251y0431d 5/03/04
ECO251 QBA1
THIRD HOUR EXAM
Apr 20, 2004
Name: ___KEY______________
Student Number: _____________________
Class Time (Circle) 1pm 2pm
Part I: 10 points.
IQs are supposedly Normally distributed with a population mean of 100 and a standard deviation of 16
x ~ N 100 ,16  . Find the following. Make diagrams! Add a vertical line at zero to the diagrams below.
.
78 .4  100 

Px  78 .4   P  z 
  Pz  1.35 
16


 P1.35  z  0  Pz  0  .4115  .5  .9115
Colors in diagram are reversed.
0.4
0.3
f
1.
0.2
0.1
0.0
-4
-3
-2
-1
0
1
2
3
4
1
2
3
4
1
2
3
4
x
2.
0.4
0.3
f
121 .6  100 
 78 .4  100
P78 .4  x  121 .6  P 
z

16
16


 P 1.35  z  1.35 
 2P0  z  1.35   2.4115   .8230
0.2
0.1
0.0
-4
3.
P47 .52  x  84 .2
-2
-1
0
x
0.4
0.3
f
84 .2  100 
 47 .52  100
 P
z

16
16


 P3.28  z  0.99 
 P 3.28  z  0  P 0.99  z  0 
 .4995  .3389  .1606
(The value for 3.28 comes from the
lower part of the Normal table.)
-3
0.2
0.1
0.0
-4
-3
-2
-1
0
x
251y0431d 4/20/04
4.
P111 .04  x  121 .6
 .1566
0.3
f
121 .6  100 
111 .04  100
 P
z

16
16


P0.69  z  1.35 
 P0  z  1.35   P0  z  0.69   .4155  .2459
0.4
0.2
0.1
0.0
-4
-3
-2
-1
0
1
2
3
4
x
5. To get into Mensa you must be in the top 2% 0f the population. What IQ do you need? (Hint: find x.02 )
the 98th percentile. Make a diagram: Show a Normal curve with a mean at zero. Label the area below zero
‘50%.’ By definition z .02 is the point with a probability of 2% above it, which means it must have 98%
below it and 98% - 50% = 48% between it and zero. We can try to find a point of the Normal table with
P0  z  z.02   .4800 . The closest we can come to .4800 is P0  z  2.05   .4798 or
P0  z  2.06   .4803 . Either is acceptable, but a good compromise might be z.02  2.054 . So, if you use
the formula, x    z , your answer could be x.02  100  2.0516   132 .8 , x.02  100  2.054 16 
132 .9  100 

 132 .9 or x.02  100  2.0616   133 .0 . Check: Px  132 .9  P  z 
  Pz  2.06 
16


 Pz  0  P0  z  2.06   .5  .4803  .0197  .02
2
251y0431 4/20/04
Part II: (20+ points) Do all the following: All questions are 2 points each except as marked. Exam is
normed on 50 points including take-home. (Showing your work can give partial credit on some problems!
In open-ended questions it is expected. Please indicate clearly what sections of the problem you are
answering and what formulas you are using. Neatness counts!)
1.
Thirty-six of the staff of 80 teachers at a local intermediate school are certified in CardioPulmonary Resuscitation (CPR). In 180 days of school, about how many days can we expect that
the teacher on bus duty will likely be certified in CPR?
a) 5 days
b) 45 days
c) 65 days
d) *81 days
36
 .45
Explanation: If you want to get technical this is a Binomial distribution with p  80
and n  180 . So   np  .45180   81 .
2.
What type of probability distribution will most likely be used to analyze the number of cars with
defective radios in the following problem?
From an inventory of 48 new cars being shipped to local dealerships, corporate reports
indicate that 12 have defective radios installed. The sales manager of one dealership wants to
predict the probability that out of the 8 new cars it just received, when each is tested, no more
than 2 of the cars have defective radios.
a) binomial distribution.
b) Poisson distribution.
c) *hypergeometric distribution.
d) none of the above.
Explanation: This is hypergeometric and not binomial because there are only 12 cars with
defective radios. On the first pick the probability of getting a defective radio is p  12 48 , but
on the second pick p  11 47 or 12 47 . Without a constant value for p , we cannot have a
binomial distribution.
A company has 125 personal computers. The probability that any one of them will require repair
on a given day is 0.025. To find the probability that exactly 20 of the computers will require
repair on a given day, one will use what type of probability distribution?
a) *binomial distribution.
b) Poisson distribution.
c) hypergeometric distribution.
d) none of the above.
Explanation: This is binomial and not hypergeometric because the probability that any pc
needs repair is a constant .025. Though ‘binomial’ is the correct answer , the binomial
distribution can be approximated by the Poisson because np  .125
 5000  500.
025
3.
4.
The probability that a particular type of smoke alarm will function properly and sound an alarm in
the presence of smoke is 0.8. You have 2 such alarms in your home and they operate
independently. The probability that both sound an alarm in the presence of smoke is __.64____.
Explanation: If you want to get technical this is a Binomial distribution with p  .8 and
n  2. So Px  C xn p x q n x and P2  C22 .8 2.2 0  .64. . Alternatively, let A1 be ‘First
alarm works’ and A2 be ‘Second alarm works.’ If the events are independent, P A1  A2 
 P A1  P A2   .8 2  .64.
3
251y0431 4/20/04
5.
The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6
outages per year. The probability that there will be at least 1 power outage in a year is _0.9975___.
Explanation: Px  1  1  P0  1  .00248  .99752 .
6.
An Undergraduate Study Committee of 6 members at a major university is to be formed from a
pool of faculty of 18 men and 6 women. If the committee members are chosen randomly, what is
the probability that all of the members will be men? (3)
Solution: This is hypergeometric because we are taking a sample of 6 that is more than
5% of a population of 18 + 6 = 24. The formula is Px  
C nNxM C xM
C nN
, which gives the
probability of x  6 successes in a sample of n  6 taken from a population of N  24
in which there are M  18 successes.
18!
C 06 C 618 12! 6! 18 17 16 15 14 13
P6 


 .1379 , but you could also argue that the
24!
24  23  22  21  20 19
C 624
18! 6!
probability of a male on the first try is
18
17
, on the second try is
etc.
23
24
7.
The quality control manager of Marilyn’s Cookies is inspecting a batch of chocolate chip cookies.
When the production process is in control, the average number of chocolate chip parts per cookie
is 6.0. What is the probability that any particular cookie being inspected has at least 6.0 chip parts?
Solution: Use the Poisson table. Px  6  1  Px  5  1  .44568  .55432
8.
(Dummeldinger) The amount of time between pauses on a full-screen edit terminal is uniformly
distributed between 0.2 and 0.8 seconds. What is the expected (mean) pause time for the editor?
[16]
a) 0.4 seconds
b) *0.5 seconds
c) 0.6 seconds
d) 0.8 seconds
e) 1.0 seconds
f) 1.67 seconds
Solution: From the outline for the continuous uniform distribution
1
f x  
for c  x  d and f x   0 otherwise.
d c
d  c 2 . In this case c  .2 and d  .8, so
cd
and  x2 
  E x  
12
2
.2  .8
  E x  
 0.5.
2
9.
I am a real estate agent who gets an average of 1 customer every three days ( p  13 ). If I go
away for three days, what is the chance that I miss a (one or more)customer?
Solution: This was Problem L5a) (geometric distribution) p  13 , so q  2 3 and
3
F x  1  q x . So Px  3  F 3  1  2 3   1  8 27  .70370 . There is another way to
4
do this. You could consider the problem Binomial with p 
Px  0  1  P0
 1  C03 p 0 q 3
1
3
and n  3.
 1  8 27  .70370.
251y0431 4/20/04
10. In problem 9, what is the chance that my first customer comes in on the third through the tenth day
after I go away? [20]
Solution: In the previous problem p  13 , so q  2 3 and F x  1  q x . So


P3  x  10   F 10   F 2  1  2 3   1  2 3   2 3   2 3 
10


 4 9  1  2 3   4 9 .9609815   .4271
8
2
2
10
11. From an inventory of 900 new cars being shipped to local dealerships, corporate reports indicate
that one-fourth have defective radios installed. What is the probability that out of the 8 new cars it
just received, when each is tested, no more than 2 of the cars have defective radios?
Solution: Hey! Doesn’t this sound a lot like problem 2. The difference is that the
population of 900 is more than 20 times the sample size of 8. Use the binomial table.
p  .25 , n  8 and we want Px  2  .67854 .
12. A concert hall has a capacity of 98 people and 100 tickets have been sold. There is a 3%
probability that any given ticket-buyer will not use his/her ticket. What is the chance that everyone
who shows up will get a seat? Hint: you need the probability that 2 or more people out of n  100
do not come. Since you do not have tables for p  .03 , show that a Poisson distribution applies
and use it for your answer. (3)
n 100

 3333 .33  500   np  100 .03  3 . Use the Poisson table with a
Solution:
p .03
parameter of 3. Px  2  1  Px  1  1  .11915  .88085 .
13. Extra credit: The geometric distribution is a special case of the negative binomial distribution. The
probability that the m th success occurs on try x when the probability of success on any one try is
p is Px  Cmx11 p m q xm . Use this to find a) the probability of the first success on the 5 th try and
b) the probability of the second success on the 5th try. Show that your first result is identical to the
geometric distribution. (3) [28] p  .1
Solution: a) For the geometric distribution, p  .1, q  1  p  .9 P( x)  q x 1 p , so
P(5)  q 4 p  .9 4.1  .06561. For the negative binomial m  1, so
P5  C051 p1q 4  q 4 p  .9 4.1  .06561
b) For the negative binomial with p  .1 and m  2,
P5  C14 p 2 q 3  4.12 .93  .0090496 .
5
251y0431 4/20/04
ECO251 QBA1
THIRD EXAM
Apr 20, 2004
TAKE HOME SECTION
Name: ________KEY_____________
Student Number: _________________________
Throughout this exam show your work! Please indicate clearly what sections of the problem you are
answering and what formulas you are using. Though I do not want typed answers, neatness counts!
Part II. Do all the Following (20 Points) Show your work!
Before you start find the number that we shall call a . It is the third to last digit of your student number. For
example Seymour Butz’s student number is 976500, so a  5 . a can be zero.
1. The table below represents the returns of two stocks. Change the table by adding .01a to .2 and
subtracting .01a from .6. (Seymour changes .2 to .25 and .6 to .55.)
x
200
y
0  100
100
90
.2
0
.1
50
0
0
0
.6
0
 180
0
.1
0
0
a) Ex    x (1), b) E  y  (1), c)  x (1), d)  y (1), e)  xy (1) Returns are
Now find the following.
measured by E
and risk by the coefficient of variation, which is

. Now create portfolios by letting w

vary from zero to 1 by steps of .1 and saying that the return is R  wx  1  wy . (This stuff is done starting
on page 65 of the supplement.) Compute the mean, standard deviation and coefficient of variation for each
portfolio. (9) What portfolio would you recommend for a person who doesn’t care about risk, for a super
cautious individual, for you, for me? Why? Write a useable report on your results. (2) [16]
Two versions of the problem are below.
Solution 1: Looking below, we find a) E x   40 , b) E  y   39, c)  x  91 .6515 , d)  y  75.1598 and
e)  xy  660
x
90
y
50
 180
Px 
xPx 
200 100 0  100
0.2
0
0 0.1


0 0.6
0
 0
 0 0.1
0
0


0.2
0.1 0.6
0.1
40
10
x Px  8000 1000
2
yP y  y 2 P y 
27
2430
0.6
30
1500
0.1
 18
3240
 1.0
39
7170
 10
 40
0 1000
 10000
0
 Px  1 ,   Ex   xPx  40 E x    x
 E  y    yP y   39 and E y    y P y   7170
To summarize
y
P y 
0.3
2
x
2
2
Px   10000
 P y   1 ,
2
6
251y0431 4/20/04
 0100 90 
 0090   0.1 100 90 
 0.2200 90 

E xy  
xyPxy     0200 50 
 0100 50   0.6050 
 0 100 50 
 0200  180   0.1100  180   0 180   0 100  180 
 0  0  900 
3600

  0
0 0
 0  900
  0  1800  0
 0

 xy  Covxy  Exy   x  y  900  4039  660 ,
 
 
 x2  E x 2   x2  10000  40 2  8400 and  y2  E y 2   y2  7170  392  5649 . So that
 xy
 0.660
660 2  
8400 5649 
.009180  0.09581 . (  x  91.6515,  y  75.1598 )
8400 5649
The correlation and covariance are negative, indicating a tendency of y to fall when x rises.
 xy 
 x y


2
 xy
 .009180 hardly exists on a zero to one scale, indicating that the relationship is very weak.
f) From the posted solution to Exercise 5.10. -- We find that E ( R)  wE ( x)  1  wE ( y) and
Var ( R)  w 2Var ( x)  1  w2 Var ( y)  2w1  wCov( x, y) or  R2  w 2 x2  1  w2  y2  2w1  w xy .
So if w=.1 E ( R)  .1E ( x)  .9E ( y)  .140   .939   39.1
 R2  .12 8400  .92 5649  2.1.9660  84 .00  4575 .69  118 .80  4540 .89 and
 67 .3861
 R  4540 .89  67 .3861 C  
 1.72

39 .1
Obviously all the expected returns are almost same. My results for the proportions are below.
The largest return comes from buying x alone. The safest package is 40% x and 60% y . Any
portfolio that someone would actually buy would be between these. A risk-loving investor would buy
only x . A cautious investor would take the safest package. The rest of us would fall between. I might
advise younger investors to put 80% or 90% in x , but older investors ought to decrease the part of
their portfolios in x down to 50% or 60%.
w
1  w
0
.1
.2
.3
.4
.5
.6
.7
.8
.9
1.0
1.0
.9
.8
.7
.6
.5
.4
.3
.2
.1
0
w40  1  w39   E R  w 2 8400 + 1  w2 5649 2w1  w660   R2
0
4
8
12
16
20
24
28
32
36
40
39.0
35.1
31.2
27.3
23.4
19.5
15.6
11.7
7.8
3.9
0.0
39.0
39.1
39.2
39.3
39.4
39.5
39.6
39.7
39.8
39.9
40.0
0
84
336
756
1344
2100
3024
4116
5376
6804
8400
5649.00
4575.69
3615.36
2768.01
2033.64
1412.25
903.84
508.41
225.96
56.49
0.00
0.0
-118.8
-211.2
-277.2
-316.8
-330.0
-316.8
-277.2
-211.2
-118.8
0.0
5649.00
4540.89
3740.16
3246.81
3060.84
3182.25
3611.04
4347.21
5390.76
6741.69
8400.00
R
75.1598
67.3861
61.1568
56.9808
55.3249
56.4114
60.0919
65.9334
73.4218
82.1078
91.6515
C
1.93
1.72
1.56
1.45
1.40
1.43
1.52
1.66
1.84
2.06
2.29
7
251y0431 4/20/04
Solution 2: Looking below, we find a) E x   58 , b) E  y   42 .6, c)  x  101 .173 , d)  y  76.5457 and
e)  xy  49.2
x
90
y
50
 180
Px 
xPx 
200 100 0  100
0.29
0
0 0.1


0 0.51
0
 0
 0 0.1
0
0


0.29
0.1 0.51
0.1
58
x 2 Px  11600
E xy  

10
1000
P y 
0.39
0.51
25 .5
1275
0.10
 18 .0
3240
 1.00
42 .6
7674
 10
 58
0 1000
 13600
0
yP y  y 2 P y 
35 .1 3159
 Px   1 ,
  E x    xPx   58
E x    x Px   13600
 P y   1 ,
  E  y    yP y   42 .6
and E y    y P y   7674
To summarize
x
2
2
y
2
2
 0100 90 
 0090   0.1 100 90 
 0.29 200 90 

xyPxy     0200 50 
 0100 50   0.51050 
 0 100 50 
 0200  180   0.1100  180   0 180   0 100  180 
 0  0  900 
5220

  0
0 0
 0  2520
  0  1800  0
 0
 xy  Covxy  Exy   x  y  2520  5842.6  49.2 ,
 
 
 x2  E x 2   x2  13600  58 2  10236 and  y2  E y 2   y2  7674  42.62  5859.24 . So that
 xy 
 xy
 x y

49 .2
10236 5859 .24
(  x  101.173,  y  76.5457 )

49 .22

10236 5859 .24 
.00004036  0.006353 .
The correlation and covariance are positive, indicating a tendency of y to rise when x rises.
2
 xy
 .00004036 hardly exists on a zero to one scale, indicating that the relationship is very weak.
f) From the posted solution to Exercise 5.10. -- We find that E ( R)  wE ( x)  1  wE ( y) and
Var ( R)  w 2Var ( x)  1  w2 Var ( y)  2w1  wCov( x, y) or
So if w=.1 E ( R)  .1E ( x)  .9E ( y)  .158   .942.6  44.14 ,
 R2  w 2 x2  1  w2  y2  2w1  w xy .
 R2  .12 10236  .92 5859 .24  2.1.949.2  102 .36  4745 .98  8.86  4857 .20 and
 69 .6936
 R  4857 .20  69 .6936 . C  
 1.58

44 .14
8
251y0431 4/20/04
My results for the proportions are below. The largest return comes from buying x alone. The safest
package is 40% or 50% x and 60% or 50% y . Any portfolio that someone would actually buy would
be between the lowest risk and the highest return. A risk-loving investor would buy only x . A
cautious investor would take the safest package. The rest of us would fall between. I might advise
younger investors to put 80% or 90% in x , but older investors ought to decrease the part of their
portfolios in x down to 60% or 70%.
w
1  w
0 1.0
.1 .9
.2 .8
.3 .7
.4 .6
.5 .5
.6 .4
.7 .3
.8 .2
.9 .1
1.0
0
w58  1  w42.6  E R  w 2 10236 + 1  w2 5859 .24 2w1  w49.2   R2  R
0.0
5.8
11.6
17.4
23.2
29.0
34.8
40.6
46.4
52.2
58.0
42.60
38.34
34.08
29.82
25.56
21.30
17.04
12.78
8.52
4.26
0.00
42.60
44.14
45.68
47.22
48.76
50.30
51.84
53.38
54.92
56.46
58.00
0.0
102.4
409.4
921.2
1637.8
2559.0
3685.0
5015.6
6551.0
8291.2
10236.0
5859.24
4745.98
3749.91
2871.03
2109.33
1464.81
937.48
527.33
234.37
58.59
0.00
0.000 5859.2 76.546
8.856 4857.2 69.694
15.744 4175.1 64.615
20.664 3812.9 61.749
23.616 3770.7 61.406
24.600 4048.4 63.627
23.616 4646.1 68.162
20.664 5563.6 74.590
15.744 6801.2 82.469
8.856 8358.6 91.425
0.0
10236.0 101.173
C
1.80
1.58
1.41
1.31
1.26
1.26
1.31
1.40
1.50
1.62
1.74
9
251y0431 4/20/04
2. (McClave et al.) You are an inspector for the EPA and want to compare miles per gallon that cars get
with those projected in EPA’s mileage guide. On the basis of previous experience, you believe that the
percent that are accurate (within 2 mpg) is b per cent, where b  50  5a use 55% if your a is zero. (Since
Seymour has a  5 , he uses b  50  5 5  75 per cent.) Tell what distribution you are using and show
your work.
a. If you believe that b per cent applies to a garage with 100 cars in it, what is the chance that at least one
of a sample of 20 deviates from the mpg in the mileage guide. (2)
b. If the same percent applies to a large fleet of cars, what is the chance that more than half of a sample of
20 do not deviate from the mpg in the mileage guide? (1)
c. If you are sampling from this large fleet, what is the chance that the first car that you find that has mpg as
in the mileage guide is the first you look at? The second? The third? (1.5)
d. How large a sample would you have to use to be able to answer b) with the Poisson distribution? (1.5)
e. (Extra credit) The number of serious accidents in a plant has a Poisson distribution with a mean of 2, or
we could say that the average waiting time between accidents is 0.5 months. Show that if a serious accident
occurs today the probability that the next serious accident will not occur in the next month can be found by
i) finding the Poisson probability of no serious accidents over a month and ii) finding the exponential
probability that we will wait more than a month before a serious accident (This cannot possibly be 1  P0 .
Why?) and (iii) noting that they are the same. What is the mathematical reason for this? (3) [24]
Solution 1: a  0 , b  55 % are accurate, 45% are inaccurate
a) Hypergeometric Distribution: N  100 , M  np  .45100   45 deviate, n  20. Px  1  1  P0
45!
25!20!
55  54  53  52  51  50  49  48  47  46  45  44  43  42  41  40  39  38  37  36
 1
 1
 1
100
100!
100  99  98  97  96  95  94  93  92  91  90  89  88  87  86  85  84  83  82  81
C 20
20!80!
 1  .000001595  .999998
b) p  .55 , n  20 . Px  11  1  Px  10  But if 55% are accurate, 45% are inaccurate and in the sample
11 or more are accurate, which means 10 or fewer are inaccurate. From the Binomial table with p  .45
and n  20 , Px  10   .75071 .
55
C 045 C 20
c) On the first try .55, on the second try .45.55   .02475 and on the third try .452 .55  .1114
n
n

 500 This implies n  .45500   225 or larger. We have a choice of using .45 or .55 in this
p .45
problem. Since the Poisson distribution is supposed to be used to deal with large counts and small
probabilities, .45 is the better choice for p . If you used .55 you should have gotten 275. In any case, to
actually do this, we would have to jump to the Normal distribution. I did not ask for this but if n  225 , the
d) If

10 .5  101 .25 
mean would be np  .45225   101 .25 Px  10   P  z 
  Pz  9.02   0
101 .25 

10
251y0431 4/20/04
e m m x
gives probability of x successes in an interval in which the average
x!
1
number of successes is m . For the exponential, F x  1  ecx , where the mean time to a success is ,
c
e) For the Poisson, Px  
1
e 2 2 0
 .5 implies that c  2 . Thus (i) for the Poisson with m  2 P0 
 e  2 , and (ii) for the
c
0!
exponential Px  1  1  Px  1  1  1  e 21  e 2  .13534 . Note that Px  1  Px  1  1  P0


because P0 is meaningless in a continuous distribution.
Solution 2: a  9 , b  95 % are accurate, 5% are inaccurate.
a) Hypergeometric Distribution: N  100 , M  np  .05100   5 deviate, n  20. Px  1  1  P0
95!
75!20!
95  94  93  92  91  90  89  88  87  86  85  84  83  82  81  80  79  78  77  76
 1
 1
 1
100!
100  99  98  97  96  95  94  93  92  91  90  89  88  87  86  85  84  83  82  81
20!80!
 1  .31931  .6807
b) p  .95 , n  20 . Px  11  1  Px  10  But if 95% are accurate, 5% are inaccurate and in the sample
11 or more are accurate, which means 10 or fewer are inaccurate. . From the Binomial table with p  .05
and n  20 , Px  10   1.00000 .
95
C 05 C 20
100
C 20
c) On the first try .05, on the second try .95.50   .0475 and on the third try .952 .05  .0451
n
n

 500 This implies n  .05500   25 or larger. We have a choice of using .05 or .95 in this
p .05
problem. Since the Poisson distribution is supposed to be used to deal with large counts and small
probabilities, .05 is the better choice for p . If you used .95 you should have gotten 475. I did not ask for
this but if n  25 , the mean would be np  .0525   1.25. We do not have a Poisson table for a parameter
of 1.25, but Px  10   1.00000 for parameters of both 1.2 and 1.3, so it should also be true for 1.25.
e) Same as above.
d) If
11