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ECO 251 QUANTITATIVE BUSINESS ANALYSIS I
SECOND EXAM
OCTOBER 24, 2000
NAME: ____KEY___________
SECTION ENROLLED: MWF TR 10 11 12:30
(Circle both days and time separately or just write down days and time)
Part I. Multiple Choice (20 Points)
Note: It seems that about a third of you are missing 3 basic facts!
1. The Addition Rule: P A  B   P A  PB   P A  B  Always!
 
2. The Multiplication Rule: P A  B  P A B PB Always!
3. 'Independent' and 'Mutually Exclusive' are extreme opposites!
Employees at the Mason and Lind Corporation are classified by gender and status.
Status
Male
Female
Total
Executive
80
20
100
Supervisor
100
300
400
Production
150
250
400
Clerical
40
60
100
Total
370
630
1000
Define the following events: M  Male, F  Female , E  Executive, S  Supervisor ,
P  Production and C  Clerical. All probabilities in questions 1-4 are given to only 3 places.
M F
PC  F   .06
E .08 .02  .10


PC  F  .06
S .10 .30  .40
P C F  

 .095
Use the joint probability table:
P F 
.63
P .15 .25  .40


PC  F   PC   PF   PC  F 
C .04 .06  .10
 .10  .63  .06  .67
.37 .63 1.00
1.
From the table above find the probability that a randomly picked female is a clerical worker.
a.
PC  F   .060
b.
c.
d.*
PC  F   .600
PF C   .600
PC F   .095
PC F   .600
From the same table find the probability that, if we pick an employee at random, we will get a
female clerical worker.
a.*
PC  F   .060
b.
PC  F   .600
e.
2.
c.
d.
e.
PF C   .600
PC F   .095
PC F   .600
251y0022 10/18/00
3.
From the above table find the probability that a worker picked at random is female or a clerical
worker (or both).
a.
PC  F   .670
b.
PC  F   .060
c.
d.*
e.
4.
5.
PC  F   .060
PC  F   .670
None of the above. Write in answer.
From the table above, the events
P and C are mutually exclusive
a.
b.
M and F are independent
c.*
M and F are mutually exclusive
d.
Both b and c are true.
e.
None of the above are true.
(MBS 9) The top speeds in miles per hour for a sample of five new automobile brands are listed
below. Calculate the standard deviation of the speeds.
100, 125, 115, 175, 120
a.*
28.4165
b.
807.5000
c.
25.4165
d.
646.0000
e.
None of the above - supply answer and show your work.
Index
x
x2
x 635
2
n

5
,
x

635
,
x

83875
,
x


 127 ,
1 100 10000
n
5
2 125 15625
x 2  nx 2 83875  5127 2
2
s 

 807 .5
3 115 13225
n 1
4
4 175 30625
s  807 .5  28 .4165
5 120 14400




635 83875
z is a standardized random variable. y  9z  4 . The standard deviation of y is:
a.
-5
b.
-9
This is essentially problem J6!
c.*
9
d.
81
e.
None of the above - supply answer and show your work.
Solution: Our rule says that if y  ax  b , Var  y    y2  a 2Var x   a 2 x2 . If z is a standardized
6.
variable, E z    z  0 and Varz    z2  1 . So, in the rule, replace x by z and let a  9 . Then
Var y    y2   92 Varz   811  81 and  y  49  7 .
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7.
The firm of Mason and Lind has 12 directors, four of which are women and 8 of which are men. A
four-member search committee is selected from the directors at random. Define the following
events: 4M  The committee is all male and 4F  The committee is all female.
Then the following is true (Answers are rounded to three decimal places):
a.*
P4M   .141
b.
P4 F   .141
c.
d.
e.
Solution:
P4M   .002
P4M   P4F   1
More than one of the above are true.
P4M  
8.
C 48 C 04
C 412
 87 65 


C 4C 8
1 1
 4  3  2 1 
 .00202

 .14141 P4 F   4 12 0 
12 11 10  9
12 11 10  9
C4
4  3  2 1
4  3  2 1
(MBS 32)If the events A and B are independent and P A  .8 and PB   .2 , which of the
following is correct?
a.
P A  B   0
b.
P A  B   .16
P A  B   1.00
d.*
P A  B   .84
e.
P A  B   1.00
Solution: If A and B are independent then P A  B   P APB  .8.2  .16 . Thus
P A  B   P A  PB  P A  B   .8  .2  .8.2  .84
c.
9.
(ML 193) A manufacturer buys 30% of its microchips from Firm A, 20% from Firm B and 50%
from Firm C. The manufacturer knows that 3% of chips from Firm A are defective, 5% from Firm B are
defective and 4% of chips from Firm C are defective. If a worker picks a chip at random and finds that it is
defective, what is the probability that it comes from Firm C? Choose the answer that is closest to the correct
one.
a.
.67
This is a Bayes' Rule problem. If we create a
b.
.50
joint probability table, we find
c.
.04
P A  .30 , P D A  .03 , so
d.
.02
PD  A  (.03)(. 30)  .009 . This appears in the
e.*
.51
upper left corner of the table. We get the other
items in the top row the same way. The 30%,
20% and 50% go at the bottom, so we can add to
get PD  .039 Then
 
PC  D .02

 .513
PD
.039
A
B
C
.009 .010 .020 .039




.300 .200 .500 1.00
PC D 
D
D
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251y0022 10/18/00
10.
A satellite has four independent power sources. Its camera will function as long as one of the
power sources is functioning. If each of the sources has a 40% probability of failing in the next
month, what is the chance that the camera will not work (rounded to two decimal places)?:
a.
.40
b.
.99
c.*
.03
d.
1.00
e.
None of the above . Write down your own answer.
Solution: Since the event that any given power source fails is independent of the failure of any other, the
probability that all four fail is .40 4  .0256 . This is important - Why would anyone ever use a backup
power source if it didn't improve the chance that a device will not fail?
Every time I ask people to compute the probability of an event occurring, and then ask them to tell
me the probability of it happening twice, half the people tell me that the two probabilities are the
same. Is this refusing to think or what? This is a warning for the next exam.
Part II. Show your work!
(ML 223) The number of cars you expect to sell on Saturday is represented by the probability distribution
below. (This is a probability distribution, not a sample!) Verify that this is a valid distribution and compute
the standard deviation of the following data (5 Points - 2 Point Penalty for not trying.)
Number of Probability
x P x 
x 2 Px 
Cars Sold
0
.10
0.00
0.00
1
.10
0.10
0.10
2
.20
0.40
0.80
3
.30
0.90
2.70
4
.30
1.20
4.80
1.00
2.60
8.40
We have a valid distribution since all the probabilities are between one and zero and add to 1. The formula
xPx  and
for a population variance for a distribution is  x2  E x 2   x2 , where   E x  
   x
Ex
2
 
2
Px  . So
 x2

 8.40  2.60   1.64 and  x  1.64  1.2806 . A third of you computed a
2
sample variance here - perhaps because that was what I asked for on last year's exam. I guess that
this proves that you can't answer a problem that you haven't read!
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251y0022 10/18/00
Part III. Do at least 2 1/2 (25 Points) of the four problems on the next four pages. Show your work! You
receive extra credit for extra work!
1. You are dealt seven cards from a deck.
a. How many possible hands are there? (2) Work out answers to a) and b).
b. What is the probability of getting three kings and four aces? (2)
You may leave the answers to the remaining sections of this question in factorial form.
c. What is the probability of getting two kings and two aces? (2)
d. What is the probability of getting four kings and four aces? (1)
e. What is the probability of getting two kings and at least two aces? (3)
Solution: a) C rn 
b) P3K ,4 A 
52!
52  51  50  49  48  47  46
n!
, C752 

 133784560
45!7!
7  6  5  4  3  2 1
n  r ! r!
C 34 C 44
C 752
4!  4! 


1!3!  4! 0! 
41
4



 .00000003  2.99 10 8
52!
52  51  50  49  48  47  46 133784560
45! 7!
7  6  5  4  3  2 1
4!  4!  44! 
4  3  4  3  44  43  42





2! 2! 2! 2!  41!3! 
2 1  2 1  3  2 1
6  6 13244
c) P2 K ,2 A 



 .00356
52!
52  51  50  49  48  47  46 133784560
45! 7!
7  6  5  4  3  2 1
d) Impossible P4K ,4 A  0
e) P2K ,2 A  .00356
C 24 C 24 C 344
C 752
P2 K ,3 A 
C 24 C 34 C 244
C 752
P2 K ,4 A 
4!  4!  44! 
43



4 44  43
2! 2! 3!1!  42! 2! 
6  4  946
2

1
2 1



 .0001697
52!
52  51  50  49  48  47  46 133784560
45! 7!
7  6  5  4  3  2 1
C 24 C 44 C144
C 752
4!
43
144
1 44
2! 2!
6 1 44
2

1



 .00000197
52!
52  51  50  49  48  47  46 133784560
456! 7!
7  6  5  4  3  2 1
So what we want is:
P2K ,2 A  2K ,3 A  2K ,4 A  P2K ,2 A  P2K ,3 A  P2K ,4 A  .00356  .00017  .00000
 .00373 . How many of you know what 'at least' means?
5
251y0022 10/18/00
2. I toss a coin 6 times. What is the probability that the result will be:
a. All tails? (1)
b. At least one head (2) ?
c. If x is the number of tails, make a table showing the values of x and their probabilities. (6)
d. Using your table, if possible, find the Expected Value and Standard deviation of x . (3)
e. Using your table, if possible, find F 2 . (2)
f. Find the conditional probability of getting all heads, given that you got at least one head. (3)
Note - it is possible to deduce some of these answer from tables in the book that we have
Not yet covered. I expect you to show me that you got these answers yourself.
6
1
1
1
Solution: a) PTTTTTT      6 
 .0015625
64
2
2


6
1
1
1
 .984375 . You could
b) P At least one Head   P TTTTTT  1  PTTTTTT   1     1  6  1 
64
2
2
also get this by computing P1  P2  P3  P4  P5  P6 from the table below but it's too much
work.
c, d)
x
Px 
x P x 
x 2 P x 
0
6! 1
1
6
0.00000
0.00000
C 06 12 

 .015625
6! 0! 64 64
1
6! 1
6
6
1(.093750)=0.09375
1(.09375)=0.09375
C16 12 

 .093750
5!1! 64 64
2
6! 1 6  5 1 15
6
2(.234375)=0.46875
2(.46875)=0.93750
C 26 12 


 .234375
4! 2! 64 2 1 64 64
3
6! 1 6  5  4 1 20
6
3(.312500)=0.93750
3(.93750)=2.81250
C 36 12 


 .312500
3!3! 64 3  2 1 64 64
4
6! 1 6  5 1 15
6
4(.234375)=0.93750
4(.93750)=3.75000
C 46 12 


 .234375
2! 4! 64 2 1 64 64
5
6! 1
6
6
5(.093750)=0.46875
5(.46875)=2.34375
C 56 12 

 .093750
1!5! 64 64
6
6! 1
1
6
6(.015625)=0.09375
6(.09375)=0.56250
C 66 12 

 .015625
0! 6! 64 64
3.00000
10.50000
Total
64
1
64







 Px  1.0 (a check),   Ex   xPx  3.0 and E x    x Px   10.5 .
Finally   Varx  Ex     x Px    10.5  3.0  1.5   1.5  1.225
2
So
2
2
2
2
2
2
2
e) This is the cumulative distribution F 2  Px  2  P0  P1  P2 
1
6 15 21



 .328125625
64 64 64 64
6
251y0022 10/18/00
6
1
1
1
1
 .984375 , P All Heads   PHHHHHH      6 
 .015675
64
2
64
2
 
P All Heads  At least one Head 
P All Heads 
So P All Heads At least one Head 

P At least one Head 
P At least one Head 
1
1
 64 
 .01587
63
63
64
f) P At least one Head   1 


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251y0022 10/18/00
3. Assume that P A  .2 and PB   .4
a. What is P A  B  if
(i) A and B are independent? (1)
(ii) A and B are mutually exclusive? (1)
(iii) P B A  .9 ? (1)
 
 
b. What is P A B if
(i) A and B are independent? (1)
(ii) A and B are mutually exclusive? (1)
(iii) P B A  .9 ? (1)
 
 
c. Make a joint probability table of A, B, A and B , assuming that P B A  .6 (4)
Solution:
a. (i) P A  .2 and PB   .4 . Independence means P A  B   P APB  .2.4  .08 . So
P A  B  P A  PB  P A  B  .2  .4  .08  .52
(ii) Mutual exclusiveness means P A  B   0 . So P A  B  P A  PB  P A  B   .2  .4  0  .6
 
 
(iii) P B A  .9 means that P A  B  PB  A  P B A P A  .9.2  .18 . So
P A  B  P A  PB  P A  B  .2  .4  .18  .42
 
P A  B 
(ii) Mutual exclusiveness means P A  B   0 . So PA B  
 0.
PB 
PB AP A 9.2

 .45 .
(iii) PB A  .9 means that PA B  
P B 
.4
c. P A  .2 and PB   .4 . PB A  .6 means that P A  B  PB  A  PB AP A  .6.2  .12 .
b. (i) P A  .2 and PB   .4 . Independence means P A B  P A  .2
A
Our table shows
B
A A
.12
B
B
.4
B
.2
1.0
A
  PB  and, with the information above, reads
PA  B  PA  B  PB 
P  A
P A 
1
P A  B  P A  B
. If we just fill in the blanks to make it add up, we get
B
B
A A
.12 .28
.08 .58
.20 .80
.4
.
.6
1.00
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251y0022 10/18/00
4. 25% of a firm's employees intend to retire within 5 years, 30% within 6 to 15 years and the remainder
after 15 years - all are in a profit-sharing plan. Of the employees who will retire soonest, 70% favor a
buyout of the company. In the 6-15 year group 38% favor the buyout. Of the remaining employees only
28% favor a buyout.
a. Using these numbers construct a joint probability table with events describing the three
retirement classes and whether they favor or do not favor a buyout. (3)
b. Find the joint probability that someone will retire within 5 years and favors a buyout. (1)
c. If we pick someone who favors a buyout, what is the chance that that person will retire within 5
years? (3)
d. What proportion of the company's employees are opposed to a buyout? (2)
e. An automobile insurer knows that on a $50000 sports car there is, in the next year, a 0.1% chance of a
total loss, a 2% chance of a 50% loss, a 5% chance of a 25% loss and a 10% chance of a 10% loss. If there
is no deductible, how much should the firm charge for an annual policy if they wish to make an average of
$300 on each policy sold? (3)
Solution: Define the following events: R5  Will retire within 5 years,
R6  15  Will retire within 6 to15 years, RL  Will retire later and F  Favors buyout.
a) Then the problem tells us that PR5  .25 , PR6  15   .30 , PRL   .45 (since it’s a leftover)
PF R5  .70 , PF R6  15  .38 and PF RL  .28 . Then, Using the Multiplication Rule,
PF  R5  PF R5PR5  .70.25  .175 , PF  R6  15  PF R6  15PR6  15  .38.30  .114


and PF  RL  P F RL PRL  .28.45  .126 . If we put these numbers into a table, we find
F
F
R5
.175
.25
R6  15
.114
RL
.126
.30 . If we fill in the blanks to make it add up we get R6  15
RL
.45
R5
1.00
F F
.175 .075
.25
.114 .186
.30 .
.126 .324
.45
.415 .585
1.00
b) We have already found PF  R5  .175 .


c) If we use the table and the Multiplication Rule, P R5 F 
 
PR5  F  .175

 .422 .
PF 
.415
d) From the table P F  .585 .
e. Using the numbers above we can construct a distribution of the losses and find the expected loss.
% Loss
Amount Lost
P x 
xPx 
(thousands)
(thousands)
100%
1.00(50)= 50
.001
0.050
50%
.50(50)= 25
.020
0.500
25%
.25(50)= 12.5
.050
0.625
10%
.10(50)= 5
.010
0.050
0%
0
.919
0.000
1.000
1.225
If the expected loss is $1,225 and we need a $300 profit, we must charge $1,525 (plus any administrative
expenses).
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