
Explanation: 0.05 + .01 + .005 + .001 + .0005 + .0001 = .0666.

19. Referring to Table 4-7, the probability that you win at least \$100.00 is .0066.
Explanation: The probability that you win less than \$100 is 0.05 + .01 = .06. So the probability
that you both won and got at least \$100 is .0666 - .06 = .0066.

20. Referring to Table 4-7, if you have a winning ticket, the probability that you win at least \$100.00 is
.099. (I will accept .1)
Explanation: Let W be the event that you win and let H be the event that you win at least
PH  W  .0066
\$100. By the multiplication rule, PH W                         .099 .
PW        .0666

Another way to think about this is to say that 10000 people bought chances. Then .05(10000) =
500 got \$2, 100 got \$25, 50 got \$100, 10 got \$500, 5 got \$5000 and one got \$10000. This is a total of 666
winning tickets. Of these 66 got \$100 or more. 666 out of 10000 is .0666. 66 out of 10000 is .0066. 66 out
of 666 is .099.

5
251y0321 3/20/03                             ECO251 QBA1
FIRST EXAM
February 21, 2003
TAKE HOME SECTION
-
Name: _________________________
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
and use your Social Security Number to provide the first digit, so that , for me, the numbers would become
210, 623, 517, 316, 932, 835, 244, 445, 833. Compute a sample standard deviation for the resulting
numbers. (3- 2point penalty for not doing)

Solution: Of course, I faked this using Minitab. Here’s how I did it. I used the ‘edit’ menu to copy the row
of numbers, opened Minitab, and used the ‘set’ instruction to copy in the numbers. My command sheet
—————    3/20/2003 8:13:29 PM        ————————————————————

Welcome to Minitab, press F1 for help.
MTB > set c1
DATA> 210, 623, 517, 316, 932, 835, 244, 445, 833. #These numbers copied from exam.
DATA> end
MTB > let c2=c1*c1
MTB > print c1 c2             #It printed out two columns. These are cut and pasted below.
MTB > sum c1

Sum of C1

Sum of C1 = 4955.0
MTB > sum c2

Sum of C2

Sum of C2 = 3316673
MTB > describe c1   #Just to check my work.

Descriptive Statistics: C1
Variable             N             Mean       Median       TrMean         StDev      SE Mean
C1                   9            550.6        517.0        550.6         271.3         90.4

Variable          Minimum      Maximum            Q1           Q3
C1                  210.0        932.0         280.0        834.0

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251y0321 3/20/03

Row      x            x2
1      210         44100
2      623        388129
3      517        267289
4      316         99856
5      932        868624
6      835        697225
7      244         59536
8      445        198025
9      833        693889
4955       3316673

 x 4955
So    x  4955,  x  3316673 and n  9 . x  n  9  550.556
2

s2   
 x  nx  3316673 9550.556  588670.22  73583.778 s 
2      2                        2
73583.778  271.26
n 1                  8                  8

Another way to speed up your work is to find out where your calculator keeps   x    and   x    2

and to use the calculator to compute the standard deviation. You still have to write down the values of x
and to compute x 2 , but the place where people spend the most time and make the most mistakes is in
summing x 2 . You, of course, get the variance by squaring the standard deviation, and I’ll never know that
you didn’t really do it.

7
251y0321 3/20/03

2. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that the
jorcillator only works as long as both components work. The probability of the phillinx failing in the
first month of service is .1, and the probability of the phillinx failing in the second month is .3, while
the probability of the flubberall failing in the first month is .7 and probability of the flubberall failing in
the second month is .2. Failure of components is assumed to be independent, so the probability of both
components failing in the first month is (.1) (.7) =.07 (This is not the probability that the jorcillator will
fail in the first month!)
a) What is the probability that the jorcillator will fail in the first month?
b) What is the probability that the jorcillator will fail in the second month?
c) What is the probability that the jorcillator will last beyond 2 months?

Solution: This is a version of Problem H4.

The probabilities you have are:                               Second, write down the joint events and their
Month     Component 1        Component 2                    probabilities.
1              .1              .7                          Joint Event Probability         Machine fails
2              .3              .2                                                          on day
3+             .6              .1                           A D          .1.7  .07          1
A E             .1.2    .02          1
.1.1 
Let the table below define events A through E .
A F                        .01          1
Month       Component 1      Component 2
Fails            Fails
BD              .3.7     .21         1
1              A                D                         BE              .3.2    .06          2
2              B                E                         BF              .3.1    .03          2
.6.7 
3+             C                F                         CD                                      1
.42
There are two ways to do this. First, remember                CE              .6.2     .12         2
that if one component fails, the whole shebang                CF              .6.1    .06          3
fails. Thus                                                   Sum                     1.00
P1  P A  B   P A  PB   P A  B 
 .1  .7  .1.7  .73 .                                      If we handle this with a joint probability table
The machine only lasts to period 3 if both                    we can write it as:
components last to period 3.
P3   PC  F   PC  PF   .6.1  .06 .            Events      D            E          F         Total
Since the machine has to fail sometime,                       A          .07          .02        .01         .10
P2  1  P1  P3   1  .73  .06  .21 .
B          .21          .06        .03         .30
C          .42          .12        .06         .60
Total      .70          .20        .10        1.00

If we add together probabilities for each day,

Day            Probability of Failure
1              .07  .02  .01 .21 .42  .73
2                        .06  .03 .12  .21
3+                                       .06
Sum                                       1.00

8

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