251y0224 10/30/02
ECO 251 QUANTITATIVE BUSINESS ANALYSIS I
SECOND EXAM
OCTOBER 29, 2002
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 (40 Points) Remember that 'At least one' means 'one or two or three or four or……'.
It does not mean 'exactly one.'
1. If A and B are complementary events: a. b. c.
P
P
P


A
A
 

B

B

1





P
0
1
  d.* All of the above. e. None of the above.
2.
The following table shows the probabilities of the Whizbang Corporation’s stock movement:
Total Economic
Conditions
Good
Fair
Poor
Total
Stock Price
Rises
.30
.20
__
__
Stock Price
Falls
.04
__
.20
.44
__
__
__
Complete the table and then find the probability that the stock price goes up given that economic conditions are poor. a. .0600 b.* .2308 c. .7600 d. .1456 e. .1075
Solution: Here is the completed table.
Economic
Conditions
Good
Fair
Poor
Total
Using the notation in the next problem,
Stock Price
Rises
.30
.20
.06
.56
P

U Po


Stock Price
Falls
.04
.20
.20
P

U
P

 
.44
Po


.
06
.
26

.
2308
Total
.34
.40
.26
1.00

251y0224 10/30/02
3.
Let us name the events in the previous problem as follows: G Economic Conditions Good; F
Economic Conditions Fair; Po Economic Conditions Poor; U Stock price goes up and D Stock price goes down. Then the probability that conditions are poor if the stock price is going down is a. P

D

Po
 b.

Po
 c.* d.
P
P
P


D
Po
D
D


Po
 e. None of the above.
4.
In the two problems above, the joint probability of a rising stock price and poor economic conditions is: a.* b. c. d. e.
P
P
P
P
P





U
U
U
U
U

Po


Po


Po


Po


Po


.
06

.
76

.
23

.
06

.
76
5.
If the probability that a child is a boy or a girl is equally likely and I have 4 children, what is the probability that at least one of the first three is a girl? a. .375 b.* .875 c. .625 d. .667 e. .750
Solution: Forget about the fourth child! The probability of at least one girl in 3 tries could be the sum of the probabilities of 1, 2 and 3 girls. To see how to do it this way look at the examples where we found the probabilities of 1, 2 or three heads. To do it right, find
P
 

1

P

BBB


1

.
5

.
5

.
5

1

.
125

.
875 .
6.
A factory has three veeblefetzers. Veeblefetzer A produces defective items 4 percent of the time is only used for 10% of the output. Veeblefetzer B produces 20% of the output, and 3% of its output is defective. Because only 2% of the output of veeblefetzer C is defective, 70% of the output is produced on veeblefetzer C. If a defective item is found, the probability that it comes from veeblefetzer C is closest to a. 70% b.* 60% c. 50% d. 40% e. 30%
2

251y0224 10/30/02
Solution: Here is the completed table.
A B
D 0.4
D
0.6
9.6 19.4
Total 10
C
1.4
68.6
20 70
Total
2.4
97.6
100
To get the ‘total’ row, take 100 and multiply it by the percents given for production by each machine.
To get the D row, multiply the total by the percent defective. For example, 4% of 10 is 0.4. Get the
total defective by adding the row. Then the fraction of defective items that come from
of 1.4 or 50%. Formally, you were given the following: P
 

.
10 , P
 

.
20 ,
P
 

.
04 , P
 

.
03 and P
 

.
02 . You were asked for P
C is 0.7 out
P
 

.
70 ,
 
. By Bayes’ rule,
P
P

   
 
.
04

.
P

10
D

P

A
.
03
P

P
.
20

D


.
024
B
.
01

P
.
70

D
.
014
.
024


C

.
003



0 .
5833
P
 
.
004
A

, where we found
P
 
.
007


P
 
.
014
B P
P
 
using
 

P D C P
 
7.
In problem 6, the proportion of output that is defective is closest to a.* 2% b. 3% c. 3.5% d. 4% e. 1.5%
8.
The event A has a probability of p ; the event B also has a probability of p . If the two events are independent a.* b. c. d.
P

A

B

P
P
P
P




A
A
A
A



B
B

B
B









2 p
 p
2
2 p
 p
0 p
2 p
2
2 e.
 
Solution: From the addition rule P

A

B
     
P B

P

A

B p

and, since
 p
 p

A and B are independent, p
2 
2 p
 p
2
.
P

A

B


P
A
A

P
B

. P

  p
2
 p
. So
,
3

251y0224 10/30/02
9.
If the events independent, a. b. c.
P
P



A
A


A and B each have probabilities above zero and below 1, A and B are not
P
B
B



A B
is 1.

.
75 and P
 

is below .6 and above zero.
P A

B is between .6 and .75.
A .
6 , then the following must be true: d.* Either b. or c. could be true. e. There is not enough information to answer this question.
Solution: From the multiplication rule,
P

A

B
 
B

A


P
 
P

A

B

must be less than both P
P
A
P
 
 

A

and
B
. Since
P


B
P
 
 
P
 
and
.
P
 
and
P
 
are both less than 1,
10.
The following table shows the number of days absent among your employees in a month and the probability.
Number of days absent Probability
0 .50
1 .30
2 .11
3 .05
4 .04
5 0
What is the expected value (mean) of the number of days absent? a. 1.00 b.* 0.83 c. 2.00 d. 0.166 e. 10.0
Solution: Here is the completed table for both the definitional and the computational formula as generated by Minitab. There is no reason to bother with the definitional method.
row x P
  xP
  x
2
P
  x
 
 x
      x
    
1 0 0.50 0.00 0.00 -0.83 -0.4150 0.344450
2 1 0.30 0.30 0.30 0.17 0.0510 0.008670
3 2 0.11 0.22 0.44 1.17 0.1287 0.150579
4 3 0.05 0.15 0.45 2.17 0.1085 0.235445
5 4 0.04 0.16 0.64 3.17 0.1268 0.401956
1.00 0.83 1.83 0.0000 1.1411
We have a valid distribution since all the probabilities are between one and zero and add to 1. The formula for a population variance for a distribution is
E
 

 x
2
P

1 .
83 . So

2 x

1 .
83


2 x

E
 

0 .
83

2 
  2 x
, where
1 .
1411 and


 x
E

 

 xP
1 .
1411
 


0 .
83
1 .
06822
and
.
4

251y0224 10/30/02
11.
What is the value closest to the standard deviation of the number of days absent in problem 10? a.* 1.07 b. 1.14 c. 0.69 d. 2.50 e. 1.58
12.
What is F
  using the table in problem 10? a. 1.00 b. .08. c.* .96. d. .04 e. None of the above. For credit, write in an answer, showing your work.
Solution: You have been asked for the probability that x is less than or equal to 3.
F

 
P
 x


P
0
 x


3
P
 x

1
  x

2
  x

3


.
5

.
3

.
11

.
05

.
96
13.
If I know that seven workers are available to fill four openings, in how many ways can I fill them? a. 2401 b. 840 c. 28 d.* 35 e. 65536
Solution: Since order is not important and C r n  n !
 n
 r

!
r !
, C
4
7 

7

7 !
4

!
4 !

 
7 !
!
4 !

7

6

5
3

2

1

210
6

35 .
14.
The number of ways that 4 items can be taken from 13 if order is important and replacement is not allowed is a. 0 b. 1 c. 1365 d. 50625 e.* None of the above. For credit, write in an answer, showing your work.
Solution: Since order is important and P r n  n !
 n
 r

!
, P
4
13 

13 !
13

4


13 !
9 !

13

12

11

10

17160 .
5

251y0224 10/30/02
15.
If x is a standardized variable (z-score) and y

3

6 x , the expected value of y is a.* 3 b. -6 c. -3 d. 0 e. None of the above. For credit, write in an answer, showing your work
Solution: Since if y
 ax
 b ,
 y

E
 
 aE
 
 b and

2 y

Var
 
 a
2
Var and a standardized variable has a mean of zero and a standard deviation of 1, let compute
 y

E
 
 
6 E

3
 
6
 

3

3 and

2 y

Var y a



6

and b

3 and

36
16.
If x is a standardized variable (z-score) and y

3

6 x , the standard deviation of y is a. 36 b. -6 c. 1 d.* 6 e. None of the above. For credit, write in an answer, showing your work
Solution: From the above

2 y

Var
      
36 , so
 y

36

6 .
17.
As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that as long as one of the two components is working, the jorcillator works. If the probability of the phillinx failing in the first month of service is .6, 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 .5 and probability of the flubberall failing in the second month is .3, what is the probability that both components fail in the first month? Assume that the failure of one component is independent of failure of the other. a.* .30 b. .80 c. 1.00 d. 0 e. None of the above. For credit, write in an answer, showing your work
Solution: The problem asks nothing about the jorcillator. Let A
1
be the probability that the phillinx fails in the first period, let A
2
be the probability that the phillinx fails in the second period, and let A
3 be the probability that the phillinx fails in the third period (after the first and second period). Let B
1
be the probability that the flubberall fails in the first period, let B
2
be the probability that the flubberall fails in the second period, and let events are independent. Then we have been asked for

.
6
 

.
30 .
B be the probability that the flubberall fails in the third period. The
3
P

A
1

B
1

A
1
B
1
6

251y0224 10/30/02
18.
As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that as long as one of the two components is working, the jorcillator works. If the probability of the phillinx failing in the first month of service is .6, 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 .5 and probability of the flubberall failing in the second month is .3, what is the probability that the jorcillator will last beyond 2 months? Assume that the failure of one component is independent of failure of the other. a. .02 b. .30 c.* .28 d. .32 e. None of the above. For credit, write in an answer, showing your work
Solution: The problem asks for the probability that the jorcillator lasts beyond 2 months. We already know that P

A
1

B
1
A
1
B
1

.
6
 

.
30
A
1
. Make the following table.
A
2
A
3
Total
.15 .05 0.50
B
1
.30
B
2
.18
.09 .03 0.30
B .12
3
.06 .02 0.20
The event
Total .60 .30 .10 1.00
A
1

B
3
, for example will down the jorcillator in the third period because as long as one of the two components is working, the jorcillator works. If we add together all the events that down the jorcillator in the third period, we find
P

A
1

B
3
A
2

B
3
A
3

P

.
12
A
3


.
06
B
3



.
02
P

.
3
03


P
.
05
B
3

.
28

B
3
A
3

B
2
A
3
. This turns out to be the same as
A
3

B
3


.
1

.
2
 

B
1
.
28

19.
For the joint probability table below, the probability of the union of
B
B
A
.10
(not A ) and B is:
A .30
.10
A
.50 a. .50 b. .80 c.* .70 d. Above 1 e. None of the above. For credit, write in an answer, showing your work
Solution: By the addition rule could also say that P

A

B

P


A
P



A
B

  
B



1


P
P

 
A


B
P



A
1


B
.
30



.
6
.
7

.
.
6

.
5

.
7 . You
20.
In problem 19, A a. Complements
and B are b. Mutually exclusive c. Collectively exhaustive d. Independent e.* None of the above.
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251y0224 10/30/02
Part II. Show your work!
The number of days of absence last month for a sample of 6 employees appears below. (This is a sample not a probability distribution!) Compute the sample standard deviation (5 Points - 2 Point Penalty for not trying.)
So s
2
 x

24 ,

 x
2  n x
2 n

1
 x

2
Number of
Days
0
3
0
5
15
1
24
 

260 and n

6 . x
260

6

4 .
00

2
5
0
9
0
25
225
1
260

2 x

 x n
32 .
80 s


24
6

4 .
00
32 .
80

5 .
72713
8

251y0224 10/30/02
Part III. Do at least 1 (5+ Points) of the two following problems: Show your work! You receive extra credit for extra work!
1. You are dealt three cards from a deck. Remember that there are 4 cards of each denomination (Aces. 2's, etc.) but 13 cards of each suit (Hearts, Clubs etc). Let A be the event that you get two diamonds and B the event that you get one hearts. a. How many possible hands are there? (2) Work out answer to a) . You may leave the answers to the remaining sections of this question in factorial form.
Solution: C r n 
 n
 n r
!

!
r !
, C
3
52 

52
52

!
3

!
3 !

52
 
!
3 !

52
3


51
2


1
50

132600
6

22100 b. P
 
(2)
Solution: P
 

C
2
13
C
1
39
C
3
52



13 !
11 !
2 !




39 !
38 !
1 !




52 !
49 !
3 !





13

12
2

1


22100
39
1

3042
22100

.
13765 c. P

A

B

(2)
Solution: P

A

B


C
2
13
C
1
13
C
3
52



13 !
11 !
2 !




13 !
12 !
1 !





52 !
49 !
3 !




13
2


12
1


22100
13
1

1014
22100

.
04588
9

251y0224 10/30/02
2. We are trying to sell residential real estate. Let least 98% of the asking price. P
 

.
30 . Let
A be the event that the selling price of the property is at
B be the event that the property takes more than 3 months to sell. If the selling price is at least 98% of the asking price, there is a 20% probability that the property takes more than 3 months to sell. If the selling price is less than 98% of the asking price, there is a
40% probability that the property takes more than 3 months to sell. a. Find
Solution: Since
P
P
 
 

.
30 , P
 

.
20 and P
 

.
40 , it must be true that P

.
80 and for 100 properties, we can make the following table
B
P

A

B

P

P
 
P
A 6
28
A
34
 

.
20
 

.
06 , P

A

B

  
P

A

B
 
P

A

B


.
06

.
28

.
34 .

P
B
24
42
30
70
B
66
A P
 

.
40 .
100
 

.
28 , so
b. If the property is on the market for more than 3 months, what is the probability that it will sell for more than 98% of the asking price?
Solution: From the table,
P
   
P
A P
 
 
P

 
.
20 .
.
34

 
6
34


.
17647
.
17647
, or by Bayes’ rule,
10