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ECO251 QBA1
Name
key
FINAL EXAM
Class ________________
DECEMBER 12, 2000
Hint: Though you should never give 2 answers to a problem, never cross out the answer to a problem that
you have given up on. I may see something worth partial credit in the answer.
Part I. Do all the Following (20 Points) Make Diagrams!
A. z ~ N (0,1)
1. Pz  2.57   Pz  0  P0  z  2.57   .5  .4949  .0051 Diagram! These are expected and will be
provided in class for version 2.
Half of you didn't read this question: You assumed that it said Pz   , because that was on last year's
exam.
2. P3.12  z  0.24   P3.12  z  0  P0.24  z  0  .4991  .0948  .4043
3. P0.97  z  0.97   2P0  z  0.97   2.3340   .6680
4. P0.97  z  2.57   P0  z  2.57   P0  z  0.97   .4949  .3340  .1609
5. The 87th percentile of the distribution. z .13 is the point with a probability of .13 above it or .87 below it.
Since 87% is below this point and 50% is below zero, from the diagram P0  z  z.13   .3700 . the closest
we can come to this probability using the Normal table is P0  z  1.13  .3708 . So z .13  1.13 .
Note that probabilities cannot be above 1 or below 0.
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B. x ~ N 1.1, 5 .
Remember z 
x
. As usual, people made diagrams of x with

zero in the middle. Make up your mind! If you are diagramming x , put the mean in the middle; if
you are diagramming z put zero in the middle.
2.57  1.1 

1. Px  2.57   P z 
  Pz  0.29   Pz  0   P0  z  0.29   .5  .1141  .3859
5


 0.24  1.1 
  3.12  1.1
z
2. P3.12  x  0.24   P
  P 0.84  z  0.27 
5
5


 P0.84  z  0  P0.27  z  0  .2995  .1064  .1931
0.97  1.1 
  0.97  1.1
z
3. P0.97  x  0.97   P
  P 0.41  z  0.03 
5
5


 P0.41  z  0  P0.03  z  0  .1591  .0120  .1471
2.57  1.1 
 0.97  1.1
z
4. P0.97  x  2.57   P
  P 0.03  z  0.29 
5
5


 P0.03  z  0  P0  z  0.29   .0120  .1141  .1261
5. The 87th percentile of the distribution. x.13 is the point with a probability of .13 above it or .87 below it.
Because it is above the 50th percentile, which is the mean for the Normal distribution, the point we want will
be above 1. On the previous page we found z .13  1.13 . So x.97    z.97
 1.1  1.135  6.75 .
6.75  1 

To check to see if this is correct: Px  6.75   P z 

5 

 Pz  1.13  Pz  0  P0  z  1.13  .5  .3708  .8708
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II. (4 points-2 point penalty for not trying .) Show your work!
We are investigating the cost of a business trip for people in the financial industry. The
data for a sample of 6 are below.
x
1430
1225
1430
1570
1330
941
Compute the sample standard deviation, s . Show your work. (4)
Solution:
x
1430
1225
1430
1570
1330
941
7926
1
2
3
4
5
6
x
 x  7926  1321
n
6
s x2 
x
2
 nx 2
n 1

x2
2044900
1500625
2044900
2464900
1768900
885481
10709706
10709706  61321 2 239460

 47892
5
5
s x  47892  218 .8424 .
How about those jokers who are still trying to compute
 x  x 
2
by computing
 x  x  !
2
It
didn't work last term and it won't work this term. I don't encourage using definitional formulas, but, if you
insist on using them, I will be happy to give you a personal tutorial on how to do it. They are generally
harder to use than computational formulas.
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III. Do at least 4 of the following 6 Problems (at least 12 each) (or do sections adding to at least 48 points Anything extra you do helps, and grades wrap around) . Show your work! Please indicate clearly what sections of the
problem you are answering! If you are following a rule like E ax  aEx  please state it! If you are using a formula, state it! If
you answer a 'yes' or 'no' question, explain why! If you are using the
Poisson or Binomial table, state things like n , p or the mean. Avoid crossing out answers that you think are inappropriate - you
might get partial credit.
1.
a. Find z .065 (2)
b. If x ~ N 15 ., 5 , find a symmetrical interval around the mean with a probability of 87%. (2)
c. If x ~ N 15 ., 5 , find F 22 .6 (2)
d. If computer parts last an average of fifteen months, with a standard deviation of 5, what is the probability
that 15 of them will all last less than 22.6 months. (2)
e. If a sample mean is taken from a sample of 15 from a normal population with a mean of 15 and a
standard deviation of 5, what is the standard deviation of the sample mean? (2)
f. What is the probability that the sample mean in 5) above is less than 22.6? (2)
g. If sample mean of 14.3 comes from a sample of 15 from a normal population with a population standard
deviation of 5, do an 87% confidence interval for the population mean. (3)
Solution: From the problem statement x ~ N 15 ., 5 , i.e.   15 and   5 .
a) z .065 is the point with a probability of .065 above it or .935 below it. It is the 93.5 percentile of z. From
the diagram P0  z  z.065 0  .4350 . the closest we can come to this probability using the Normal table is
P0  z  1.51  .4350 . So z.065  1.51 . Make a diagram!
b) A symmetrical interval around the mean with a probability of 87%: We can split the 87% into two halves
of 43.5%, on either side of zero. Make a diagram! From the diagram, we want two points z.065 and
z.935   z.065 so that Pz.935  z  z.065   .8700 . And, since we found z.065  1.51 in part a), our interval
for z is -1.51 to 1.51.
The interval for x can then be written x    z.065  15  1.515  15  7.55 or 7.45 to 22.55.
22 .55  15 
 7.45  15
z
To check this P7.45  x  22 .55   P 

5
5


 P 1.51  z  1.51  2P0  z  1.51  2(.4345 )  .8690  87%
22 .6  15 

c) F 22 .6  Px  22 .6  P z 
  Pz  1.52   Pz  0  P0  z  1.52   .5  .4357  .9357
5


15 15 0
d) Binomial n  15 , p  .9357 -- P15   C15
p q  .9357 15  .3690
e)   15 , n  15 ,   5 . So  x 
x
n

5
 1.2910
15
22 .6  15 

f) Px  22 .6  P z 
  Pz  5.89   Pz  0  P0  z  5.89   .5  .5000  1
1.2910 

g) Given: x  14.3 , n  15 ,   5 . Step 1: Confidence level is 87%.; significance level is

5
 1.2910 .
  1  .87  .13 . Step2: z 2  z.065  1.51 .
Step 3:  x  x 
n
15
Step 4:   x  z  x  14.3  1.511.2910   14.3  1.95 or 12.35 to 16.25. More formally
2
 P12.35    16.25   87%
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2. According to Ken Black, a survey was taken asking how long a group of firms has done business with
India. A random sample of 44 responses gave a mean of 10.455 years.
a. Assume that the population standard deviation is known to be 7.7, construct a 90% confidence interval
for the population mean.(4)
b. Assume that the population standard deviation is unknown and that 7.7 is the sample standard deviation
instead. Construct the 90% confidence interval again. (4)
c. Assume that once again the population standard deviation is unknown and that the sample mean of
10.455 and the sample standard deviation of 7.7 are taken from a sample of 15 that is part of a population of
100. Construct the 90% confidence interval again. (4)
d. Assume that we use a smaller sample and a higher confidence level. What would happen to the
confidence interval and why? (2)
Solution: x  10.455 . For all these sections the confidence level is 90% and the significance level is
  1  .90  .10 . n  44 .
Repeat after me! " z goes with  (sigma - population variance); t goes with s (sample variance)!"

7.7
 1.161 .
a) From the t-table, z  z.05  1.645 .  x  x 
2
n
44
  x  z  x  1`0.455  1.645 1.161  10.455  1.910 or 8.545 to 12.365. More formally
2
 P8.545    12.365 .98   90%
s
7.7
43
 1.161 .
b) The degrees of freedom are n  1  44  1  43 . t n1  t .05
 1.681 s x  x 
2
n
44
Putting this all together   x  tn1 s x  10.455  1.6811.161   10.455  1.952 or 8.503 to 12.407. More
2
formally, P8.503    12.407   .90 .
c) The population size, N  100 , is less than 20 times the sample size n  15 (Sorry! This was supposed to
N n
7.7 100  15

 1.9881 0.92660   1.8422 ,
n N 1
15 100  1
14
tn1  t.05
 1.761 and   x  tn1 s x  10.455  1.7611.8422   10.455  3.244 or 7.211 to 13.699.
have stayed at 15.), so s x 
2
sx
2
More formally, P7.211    13.699   .90 .
d) The effect of a smaller sample is to decrease the denominator in the standard error of the
s
mean, s x 
. This makes the standard error larger and since the size of the confidence interval is
n
proportional to the standard error, the interval is also larger. The effect of a higher confidence level is to
lower the significance level and thus to require a larger value of t or z . This also makes the confidence
interval larger.
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3. Describe the meaning and give the probabilities for the problems below.
Example: For the Hypergeometric Distribution, P1  x  2 when N  18, n  5, and M  9 is the
probability of between 1 and 2 successes when a sample of 5 is taken from a population of 18 where there
9! 9!
9! 9!
C19 C 49 C 29 C 39 1!8! 4!5! 2!7! 3!6!
are 9 successes in the population. P1  x  2 
 .48529 . Use tables
 18 

18!
18!
C 518
C5
5!13!
5!13!
where appropriate!
a. Binomial P3  x  8 , when p  .30 and n  13 (2)
b Binomial P5  x  11 , when p  .80 and n  13 (2)
c Poisson P5  x  11 , when m  17 .5 (2)
d Geometric P5  x  11 , when p  .30 (2)
e. No meaning is needed for the following:
(i) Binomial Px  7 , when p  .30 and n  13 (1)
(ii) Geometric Px  7 , when p  .30 (1)
(iii) Binomial Px  7 , when p  .30 and n  13 (1)
(iv) Geometric Px  7 , when p  .30 (1)
(v) Poisson Px  7 , when m  17 .5 (1)
(vi) Continuous Uniform P5  x  11 when c  0 and d  10 (2)
(vii) Normal x ~ N 15 ., 5 . P
 x  400  when n  25 (2)
Solution:
a. Binomial Distribution with p  .30 and n  13
The probability of between 3 and 8 successes in 13 tries when the probability of success on a
single try is .30 is P3  x  8  Px  8  Px  2  .99597  .20248  .79349
b. Binomial Distribution with p  .80 and n  13 .
The probability of between 4 and 10 successes in 13 tries when the probability of success on a
single try is .80 is P5  x  11 . It 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 - .80 = .20. 5 successes correspond to 8
failures out of 13 tries. 11 successes correspond to 2 failures. So try 2 to 8 successes when
p  .20 and n  13 . P2  x  8  Px  8  Px  1  .99983  .05498  .944485
c. Poisson Distribution with parameter of 17.5 (parameter of 17.5 means 17.5  m     2 )
The probability of between 5 and 11 successes in a unit of space or time when the average number
of successes is 17.5 is P5  x  11  Px  11  Px  4  .06840  .00012  .06828
d. Geometric Distribution with p  .30
The probability that the first success will occur between try 4 and try 10 when the probability of
success on any one try is .30 is P5  x  11 . Remember that F c  Px  c  1  q c , because
success at try c or earlier implies that there cannot have been failures on the first c tries.
q  1  p  1  .30  .70 . P5  x  11  Px  11  Px  4  F 11  F 4  1  .7011  1  .704
 .70  .70  .24010  .01977  .22033 Note: F 11  .98023 , F 4  .75990 .
4


11
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12/11/00 251y0041
e. No meaning is needed for the following:
(i) Binomial Distribution with p  .30 , n  13 . Px  7  1  Px  6  1  .93762  .06238


(ii) Geometric Distribution with p  .30 Px  7  1  Px  6  1  1  .706  .706  .117649
(iii) Binomial Distribution with p  .30 , n  13 .
Px  7  Px  7  Px  6  .98178  .93762  .04416 or
P7  C713 p 7 q 6  C713.30 7 .70 6  .04416
(iv) Geometric Distribution with p  .30

Px  7  q x 1 p  .70 6 .30   .035295 or
 

F 7  F 6  Px  7  Px  6  1  .707  1  .706  .917645  .882351 .03529
(v) Poisson Px  7 , when m  17 .5 Px  7  .0025043 (directly from table).
(vi) Continuous Uniform P5  x  11 when c  0 and d  10 Make a diagram!
f x  
1
1
1


 0.1 P5  x  11  P5  x  10   50.1  .5
d  c 10  0 10
 x  400  when n  25 ,   15 ,   5 . So
 x  400  16 , P x  400   Px  16
and since x 
n
25
(vii) Normal x ~ N 15 ., 5 . P
x 
x

5
 1.2910
n
15
16  15 

 P z 
  Pz  0.77   Pz  0  P0  z  0.77   .5  .2794  .7794
1.2910 

f. Extra Credit! A company designs washing machines to go an average of six years without a major
breakdown. (exponential distribution).
(i) What is the probability that the machine lasts more than 2 years? (2)
(ii) What is the probability that the machine lasts longer than 1 year? (1)
(iii) What is the standard deviation of the time that it lasts? (1)
(iv) ( Difficult!!!) If the company offers a warranty against major breakdowns and wants to be
sure that the no more than 20% of the machines break down in the warranty period, what is the
longest time for which the company could offer the warranty? (4)
I'll do this if you do! For one answer, see 251y0042.
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12/11/00 251y0041
4. Assume that P A  .2 and PB   .4


b. If A and B are mutually exclusive what is (i) P A  B  , (ii) PA B , (iii) PA  B  ? (3)
c. If PB A  .5 what is (i) P A  B  , (ii) PA B , (iii) PA  B  ? (3)
 
a. If A and B are independent what is (i) P A  B  , (ii) P A B , (iii) P A  B ? (3)
 
d. Make a joint probability table of A, B, A and B , assuming that P B A  .4 (3)
e. One of five employees is believed to have revealed a company secret. All five denied giving
away the secret and are considered equally likely to have done the deed. All five were given a liedetector test. George came up positive - the lie detector is implying that he is a liar. If the lie
detector comes up with a false positive 2% of the time and a false negative 1% of the time, what is
the chance that George revealed the secret? (4)
Solution: Note: A is the complement of A ; P A  1  P A .
 
 
a) Independence means P A B  P A , so that P A  B   P APB  . This implies
(i) P A  B   P APB  . What will it take to convince you that if A and B are independent,
P A  B   0 ? Since P A  .2 and PB   .4 , P A  B   P APB   .2.4  .08 . So
 
(ii) P A B  P A  .2 and
P A  B  P A  PB  P A  B  .2  .4  .08  .52

(iii) P A  B
table.
A
B
B

 1  P A  B   1  .52  .48 . One of the best ways to see this is with a joint probability
A
  PB  becomes B
B
P A  B  P A  B  P B 
P  A
P A 
1
P A  B  P A  B
it add up we get

B
B
A A
.08 .32
.12 .48
.2 .8
.4
.6
A A
.08
.4
.6
.2 .8
, because of independence. If we just make
1 .0

 

. P A  B  P A  B  P A  B  P A  B  .08  .32  .12  .52 .
1.0

P A  B  .48 can be read right off the table. And, if you really need it, PA B  
b) Mutual exclusiveness means P A  B   0 .
(i) So P A  B   P A  PB   P A  B   .2  .4  0  .6

(iii) P A  B

(ii) PA B  
P A  B 
.
P B 
P A  B 
0
PB 
 1  P A  B   1  .6  .4 . The joint probability table starts out as
B
A A
.0
.6
B
.2 .8
because of the mutual exclusiveness and becomes

 

B
B
A A
0 .4
. 2 .4
. 2 .8
.4
1.0
.4
.6
1 .0


P A  B  P A  B  P A  B  P A  B  0  .4  .2  .52 . P A  B  .4 can be read right off the
table. 9 points of this question were repeated from the hour exam with the numbers only slightly
8
changed. Nevertheless most of you made exactly the same mistakes again. You are supposed to learn
from your mistakes!
9
12/11/00 251y0041
 
 
c) P B A  .5 means that P A  B  PB  A  P B A P A  .5.2  .1 . So
(i) P A  B   P A  PB   P A  B   .2  .4  .1  .5

(iii) P A  B

 1  P A  B   1  .5  .5 . The table reads
 
 
(ii) P A B 
B
B
 
P A  B  .1
  .25
PB 
.4
A A
.1 . 3
.1 .5
.2 .8
.4
.6
1 .0
d) P B A  .4 means that P A  B  PB  A  P B A P A  .4.2  .08 . The table was done in a)
B
A A
.08 .32
.4
.6
.12 .48
.2 .8
1.0
The following laws have not been repealed:
The Addition Rule: P A  B   P A  PB   P A  B 
B
 
 
The Multiplication Rule: P A  B  P A B PB  PB  A  P B A PB
e) One of five employees is believed to have revealed a company secret. All five denied giving away the
secret and are considered equally likely to have done the deed. If R is the event that George revealed the
secret, PR   .2 . All five were given a lie-detector test. George came up positive - the lie detector is
implying that he is a liar. If the lie detector comes up with a false positive 2% of the time ( if P is the
event that the lie detector comes up with a positive, P P R  .02 ) and a false negative 1% of the
 
 
time P P R  .01 , what is the chance that George revealed the secret? (4) I somehow, don’t think that
anyone got to this without Bayes' rule. PR P  
PP R PR 
P P 
 
 
 
.99.2
.198
PR P  

 .9252
.99.2  .02.8 .198  .016

PP R PR 
    . From the above
PP R PR   P P R P R
P R  1  PR   1  .2  .8 and P P R  1  P P R  1  .01  .99 so
If you used a table for this,
R
R
P
.198
.016
.214
P
.002
.784
.786
.2
.8
1.0
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12/11/00 251y0041
5. a. A real estate office sells 450 houses in 300 working-day year. (Poisson distribution - if you
approximate it by another distribution tell why you can do it.)
(i) What is the mean and variance of the number of houses sold in a day? (2)
(ii) What is the probability that at least one house is sold in a day? (2)
(iii) What is the probability that more than 5 houses are sold in a day? (2)
(iv) What is the probability that more than 45 houses are sold in a 25 working-day month? (3)
5. b. The probability that a buyer is unable to get a mortgage is .03. In a month in which offers are made on
30 houses, what is the probability that at least one mortgage will be denied?
(i) Do this problem using the Binomial distribution (2)
(ii) Can the solution to this problem be approximated by the Poisson or Normal distribution? Test
for both and use both, one or neither distribution to solve the problem depending on the results of
your tests. (4)
Solution: Because this involves a situation in which you are given the average number of events per unit
space or time this would normally be considered to be a Poisson problem.
a) (i) Since the period involved is one working day, the mean is m    450300  1.5 . This is also the
variance since the mean and variance are identical for the Poisson distribution.
(ii) From the Poisson table for a mean of 1.5 , Px  0  1  Px  0  1  .22313  .77687
(iii) From the same table Px  5  1  Px  5  1  .99554  .000446
(iv) The mean is now m    1.525   37 .5 . Because the mean is over 20, use the Normal
distribution.

45 .5  37 .5 
PP x  45   PN x  45 .5  P  z 
  Pz  1.31  .5  .4049  .0951 (DIAGRAM?)
37 .5 

b) (i) Binomial Distribution with p  .03 , n  30 .
Px  1  1  P0  1  C 030 p 0 q 30  1  .03 0 .97 30  1  .40101  .59899
(ii) Test for Normal distribution: np  30.03   0.9  5 , nq  n  np  30  0.9  29.1  5 . We can't
use the Normal distribution because these are not both above 5.
n 30

 1000  500 . We can use the Poisson distribution because
Test for the Poisson distribution:
p .03
the ratio is above 500.
If we use the Poisson table with a parameter of 0.9 Px  1  1  P0  1  .40657  .59343
11
12/11/00 251y0041
6. a. The columns below state the Dow Jones Industrial Average(in hundreds) x and the 6-month treasury
bill rate y .
x
y
According to Minitab, the mean and standard
22
8.12
deviation for the Dow are respectively 39.75 and
27
7.51
16.33. The mean and standard deviation for the t29
5.42
bill rate are 5.390 and 1.756.
32
3.45
This was a gift! Why did so many of you
37
3.02
x2 ,
x,
y and
y2 ?
compute
45
5.51
56
5.02
70
5.07
 


(i) Compute the sample covariance between the Dow and the t-bill rate. (4)
(ii) Compute and interpret the correlation between the two quantities. (3)
(iii) If the t-bill rate was 1 point higher for all the values shown here (i.e. 8.12 was 9.12, 7.51 was 8.51
etc.), what would the values of the standard deviation for the t-bill rate, the covariance and the
correlation be? The only accepted answer here will be values found by using the formulas for the
variance, covariance and correlations of functions of x , and the values of the standard deviation,
covariance and correlation found in (i) and (ii). (3)
Solution:
x
22
27
29
32
37
45
56
70
(ii) rxy 
y
8.12
7.51
5.42
3.45
3.02
5.51
5.02
5.07
s xy
sx s y

xy
178.64
202.77
157.18
110.40
111.74
247.95
281.12
354.90
1644.70
n8
x  39 .75, s x  16 .33
y  5.39 , s y  1.756
 xy  1644 .70
 xy  nx y  1644 .70  839.75 5.39 
s 
(i)
xy
n 1
69 .32

 9.90281
7
7
 9.90281
 .3453 . Since rxy2  .119 is not very large on a 0- to 1 scale, we cannot say
16 .331.756 
that this is a strong relationship. However, it seems quite likely that interest rates and the Dow move
inversely. (iii) According to the syllabus supplement, if w  ax  b , and v  cy  d ,
 w2  Varw  a 2Varx  a 2 x2 ,  v2  Var v   c 2Var  y   c 2 y2 ,
Covw, v   wv  acCovx, y   ac xy and  wv  signac xy In this case w  x and v  y  1 , so
a  1 , b  0 , c  1 and d  1 . Since we are working with sample statistics, we can substitute s v2 for  v2 ,
s xy for  xy , rxy for  xy etc. so the new variance is s v2  Varv   12 Var y   s 2y , the new covariance is
Covw, v  s wv  11Covx, y   s xy and the new correlation is rwv  sign11rxy  rxy
12
12/11/00 251y0041
b. Find the missing values in the following joint probability table and compute the population covariance
and correlation between x and y and the variance of x  y . (5)
x
3 7


4 .6


y
.4 
9 


Solution: Since the numbers in the table already add to 1, the missing numbers must both be zeros.
x
y
Px 
xPx 
x 2 Px 
3

4
.6

9
 0
.6
1.8
 xPx  1.0 , E x    x
   y
E y
E xy  
yP y  y 2 P y 
2.4 9.6
To summarize
3.6 32 .4
6.0 42 .0
 Px  1,
 P y   1,
5.4  19 .6  25 .0
 x  E x  
2
7
P x 

0
.6

.4
.4
.4
1.0
 2.8  1.0
2
2
2
Px   25 .0 ,  y  E  y  
P y   42 .0
 .634 
 xyPxy    039
 yP y   6.0 and
 0 7 4 7.2
 0

 18 .0

 .4 7 9  0  25 .2
 
 xy  Covxy  Exy   x  y  18.0  1.06.0  12.0 ,  x2  E x 2   x2  25 .0   1.02  24 .0
 
and  y2  E y 2   y2  42.0  6.02  6.00 . So that  xy 
 xy
 x y

 12 .0
 1.00 .
24 .0 6.0
(  x  4.89898 ,  y  2.44949 )
Var x  y    x2   y2  2 xy  Var x   Var  y   2Covx, y   24 .0  6.0  212 .0  6.0
13