251y0632 4/17/06
ECO251 QBA1
THIRD EXAM
Version 2, Apr 21, 2006
Name: _______KEY__________
Student Number: _____________________
Class Time (Circle) MWF 1, MWF 2, TR 12:30, TR 2.
Part I: 16 points.
z follows the standardized Normal distribution z ~ N 0,1 .
Find the following. Make diagrams! Diagrams for both versions of this exam are in 251y063f.
1. P3.07  z  3.07   P3.07  x  0  0  z  3.07   .4989  .4989  .9978
Make a diagram. To make a diagram for z , draw a Normal curve with a center at zero. Shade the area
below zero from -3.07 to zero. Shade the area above zero from zero to 3.07. Because you have shaded an
area on both sides of the mean, you are adding.
2. P3.15  z  3.00   P3.15  x  0  0  z  3.00   .4992  .4987  .9979
Make a diagram. To make a diagram for z , draw a Normal curve with a center at zero. Shade the area
below zero from -3.15 to zero. Shade the area above zero from zero to 3.00. Because you have shaded an
area on both sides of the mean, you are adding.
3. F 1.03  Pz  1.03  Pz  0  P0  z  1.03  .5  .3485  .8485
Make a diagram. To make a diagram for z , draw a Normal curve with a center at zero. Shade the entire
area below zero. Shade the area above zero from zero to 1.03. Because you have shaded an area on both
sides of the mean, you are adding.
4. z .19 Solution: Make a diagram. z .19 is defined as a point with 19% above it and thus 100% - 19 =
81% below it, so it is the 81 percentile or the .81 fractile. The diagram for z will show an area with a
probability of 81% below z .19 . It is split by a vertical line at zero into two areas. The lower one has a
probability of 50% and the upper one a probability of 81% - 50% =31%. The upper tail of the distribution
above z .19 has a probability of 19%, so that the entire area above 0 adds to 50%. From the diagram, we
want one point z .19 so that Pz  z.19   .8100 or P0  z  z.115   .3100 . The closest we can come to
.3100 using the Normal table is P0  z  0.88   .3106 , though P0  z  0.87   .3078 is somewhat
acceptable. We can say z .19  0.88 or, with less accuracy z .19  0.87 .
251y0632 4/21/06
x follows the Normal distribution x ~ N 3, 7  .
Find the following. Make diagrams!
3.07  3 
  3.07  3
z
 P 0.87  z  0.01
5. P3.07  x  3.07   P 
7
7 

 P0.87  z  0  P0  z  0.01  .3078  .0040  .3118
Make a diagram. To make a diagram for z , draw a Normal curve with a center at zero. Shade the area
below zero from -0.87 to zero. Shade the area above zero from zero to 0.01. Because you have shaded an
area on both sides of the mean, you are adding.
You probably do not have time to make a diagram for x too. If you choose to make a diagram for x , draw
a Normal curve with a center at 3. Shade the area from -3.07 to 3. Shade the area from 3 to 3.07. Because
you have shaded an area on both sides of the mean, you are adding.
3.00  3 
  3.15  3
z
 P 0.88  z  0  .3106
6. P3.15  x  3.00   P 
7
7 

Make a diagram. To make a diagram for z , draw a Normal curve with a center at zero. Shade the area
below zero from -0.88 to zero. Do not shade the area above zero. Because you have shaded an area on one
side of the mean that starts at the mean, you can simply find the result on the Normal table.
You probably do not have time to make a diagram for x too. If you choose to make a diagram for x , draw
a Normal curve with a center at 3. Shade the area from -3.15 to 3. Do not shade any area above 3. Because
you have shaded an area on both sides of the mean, you are adding. Because you have shaded an area on
one side of the mean that starts at the mean, you can simply find the result on the Normal table after
x
making the transformation z 
.

1.03  3 

 Pz  0.28   Pz  0  P0.28  z  0
7. F 1.03  Px  1.03  P  z 
7 

 .5  .1103  .3897 .
Make a diagram. To make a diagram for z , draw a Normal curve with a center at zero. Shade only the
area below zero that is also below -0.28. Do not shade any area above zero. Because you have shaded an
area on one side of the mean that does not touch the mean, you are subtracting.
You probably do not have time to make a diagram for x too. If you choose to make a diagram for x , draw
a Normal curve with a center at 3. Shade only the area to the left of 3 that is also below 1.03. . Because
you have shaded an area on one side of the mean that does not touch the mean, you are subtracting.
8. x.19
Though you would usually need a diagram for a problem like this one, you have made all the diagram that
you need on the previous page. On the previous page, we said “The closest we can come to .3100 using the
Normal table is P0  z  0.88   .3106 , though P0  z  0.87   .3078 is somewhat acceptable. We can
say z .19  0.88 or, with less accuracy z .19  0.87 .” The transformation from z to x is x    z . So we
can say x.19  3  0.887  9.16 or, less accurately x.19  3  0.87 7  9.09 .
9.16  3 

 Pz  0.88   Pz  0  P0  z  0.88   .5  .3106  .1894  19 %
Check: Px  9.09   P  z 
7 

2
251y0632 4/21/06
Part II: (15+ 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!) Remember that you
may not be able to finish this section, so ration your time on each problem. [Numbers in brackets are a
cumulative total]
1.
If x is binomial, and n  15, find the following (If you substitute another distribution for the
binomial, justify it!): Note: There should be some credit for doing a) using the Normal
distribution. It cannot be justified on b) – d). Stating  and  gets no credit.
a) P3  x  9 when p  .40 (1.5)   6,   1.8974 
P3  x  9  Px  9  Px  2  .96617  .02711  .9391
b) P6  x  11 when p  .75 . (1.5)
P6  x  11 can't be done directly with tables that stop at p  .5 , so try to do it with
failures. (The probability of failure is 1 - .75 = .25.) 6 successes correspond to 15 – 6 = 9 failures
out of 25 tries. 11 successes correspond to 4 failures. So try 4 to 9 successes when
p  .25 and n  15 . P4  x  9  Px  9  Px  3  .99921  .46129  .5379
c) P1  x  4 when p  .02 . (2)
Since n p  15.02  750  500 , we can use the Poisson distribution. The mean is
m  np  15.02   0.3 P1  x  4  Px  4  Px  0  .99998  .74082  .2595
d) Px  1 when p  .23 . (2)
Px  1  1  P0  1  C015 p 0 q15  1  .7715  1  .01983  .9802
2.
A student calculates that P10  x  35   F 35   F 9 . The student could be working with the
a) *Binomial Distribution
b) Normal distribution.
c) Continuous Uniform Distribution
d) Exponential distribution
e) All of the above
f) None of the above.
[9]
Explanation: The equation below was crushed by Word. Click on it or print it.
 F b   F a  1 for discrete
F x 0   Px  x 0  and Pa  x  b   
 F b   F a  for continuous.
3.
If  xy  .5 ,  x  3.2 ,  y  8.5 , w  5x  7 and v  3 y  4 , find the following:
a)  xy (1)
If w  ax  b and v  cy  d , we have a  5 and c  3 .
All formulas in this section come from 251var2 or 251v2out.  xy   xy x y
 .53.28.5  13.6
b)  x y (1.5)
Var x  y    x2   y2  2 xy  3.2 2  8.5 2  213 .6  10 .24  72 .25  27 .20  109 .85
 x y  109.85  10.4809
c)  wv (1.5)
Corr (ax  b, cy  d )  SignacCorr ( x, y)  Sign 5 3.5  Sign15 .5   1.4  .5
d)  v (1.5)
Varcy  d   c 2 Var y    32 8.52  972.25   650 .25
 w  650 .25  25.5
[14.5]
3
251y0632 4/21/06
4.
What type of probability distribution will most likely be used to analyze the number of chocolate
chip parts per cookie in the following problem?
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. The manager is interested in analyzing the probability that any
particular cookie being inspected has fewer than 5.0 chip parts.
a) Binomial distribution.
b) *Poisson distribution. See Great Distributions I Have Known.
c) Hypergeometric distribution.
d) None of the above.
[16.5]
5.
The West Nantmeal police department must write, on average, 5 tickets a day to keep department
revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson
distribution with a parameter of 6.5 tickets per day. Interpret the value of the parameter. (1.5)
a) The number of tickets that is written most often is 6.5 tickets per day.
b) Half of the days have less than 6.5 tickets written and half of the days have more than 6.5
tickets written.
c) *If we sampled all days, the arithmetic average or expected number of tickets written
would be 6.5 tickets per day.
d) The mean has no interpretation since 6.5 tickets can never be written. [18]
6.
(Extra credit ) Fidel Castro has just been given an honorary Doctor of Divinity degree by an
unnamed Venezuelan University. (So now he is no longer plain Fidel, now he is Fidel DD.) In
gratitude for this attention he gives a six hour speech. Assume that the sound system has a failure
rate of 1 in 8 hours, which means that the average wait for a failure is 8 hours. Use the exponential
distribution.
a) What is the probability that the system will fail during the 6-hour speech? (1.5)
b) What is the probability, that if a back-up system with the same characteristics as the
original system is used during the three quarters of an hour that the original system is
down for repairs, it will fail before the repair time elapses?. (1.5)
c) What is the variance of the elapsed time before a failure? (1)
Solution:
1
1
1
a) F x  1  ecx , when the mean time to a success is . So if  8, c  . So
c
8
c
Px  6  F 6  1  e 0.1256  1  e 0.75  1  .4724  .5276 .
b) Px  0.75  F 0.75  1  e 0.1250.75  1  e 0.09375  1  .9105  .0895
c) The sheet says  
1
1
 8 so  2 
 64 .
c
c2
4
251y0632 4/21/06
7.
Using the joint probability table below, (a) Fill in the missing number (1) (b) check for
independence (1), (c) Compute Covx, y  , Corr x, y  (4), (d) Compute Ex  y  and Var x  y  .


(2) (e) Find Px  y  4 and P x  y  4 x  0 (2).
y
[28]
0 2 6
 2 .1 0 .2


x
4  0 .1

5 .1 0 .2


Solution: a) .30. Sorry about the switched x and y . This is Downing and Clark, Computational
Problem 3, with one column removed and I guess I was crossing x and y , as many texts do.
b) Check for independence: First you need to find Px  and P y  . Look at the upper left hand
probability below. Its value is .10 and it represents Px  2   y  0 . If x and y are
independent, we would have Px  2   y  0  Px  2 P y  0  .20.20   .04 . Since this is
not true, x and y cannot be independent. Even one place where the joint probability is not the product
of the marginal probabilities is enough. Actually the best way to prove non-independence is to look for
zeroes. If Px  4   y  0  0 and x and y are independent, then it must be true that
Px  4  0 or P y  0  0 . Notice that the second row is not proportional to the first row or any
other row.
c) Compute Covx, y  , Corr x, y 
y
0
 .10

 0
 .10

.20
2
x
4
5
P y 
yP y 
y P y 
2
2
0
Px 
.30
.10
6
.20 

.30 
.20 

.70
0.0  0.2 
4.2 
4.4
0.0  0.4  25 .2 
25 .6
.10
0
xPx  x 2 Px 
 0.6
1.2
.40
1.6
6.4
.30
1.5
7.5
1.00
2.5
15 .1
 Px   1 (a check),   Ex   xPx  2.5 , E x    x
 P y   1 ,   E  y    yP y   4.4 and E y    y P y   25.6 .
2
To summarize
x
2
2
Px   15 .1 ,
2
y
So from the above
 xPx  2.5 , Varx  Ex 2   x2   x 2 Px   x2  15.1 2.52  8.85
 E  y    yPx   4.4 and Var y   E y 2   y2   y 2 P y    y2  25.6  4.42  6.24 .
 x  E x  
y
(Note  x  8.85  2.9749 and  y  6.24  2.4980 .)
.10  20 0 22 .20  26  0.0 0.0  2.4
xyPxy     040 .10 42
.30 46  0.0 0.8 7.2   11 .6
  .10 50 .052
.20 56 0.0 0.0 6.0 
 Covxy  Exy   x  y  11.6  2.54.4  0.6 .
E xy  
 xy

So that  xy 
 xy
 x y

0.6
 0.006519  .0807 .
8.85 6.24
5
251y0632 4/21/06
2
 .0065
The correlation and covariance are positive, indicating a tendency of y to rise when x rises.  xy
is very low on a zero to one scale, indicating an extremely weak (linear) relationship. Note that
1   xy  1 always!
d) Compute Ex  y  and Var x  y  .
Ex  y   Ex  E y    x   y  2.5  4.4  6.9 and
Var x  y    x2   y2  2 xy  Var x   Var  y   2Covx, y   8.85  6.24  20.60   16.29


e) Find Px  y  4 and P x  y  4 x  0 . The table below has the sum of x and y . It is followed by
the original joint probability table with the probabilities for points with Px  y  4 starred.
y
y
0 2 6
 2   2 0 4
2


x
4  4 6 10  x
4
5  5 7 15 
5


Px  y  4  .1  0  .2  0  .3
0 2 4
.1* 0 * .2 *


 0 * .1 .3
 .1 0 .2


Px  y  4  x  0  .1  0  .2  .3 . So Px  y  4 x  0  .3  1.0000
.3
6
251y0632 4/21/06
ECO251 QBA1
THIRD EXAM
Apr 25, 2005
TAKE HOME SECTION
Name: ____Computer Solution____
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.
Part III. Do all the Following (19+ Points) Show your work! Neatness counts!
1. Ben Horim and Levy present the following actual returns for stock in Republic Steel and General Foods
over a ten year period.
Year
R1
1
22.5
2
9.3
3 -18.6
4
48.1
5
44.4
6
56.0
7 -20.1
8 -11.0
9
17.1
10
22.3
R2
-23.7
8.6
4.3
-52.2
62.4
61.4
9.8
52.6
17.1
11.7
Before you start, personalize the data below as follows. Take the third to last digit of your student number
and subtract it from -52.2 in the R2 column. This will make the results for year 4 even more disappointing
than they actually were. (Example: Seymour Butz’s student number is 123450. He subtracts the 4 from
-52.2 getting -56.2) .
a) Find the sample mean, variance and coefficient of variation for R2. Use computational formulas.
The results must look as if you calculated them by hand, although it is perfectly reasonable to
check the results using a computer or a calculator. (2.5)
b) Find the sample covariance and correlation between R1 and R2. (3)
c) Using a conventional measure of risk, explain which stock is riskier. (1).
[6.5]
d) Make a table showing 11 portfolios. Start with a portfolio with all your money invested in stock 1
(P1 = 1, P2 = 0), then move to a portfolio with 90% in stock 1 and 10% in stock 2 (P1 = .90, P2 =
.10), and move by tenths until you get a portfolio that is all stock 2. (5). The table should look as
P1
1
.9
.8
.7
.6
follows with much wider columns.
.5
.4
.3
.2
.1
0
R
s
C
Here R is the weighted average return,
s is the standard deviation of the portfolio return, and C is the coefficient of variation. [11.5]
7
251y0632 4/21/06
e)
Make a graph . Put the mean return on the y axis and the standard deviation on the x axis. (3)
Connect your points in a curve. The C-like curve you will probably get illustrates the tradeoffs
between return and risk and is called an investment opportunities frontier.
f) Looking at the graph and the coefficients, are there stock portfolios that you would never
recommend? Why? What portfolio would you recommend to an 80-year old widow? Why? What
portfolio would you pick for yourself? Why? (1.5)
[16]
Solution: Let’s assume that Fillmore Barrols has student number 123456, so that he, she or it subtracts 4
from -52.2. This is Version 4 in Computations for the Portfolio Problem.
a) Find the sample mean, variance and coefficient of variation for R2. Use computational formulas.
The results must look as if you calculated them by hand, although it is perfectly reasonable to
check the results using a computer or a calculator. (2.5)
To save a bit of space, the display for both securities is below.
Row
1
2
3
4
5
6
7
8
9
10
x 2  R12
22.5
9.3
-18.6
48.1
44.4
56.0
-20.1
-11.0
17.1
22.3
170.0
506.25
86.49
345.96
2313.61
1971.36
3136.00
404.01
121.00
292.41
497.29
9674.38
y  R2
xy  R1 R 2
y 2  R22
-23.7
561.69
8.6
73.96
4.3
18.49
-56.2 3158.44
62.4 3893.76
61.4 3769.96
9.8
96.04
52.6 2766.76
17.1
292.41
11.7
136.89
148.0 14768.40
-533.25
79.98
-79.98
-2703.22
2770.56
3438.40
-196.98
-578.60
292.41
260.91
2750.23
 y  148 .0  y  14768 .4  xy  2750 .23
 x , s   x  nx , C  std.deviation and
I copied the following formulas from Table 20: x 
So n  10,
x
x  R1
 170 .0 ,
x
2
 9674 .38 ,
2
2
 xy  nxy . Thus for
n
2
x
2
n 1
mean
148 .0
14768 .40  10 14 .80 
 14 .80 , s 2y 
 1397 .5556 ,
10
n 1
9
37 .3839
 2.5259 . While we are at it, let’s take care of x  R1 .
s y  1397.5556  37.3839 and C y 
14 .80
s xy 
y  R2 , y 
2
170 .0
9674 .38  10 17 .00 2
 17 .00 , s x2 
 753 .82 , s x  753 .82  27 .4558 and
10
9
27 .4558
Cx 
 1.6150
17 .00
x
s xy
rxy
b) Find the sample covariance and correlation between R1 and R2.
2750 .23  10 17 .00 14 .8

 26 .0256 and
9
s xy
26 .0256


 .000642931  0.025356
sx s y
753 .82 1397 .556
c) Using a conventional measure of risk, explain which stock is riskier. (1).
y  R 2 has a coefficient of variation of C y  2.5259 , which is larger than C x  1.6150 , so the second
security is riskier.
8
251y0632 4/21/06
d) Make a table showing 11 portfolios. Start with a portfolio with all your money invested in stock 1
(P1 = 1, P2 = 0), then move to a portfolio with 90% in stock 1 and 10% in stock 2 (P1 = .90, P2 =
.10), and move by tenths until you get a portfolio that is all stock 2. (5).
If we check 251varmin, we find the following. “If R  P1 R1  P2 R2 and P1  P2  1 , then
E R   P1 E R1   P2 E R2  and VarR  P12VarR1   P22VarR2   2P1 P2CovR1 , R2  is the variance of
the return. Thus if P1 and P2 are both .50 , we can say VarR =.25VarR1 +.25VarR2 +.50CovR1 R2  .
…… Since variance is a measure of risk, minimizing variance minimizes risk, though actually, the best
measure of risk is probably the coefficient of variation, the standard deviation divided by the mean, in this
R
.” The table below was created in the opposite order to that requested.  R  Var R 
E R 
P1
E R 
R
CR
case C 
Row
1
2
3
4
5
6
7
8
9
10
11
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
14.80
15.02
15.24
15.46
15.68
15.90
16.12
16.34
16.56
16.78
17.00
37.3839
33.8267
30.5437
27.6329
25.2235
23.4703
22.5272
22.4963
23.3811
25.0849
27.4558
2.52594
2.25211
2.00418
1.78738
1.60864
1.47612
1.39747
1.37676
1.41190
1.49493
1.61505
e)
Make a graph . Put the mean return on the y axis and the standard deviation on the x axis. (3)
Connect your points in a curve. The C-like curve you will probably get illustrates the tradeoffs
between return and risk and is called an investment opportunities frontier.
Notice that as the fraction of the portfolio in security 1rises, the coefficient of variation also falls until
it hits a minimum at P1  .8 It then starts rising. If you graph the standard deviation against the
expected return, from P1  0 to P1  .8 , you will get a curve with a negative slope, where the variance
falls and the return rises. After P1  .8 , the return continues to rise, but the variance rises too.
f) Looking at the graph and the coefficients, are there stock portfolios that you would never
recommend? Why? What portfolio would you recommend to an 80-year old widow? Why? What
portfolio would you pick for yourself? Why? (1.5)
There is no point in taking any portfolios between P1  0 and P1  .8 . A rational person expects an
increasing return for increasing risk, and it just isn’t there. A person who is very cautious as the widow
probably should be, will take the minimum risk package at P1  .8 . Your personal decision should lie
between P1  .8 and P1  1.0.
9
251y0632 4/21/06
Your Solution: Go to 251y063t. Find ‘Computations for the Portfolio Problem.’ The number that you
subtracted from -52.2 is your version number. The first two tables will look like the following.
Data Display
Row
x
1
22.5
2
9.3
3 -18.6
4
48.1
5
44.4
6
56.0
7 -20.1
8 -11.0
9
17.1
10
22.3
xsq
506.25
86.49
345.96
2313.61
1971.36
3136.00
404.01
121.00
292.41
497.29
y
-23.7
8.6
4.3
-52.2
62.4
61.4
9.8
52.6
17.1
11.7
ysq
561.69
73.96
18.49
2724.84
3893.76
3769.96
96.04
2766.76
292.41
136.89
xy
-533.25
79.98
-79.98
-2510.82
2770.56
3438.40
-196.98
-578.60
292.41
260.91
Data Display
sumx
170.000
sumx2
9674.38
sumy
152.000
sumy2
14334.8
sumxy
2942.63
n
10.0000
xbar
17.0000
ybar
15.2000
svarx
753.820
svary
1336.04
scovxy
39.8478
sx
27.4558
sy
36.5519
rxy
0.0397063
rxy2
0.00157659
The first column is x  R1 , the second is x 2  R12 , the third is y  R 2 , the fourth is y 2  R22 and the fifth
is xy  R1 R 2 . Accordingly sumx =
=

xy . xbar = x 
scovxy = s xy 
x
 x , ybar =
n
 xy  nxy , sx =
n 1
y
, sumx2 =
x
2
, sumy =
 y , svarx = s   x
n
s x2 , sy =
2
x
s 2y , rxy = rxy 
2
y
 nx 2
n 1
s xy
sxsy

 y and sumxy
 y  ny ,
, svary = s 
2
, sumy2 =
2
2
y
2
s xy
s x2 s 2y
2
n 1
and rxy2 = r xy2 . The
coefficients of variation are below. The next display repeats some of this material. The final display looks
like the following.
Data Display
Row Pin1
1
0.0
2
0.1
3
0.2
4
0.3
5
0.4
6
0.5
7
0.6
8
0.7
9
0.8
10
0.9
11
1.0
Pmean
15.20
15.38
15.56
15.74
15.92
16.10
16.28
16.46
16.64
16.82
17.00
Pstdv
36.5519
33.1196
29.9662
27.1890
24.9141
23.2893
22.4559
22.5023
23.4230
25.1222
27.4558
CofV
2.40473
2.15342
1.92585
1.72738
1.56496
1.44654
1.37936
1.36709
1.40763
1.49359
1.61505
The first column gives Pin1 = P1 . Since P2  1  P1 , it is not stated. Pmean = E R 
 P1 E R1   P2 ER2  . Pstdev =  s R  Var (R) , where VarR 
 P12VarR1   P22VarR2   2P1 P2 CovR1 , R2  . CofV = C 
R
E R 
is the coefficient of variation. Note
10
251y0632 4/21/06
that the first row gives the mean, standard deviation and coefficient of variation for R1 , and that the
last row gives the same statistics for R 2 .
11
251y0632 4/21/06
2. It’s time for a jorcillator problem! As everyone knows a jorcillator has two components, a Phillinx
and a Flubberall.
The life of a Phillinx, x , is described by a continuous uniform distribution between c  0  h5 and
d  30  h6 , where h5 and h6 are the last two digits of your student number .
The life of a Flubberall, y , is described by a Normal distribution with a mean of   20  h5 and a
standard deviation of   5.10 .
(Example: Seymour Butz’s student number is 123450. c  0  5  5 , d  30  0  30 and   20  5  25 ).
a)
Find the following probabilities for the Phillinx: (i) PPH 1   P0  x  10  , the probability that it
dies before the end of the tenth year, (ii) PPH 2   P10  x  20  , the probability that it dies
between the end of the tenth and the end of the twentieth year and (iii) PPH 3   P( x  20) , the
probability that it lasts beyond the twentieth year. (1.5)
b) Find the following probabilities for the Flubberall: (i) PF1   P0  y  10  , the probability that it
dies before the end of the tenth year, (ii) PF2   P10  y  20  , the probability that it dies
between the end of the tenth and the end of the twentieth year and (iii) PF3   P( y  20 ) , the
probability that it lasts beyond the twentieth year. (3)
c) Make a joint probability table using the 6 events described above. Assume independence. (1)
d) Assume that the Jorcillator will fail only if both components fail. Find the following probabilities
for the Jorcillator: (i) PJ 1  , the probability that it dies before the end of the tenth year, (ii)
PJ 2  , the probability that it dies between the end of the tenth and the end of the twentieth year
and (iii) PJ 3  , the probability that it lasts beyond the twentieth year. (3)
[23.5]
e)




(Extra credit) Find the following conditional probabilities: : (i) P J 3 F1 , (ii) P J 3 F2 (iii)
PJ 1 F3  , (iv) PF1 J 2  (2).
Solution:
a) Find the following probabilities for the Phillinx: (i) PPH 1   P0  x  10  , the probability that it
dies before the end of the tenth year, (ii) PPH 2   P10  x  20  , the probability that it dies
between the end of the tenth and the end of the twentieth year and (iii) PPH 3   P( x  20) , the
probability that it lasts beyond the twentieth year. (1.5)
All 100 solutions to this problem are available in Probabilities for the uniform distribution. Your
version number is h5 h6 . If c  0  h5 and d  30  h6 , d  c  30  h5  h6 . Make a diagram! The
1
1

. The bottom of the box is at h5 and the top is at 30  h6 .
d  c 30  h5  h6
Now divide the box by vertical lines at 10 and 20.
(i) The first division of the box will be from h5 to 10, and its area will be
height of the box is
10  h5
30  h5  h6
(ii) The second division of the box will be from 10 to 20, and its area will be
20  10
PPH 2   P10  x  20  
30  h5  h6
PPH 1   P0  x  10  
(iii) The third division of the box will be from 20 to 30  h6 , and its area will be
PPH 3   P20  x  30  
30  h6  20
30  h5  h6
Lets assume that Fillmore Barrols (Sorry, but I get my names from the Billionaires for Bush website –
many of you ought to learn their credit card song.) has student number 123456, so that h5  5 and
12
251y0632 4/21/06
h6  6. Then PPH 1   P0  x  10  
 P10  x  20  
10  h5
10  5
5


 .263158 , PPH 2 
30  h5  h6 30  5  6 19
30  h6  20
20  10
10

 .526316 and PPH 3   P20  x  30  
30  h5  h6 19
30  h5  h6
30  6  20 4

 .210526 . Since these probabilities describe the entire area of a rectangle with
19
19
1
height
and length 30  h5 h 6 , these add to one.
30  h5  h6

b) Find the following probabilities for the Flubberall: (i) PF1   P0  y  10  , the probability that it
dies before the end of the tenth year, (ii) PF2   P10  y  20  , the probability that it dies
between the end of the tenth and the end of the twentieth year and (iii) PF3   P( y  20 ) , the
probability that it lasts beyond the twentieth year. (3)
The life of a Flubberall, y , is described by a Normal distribution with a mean of   20  h5 and a
standard deviation of   5.10 .
The 10 possible sets of probabilities appear in Probabilities for the Normal distribution cataloged by
the mean. So Fillmore has a mean of 25 and the probabilities are
10  25 
 0  25
PF1   P0  y  10   P 
z
 P 4.90  z  2.94 
5.10 
 5.10
 P4.90  z  0  P2.94  z  0  .5000  .4984  .0016 , PF2   P10  y  20 
20  25 
10  25
 P
z
 P 2.94  z  0.98   P2.94  z  0  P0.98  z  0
5.10 
 5.10
 P2.94  z  0  P0.98  z  0  .4984  .3365  .1619 and PF3   P y  20 
20  25 

 Pz 
 Pz  0.98   P0.98  z  0  Pz  0  .3365  .5  .8365 . Much to my
5.10 

surprise, these are almost equal to the probabilities calculated by Minitab. My surprise is due to the
fact that the probabilities calculated by Minitab are much more accurate than those calculated from the
table, especially for numbers near zero. These probabilities add to one.
c) Make a joint probability table using the 6 events described above. Assume independence.
Let’s tick with Fillmore’s numbers.
From above - PF1  = .0016, PF2  = .1619 and PF3   .8365 ; also PPH 1   .2632 , PPH 2 
 .5263 and PPH 3   .2105 . Let’s make a joint probability table. (Columns don’t always add up
PH 1 PH 2
PH 3
F1
.0004
.0008
.0003
.0016
because of rounding) F2
.0426
.0852
.0341
.1619
F3
.2202
.4402
.1761
.8365
.2632
.5263 .2105 1.0000
d) Assume that the Jorcillator will fail only if both components fail. Find the following probabilities
for the Jorcillator: (i) PJ 1  , the probability that it dies before the end of the tenth year, (ii)
PJ 2  , the probability that it dies between the end of the tenth and the end of the twentieth year
and (iii) PJ 3  , the probability that it lasts beyond the twentieth year. (3)
Now let’s make a table of the joint events and when they down the Jorcillator. Since the Jorcillator will fail
only if both components fail, it fails only when the longer-lived component dies, so if the Flubberall fails in
period 1 and the Phillinx fails in period 3, the Jorcillator fails in period 3.
13
251y0632 4/21/06
Joint Event
Probability
F1 PH 1
.0004
Downs
Jorcillator in
period
1
F1 PH 2
.0008
2
F1 PH 3
.0003
3
F2 PH 1
.0426
2
F2 PH 2
.0852
2
F2 PH 3
.0341
3
F3 PH 1
.2202
3
F3 PH 2
.4402
3
F3 PH 3
.1761
3
(i) PF1  PH 1   PJ 1   .0004 , the probability that it dies before the end of the tenth year,
(ii) PJ 2   .0008  .0426  .0852  .1286 , the probability that it dies between the end of the tenth and
the end of the twentieth year
(iii)
PF3  PH 3   PF3   PPH 3   PF3  PH 3   .8365  .2105  .1761
 PJ 3   .0003  .0341  .2202  .4402  .1761  .8709 , the probability that it lasts beyond the
twentieth year. Note that, except for a .0001 rounding error, these add up to 1.
e)
(Extra credit) Find the following conditional probabilities:
PH 1 PH 2
PH 3
F1
.0004
.0008
.0003
.0016
F2
.0426
.0852
.0341
.1619
F3
.2202
.4402
.1761
.8365
.2632
.5263
.2105 1.0000


(i) P J 3 F1 : The only J 3 event that involves F1 is F1 PH 3 ; so the probability is
PJ 3  F 1 
.0003
 .1875 , which is probably way off because of rounding error.
 P J 3 F1  
.
0016


P F1
(ii)
PJ 3  F 2 
 P J 3 F2 : The J 3 event that involves F2 is F2 PH 3 ; so the probability is
PF2 


P J 3 F2  
.0341
 .2106
.1619
(iii) P J 1 F3 : If the Flubberall fails in period 3, the Jorcillator will not fail until period 3; so the




probability is P J 1 F3  0 .


(iv) P F1 J 2 : The J 2 events are F1 PH 2 , F2 PH 1 and F2 PH 2 . The only one that involves F1 is
F1 PH 2 . So
PJ 2  F 1 
.0008
 .0062 .
 PF1 J 2  
.1286
PJ 2 
14
251y0632 4/21/06
Your Solution: Go to 251y063t. Find ‘Probabilities for the Uniform Distribution.’ Your version
number is h5 h6 . So if h5  0 and h6  6 , you will go to Version 06 and find the following.
Data Display
version
P00-10
P10-20
P20-30
Psum
06.00000
0.416667
0.416667
0.166667
1.00000
The first line is h5 h6 , the version number; the second is P00-10 = PPH 1   P0  x  10 

10  h5
20  10
; the third is P10-20 = PPH 2   P10  x  20  
, the last is P20-30 =
30  h5  h6
30  h5  h6
PPH 3   P20  x  30  
30  h6  20
. Psum is the sum of the probabilities and should be very
30  h5  h6
close to 1.
Now find ‘Probabilities for the Normal Distribution.’ Find your mean   20  h5 . So if your version
number is h5 h6 . So if h5  0 , so that your mean is 20, you will find the following.
Data Display
Row mean var lower upper
prob
1
20 5.1
0
10 0.024908
2
20 5.1
10
20 0.475048
3
20 5.1
20
50 0.500000
Sum of prob = 0.999956
10  20  h5 
 0  20  h5
z
Row 1 gives PF1   P0  y  10   P 
 ; row 2 gives
5.10
 5.10

PF2   P10  y  20  and row 3 gives PF3   P y  20  . Note that these probabilities are exact; the
values of z used by the computer have not been rounded to hundredths, so that they may be close to
your answers, but not identical ‘Sum of prob’ is the sum of the probabilities and should be very close
to 1.
Once you have these probabilities, you should be able to follow the remainder of the solution on the
previous page.
15