# Document 15930049

```251y0831 4/18/08
ECO251 QBA1
THIRD EXAM
Apr 24, 2008
Name: ________KEY_________
Student Number: _____________________
Class time: _____________________
Part I. (16 points) Do all the following (2 points each unless noted otherwise). Make Diagrams! Show
your work! In particular you must briefly explain how you got the answer to the value of z at the bottom of
z has the standardized Normal distribution z ~ N 0, 1 for the first four problems.
1. Pz  1.23   P1.23  z  0  Pz  0  .3907  .5  .8907
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line!
Shade the entire area above -1.23. Because this is on both sides of zero, we must add the area between 0.75 and zero to the area above zero.
2. P3.25  z  0  .4994
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line!
Shade only the area between -3.25 and zero. Because this is on one side of zero and ends at zero, we can
simply look up 3.25 on the table. The area between -3.25 and zero is exactly the same as the area between
zero and 3.25 if the mean is zero. The bottom of our Normal table has the following.
For values above 3.09, see below
If z 0 is between
3.08
3.11
3.14
3.18
3.22
3.27
3.33
3.39
3.49
3.62
3.90
and
and
and
and
and
and
and
and
and
and
and
3.10
3.13
3.17
3.21
3.26
3.32
3.38
3.48
3.61
3.89
up
P0  z  z0  is
.4990
.4991
.4992
.4993
.4994
.4995
.4996
.4997
.4998
.4999
.5000
Since 3.21 is between 3.22 and 3.25 we use .4994.
3. P3.07  z  3.07   P 3.07  z  0  P0  z  3.07   .4989  .4989  .9978
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line!
Shade the area between -3.07 and 3.07. Because this is on both sides of zero, we must add the area between
-3.07 and zero to the area between zero and 3.07. These areas are identical.
4. z .135 Solution: z .135 is, by definition, the value of z with a probability of 13.5% above it. Make a
diagram. The diagram for z will show an area with a probability of 100  13 .5%  86.5% below z .135 .
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 86.5% - 50% = 36.5%. The upper tail of the distribution above z .135 has a probability
of 13.5%, so that the entire area above 0 adds to 36.5% + 13.5% = 50%. From the diagram, we want one
point z .135 so that Pz  z .135   .86 .5 or P0  z  z.135   .3650 . If we try to find this point on the
Normal table, the closest we can come is P0  z  1.10   .3646 , though P0  z  1.11  .3665 is more
or less acceptable. . So we will use z.135  1.10 , though 1.11 might be acceptable.
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x ~ N 4, 7 for problems 5 through 8. Note that all values of z are rounded to the nearest hundredth.
x

to ‘standardize’ our values of x.
 1.23  4 

5. Px  1.23   P  z 
  Pz  0.75   P0.75  z  0  Pz  0  .2734  .5  .7734
7


For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line!
Shade the entire area above -0.75. Because this is on both sides of zero, we must add the area between
-0.75 and zero to the area above zero. If you wish, make a completely separate diagram for x . Draw a
Normal curve with a mean at 4. Indicate the mean by a vertical line! Shade the area above -1.23. This
area is on both sides of the mean (4), so we add the area between -1.23 and the mean to the whole area (.5)
above the mean .
0  4
  3.25  4
z
 P 1.04  z  0.57 
6. P3.25  x  0  P 
7
7 

 P1.04  z  0  P0.57  z  0  .3508  .2157  .1351
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line!
Shade the area between -1.04 and -0.57. Because this is entirely on one side of zero, we must subtract the
smaller area between -0.57 and zero from the larger area between -1.04 and zero. If you wish, make a
completely separate diagram for x . Draw a Normal curve with a mean at 4. Indicate the mean by a
vertical line! Shade the area between -3.25 and 0. Both these numbers are to the left of the mean. This
area is on one side of the mean (4), so we subtract the smaller area between -3.25 and the mean from the
larger area between 0 and the mean.
3.07  4 
  3.07  4
z
 P 1.01  z  0.13 
7. P3.07  x  3.07   P 
7
7 

 P1.01  z  0  P0.13  z  0  .3438  .0517  .2921
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line!
Shade the area between -1.01 and -0.13. Because this is entirely on one side of zero, we must subtract the
smaller area between -0.13 and zero from the larger area between -1.01 and zero. If you wish, make a
completely separate diagram for x . Draw a Normal curve with a mean at 4. Indicate the mean by a
vertical line! Shade the area between -3.07 and 3.07. Both these numbers are to the left of the mean. This
area is on one side of the mean (4), so we subtract the smaller area between 3.07 and the mean from the
larger area between -3.07 and the mean.
8. x.135 Solution: x.135 is, by definition, the value of x with a probability of 13.5% above it. If you did not
do so on the last page, make a diagram. The diagram for z will show an area with a probability of
100  13.5%  86.5% below z.135 . 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 86.5% - 50% = 36.5%. The upper tail of the
distribution above z .135 has a probability of 13.5%, so that the entire area above 0 adds to 36.5% +
13.5% = 50%. From the diagram, we want one point z .135 so that Pz  z .135   .86 .5 or
P0  z  z.135   .3650 . If we try to find this point on the Normal table, the closest we can come is
P0  z  1.10   .3646 , though P0  z  1.11  .3665 is more or less acceptable. . So we will use
z.135  1.10 , though 1.11 might be acceptable.
You have gotten to this point with something like z.135  1.10 . We now must get back from z to x.
The opposite of z 
x

is x    z . Since x ~ N 4, 7 , we have x.135  4  1.107  11 .70 .
Check:
11 .70  4 

Px  11 .70   P  z 
  Pz  1.10   Pz  0  P0  z  1.10   .5  .3643  .1357  13 .5%
7


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Part II: (9+ 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. A small life insurance company receives an average of five death claims a day. Assume that the Poisson
distribution is correct. What is the probability that the company will receive more than 10 claims in a given
day (rounded to thousandths)? (2)
a) .986
b) .005
c)* .014
d) .032
e) None of the above (Fill in an answer!)
Explanation: From the Poisson (5) table, Px  10   .98630 . So Px  10   1  Px  10   1  .98630 =
.0137
2. A local family planning group serves 20000 teen-age girls. It costs \$50 to council each pregnant girl.
There is a 5% chance that each of the girls will become pregnant during the year. Each pregnancy can be
assumed an independent event. Based on what you know about the expected values of discrete
distributions, how much should the agency budget for counseling this year? (2)
a) \$1000
b) \$20000
c) *\$50000
d) \$100000
e) None of the above.
Explanation: The Binomial distribution applies since we have independent events with a constant
probability. n  20000 , p  .05 . If x is the number of girls pregnant, E x   np  .0520000   1000 . The
cost, however, is C  50 x , so E50 x   50 E x   501000   50000 .
3. In the agency in problem two, 20 girls are waiting to see a counselor this morning. Half of them are
pregnant. If Samantha is assigned to counsel 8 of them, what is the chance that all eight are pregnant? (I
want to see your formulas and calculations – this only took me a few minutes.) (3)
Explanation: This is a hypergeometric problem. N  20 , M  Np  10 and n  8.
10 9
10!
1
C 810 C 010 2!8!
21
P8 


 .00036 . If you need more of an explanation: there are
20
20
!
20

19

18

17
16 1514 13
C8
12!8!
87654321
10

9
10
!
C810 

 45 ways to pick 8 pregnant girls from the 10 that are pregnant and
2!8! 21
C 820 
10!
 125970 ways to pick 8 girls from the 20 waiting.
12!8!
4. How many girls would have to be waiting before we could use the Binomial distribution to solve
problem 3? (1)
[8]
Solution: There would have to be more than N  20 n  208  160
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5. OK. Assume that there are 4000 girls waiting to see a counselor and Samantha is assigned to counsel 8
of them, what is the chance that all eight are pregnant? The only answer that I will accept here is an answer
gotten from numbers in the tables. If you just write down a solution, you will get half credit. (2)
Solution: Use the binomial table with p  .5 and n  8 .
Px  8  Px  8  Px  7  1  .99609  .00391
6. A variable has the Binomial distribution . Find the following and explain how you did it.
a) Px  5 p  .02 n  200 (2)
n 200
Solution: Use the Poisson distribution because

 10000  500 . m  np  4 .
p .02
Px  5  1  Px  4  1  .62884  .37116
b) Px  60  p  .40 n  200 (2 or 2.5)
[14]
Solution: Use Normal distribution because   np  200 .40   80  5 and
nq  200 .60   200  80  120  5 .  2  npq  80.6  48 . Without continuity

60  80 
correction Px  60   P  z 
  Pz  2.89  .5  .4981  .9981 . With continuity correction
48 


59 .5  80 
Px  60   P  z 
  Pz  2.96   .5  .4985  .9945
48 

7. Find P11  x  17  for the following distribution: Continuous Uniform with c  12 , d  15 (1)
Solution: c  12 and d  15 . Make a diagram. The base of the box goes from c  12 to d  15 . Shade
the area inside the box between 11 and 17.
0
11 12
15 17
1
1
1
The height of the box is

 and the base of the shaded area is 15 – 12 = 3. So the area of
d  c 15  12 3
3
the box is P11  x  17   = 1.
3
xc
for c  x  d and
If you use the cumulative distribution method, remember F x   1 for x  d , F x  
d c
F x   0 for x  d . So P11  x  17   F 17   F 11  1  0  1 .
Never forget that all probabilities are between 0 and 1. If you get anything else, you have misused a
formula.
8. If the amount of time it takes students to find a parking place has a Normal distribution with a mean of
3.5 minutes and a standard deviation of 1 minute, 99% of students search time will fall below what
number? (For example it may be true that 99% take 3.6 or fewer minutes to find a parking place, but I
doubt it.)
(2)
Solution: According to the t-table z.01  2.327 . So x.01    z.01  3.5  2.327 1  5.827 .
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9. (Dummeldinger) You decide to invest in four independently moving risky stocks. You guess that each
stock has an independent 40% chance of becoming a total loss. What is the chance that at least one of your
stocks will tank? (2)
[19]
a) 0
a) .0256
b) .4000
c)* .8704
d) .9744
e) 1.0000
Solution: c. Binomial p  .40 n  4. 1  P0  1  .1296  .8704
10. (Extra Credit) The amount of rainfall in a 24 – hour period has an exponential distribution with a mean
of 0.2 inches. What is the chance that a randomly picked 24-hour period will have rainfall that exceeds 0.8
inches?
a)* .018316
b) .778801
c) .221199
d) .981684
e) None of the above. Produce an alternate answer.
Solution: 1  Px  0.8  1  (1  e 5.8 )  .018316 Blank page for calculations.
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ECO251 QBA1
THIRD EXAM
Apr 24, 2008
TAKE HOME SECTION
Name: _________________________
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. Write on one side of the page!
Part III. Do all the Following (25+ Points) Show your work! Neatness counts! Answers of ‘zero’ or
‘one’ especially are unacceptable without an explanation. Do not use one distribution to approximate
another without justifying the replacement!
1. Identify the distribution that you are using in each problem. If you have a number like n  20  g , make
it very clear what value of n you are using. Look at the solved problems for Section L, the solution to
Grass3 and ‘Great Distributions’ (especially the hints on the 3 rd page) before you start.
Let g be the last digit of your student number. If g  0 , change it to 2. Hint: It may help in these problems
to realize that if n  100 , 93 or more is 93 to 100 and 10 or less is 10 to zero. This is especially useful if we
count failures.
a. A basketball player makes 50  5g % of her free throws over the basketball season. In one game she
gets 20 free throws and misses 13  g  of them. The coach will investigate if the probability of doing as
badly as she did or worse is below 5%. Will the coach investigate? Do the math to justify your answer. (2)
Solution: This is a hypothesis test. We are testing the null hypothesis H 0 : p  .50  .05 g against the
alternative H 1 : p  .50  .05 g . An ‘s’ indicates that p the observed proportion (8/20 = .40 in the first
case) is significantly below p 0  .50  .05 g , so that the coach will investigate. An ‘ns’ indicates that p is
not significantly below p 0 . Note that, though in the majority of cases we cannot reject the null hypothesis,
if we had the same proportion of a larger sample we could reject it.
g  1. She misses 12. Her probability of hitting is 55%. Her probability of missing is 100 – 55 = 45%
Px  12 n  20; p  .45  1  Px  11  1  .86923 = .1308 ns
g  2. She misses 11. Her probability of hitting is 60%. Her probability of missing is 100 – 60 = 40%
Px  11 n  20; p  .40  1  Px  10   1  .87248 = .1275 ns
g  3. She misses 10. Her probability of hitting is 65%. Her probability of missing is 100 – 65 = 35%
Px  10 n  20; p  .35  1  Px  9  1  .87822 = .1218 ns
g  4. She misses 9. Her probability of hitting is 70%. Her probability of missing is 100 – 70 = 30%
ns
Px  9 n  20; p  .30  1  Px  8  1  .88667 = .1122
g  5. She misses 8. Her probability of hitting is 75%. Her probability of missing is 100 – 75 = 25%
ns
Px  8 n  20; p  .25  1  Px  7  1  .89819 = .1018
g  6. She misses 7. Her probability of hitting is 80%. Her probability of missing is 100 – 80= 20%
ns
Px  7 n  20; p  .20  1  Px  6  1  .91331 = .0867
g  7. She misses 6. Her probability of hitting is 85%. Her probability of missing is 100 – 85 = 15%
ns
Px  6 n  20; p  .15  1  Px  5  1  .93269 = .0673
g  8. She misses 5. Her probability of hitting is 90%. Her probability of missing is 100 – 90 = 10%
Px  5 n  20; p  .10  1  Px  4  1  .95683 = .0432
s
g  9 . She misses 4. Her probability of hitting is 95%. Her probability of missing is 100 – 95 = 5%.
Px  4 n  20; p  .05  1  Px  3  1  .98412 = .0159
s
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251y0831 4/18/08
b. We believe that 80% of all accidents involve alcohol. If we investigate 6  g  accidents, what is the
probability that 2  g  or fewer involve alcohol? (1)
Solution: This can also be interpreted as a hypothesis test. We are testing the null hypothesis H 0 : p  .80
against the alternative H 1 : p  .80 . An ‘s’ indicates that p the observed proportion (3/7 = .4286 in the
first case) is significantly below p 0  .80 , so that we can doubt the null hypothesis. An ‘ns’ indicates that
p is not significantly below p 0 . Note that, though in all cases except the first we cannot reject the null
hypothesis, if we had the same proportion of a larger sample we could reject it.
g  1. There are 7 accidents of which 3 or fewer involve alcohol. This is zero to 3 involving alcohol, or 4
to 7 not involving alcohol. The probability of an accident not involving alcohol is .20.
ns
P x  3 n  7; p  .80  P x  4 n  7; p  .20  1  Px  3  1  .96666 = .0333
g  2. There are 8 accidents of which 4 or fewer involve alcohol. This is zero to 4 involving alcohol, or 4




to 8 not involving alcohol. The probability of an accident not involving alcohol is .20.
P x  4 n  8; p  .80  P x  4 n  8; p  .20  1  Px  3  1  .94374 = .0563




ns
g  3. There are 9 accidents of which 5 or fewer involve alcohol. This is zero to 5 involving alcohol, or 4
to 9 not involving alcohol. The probability of an accident not involving alcohol is .20.
P x  5 n  9; p  .80  P x  4 n  9; p  .20  1  Px  3  1  .91436 = .0859




ns
g  4. There are 10 accidents of which 6 or fewer involve alcohol. This is zero to 6 involving alcohol, or 4
to 10 not involving alcohol. The probability of an accident not involving alcohol is .20.
P x  6 n  10; p  .80  P x  4 n  10; p  .20  1  Px  3  1  .87913 = .1206




ns
g  5. There are 11 accidents of which 7 or fewer involve alcohol. This is zero to 7 involving alcohol, or 4
to 11 not involving alcohol. The probability of an accident not involving alcohol is .20.
P x  7 n  11; p  .80  P x  4 n  11; p  .20  1  Px  3  1  .83886 = .1611




ns
g  6. There are 12 accidents of which 8 or fewer involve alcohol. This is zero to 8 involving alcohol, or 4
to 12 not involving alcohol. The probability of an accident not involving alcohol is .20.
P x  8 n  12; p  .80  P x  4 n  12; p  .20  1  Px  3  1  .79457 = .2054




ns
g  7. There are 13 accidents of which 9 or fewer involve alcohol. This is zero to 9 involving alcohol, or 4
to 13 not involving alcohol. The probability of an accident not involving alcohol is .20.
P x  9 n  13; p  .80  P x  4 n  13; p  .20  1  Px  3  1  .74732 = .2527




ns
g  8. There are 14 accidents of which 10 or fewer involve alcohol. This is zero to 10 involving alcohol, or
4 to 14 not involving alcohol. The probability of an accident not involving alcohol is .20.
ns
P x  10 n  14; p  .80  P x  4 n  14; p  .20  1  Px  3  1  .69819 = .3018




g  9. There are 15 accidents of which 11 or fewer involve alcohol. This is zero to 11 involving alcohol, or
4 to 15 not involving alcohol. The probability of an accident not involving alcohol is .20.
ns
P x  11 n  15; p  .80  P x  4 n  15; p  .20  1  Px  3  1  .64816 = .3518




7
251y0831 4/18/08
c. A test consists of 6  g questions and a student must get at least 70% of the questions right to pass the
course. Each question is a multiple choice question with 2 possible answers. Note that if n  16 , 70% of 16
is 11.2, so 12 or more must be right. What is the probability of passing the exam if you just guess? What
happens to the probability as the size of the exam increases to 20 questions? (2)
Assume that the professor instead offers an exam with 6  g questions with four choices and then says that
a passing grade will be 50% or more. Does this raise or lower the chance of passing for the guesser? What
happens to this probability as the size of the exam rises to 20? (2)
Solution:
g  1, n  7, p  .5 . Passing is 5 or more. P x  5 n  7, p  .5  1  Px  4  1  .77344  .2266




p  .25 . Passing is 4 or more. Px  4 n  7, p  .25  1  Px  3  1  .92944  .0706
g  2, n  8, p  .5 . Passing is 6 or more. P x  6 n  8, p  .5  1  Px  5  1  .85547  .1445


p  .25 . Passing is 4 or more. P x  4 n  8, p  .25  1  Px  3  1  .88618  .1138


g  3, n  9, p  .5 . Passing is 7 or more. P x  7 n  9, p  .5  1  Px  6  1  .91016  .0898


p  .25 . Passing is 5 or more. P x  5 n  9, p  .25  1  Px  4  1  .95107  .0489


g  4, n  10, p  .5 . Passing is 7 or more. P x  7 n  10, p  .5  1  Px  6  1  .82810  .1719


p  .25 . Passing is 5 or more. P x  4 n  10, p  .25  1  Px  4  1  .92187  .0781


g  5, n  11, p  .5 . Passing is 8 or more. P x  8 n  11, p  .5  1  Px  7  1  .88672  .1133


p  .25 . Passing is 6 or more. P x  6 n  11, p  .25  1  Px  5  1  .96567  .0343


g  6, n  12, p  .5 . Passing is 9 or more. P x  9 n  12, p  .5  1  Px  8  1  .92700  .0730


p  .25 . Passing is 6 or more. P x  6 n  12, p  .25  1  Px  5  1  .94560  .0544


g  7, n  13, p  .5 . Passing is 10 or more. P x  10 n  13, p  .5  1  Px  9  1  .95386  .0461


p  .25 . Passing is 7 or more. P x  7 n  13, p  .25  1  Px  6  1  .97571  .0243


g  8, n  14, p  .5 . Passing is 10 or more. P x  10 n  14, p  .5  1  Px  9  1  .91022  .0898


p  .25 . Passing is 7 or more. P x  7 n  14, p  .25  1  Px  6  1  .96173  .0383


g  9, n  15 , p  .5 . Passing is 11 or more. P x  11 n  15, p  .5  1  Px  10  1  .94077  .0592


p  .25 . Passing is 8 or more. P x  8 n  15, p  .25  1  Px  7  1  .98270  .0173


p  .25 . Passing is 10 or more. Px  10 n  10, p  .25  1  Px  9  1  .98614  .0139
If n  20, p  .5 . Passing is 14 or more. P x  14 n  10, p  .5  1  Px  13  1  .94234  .0577
Though the path is not even the probability of passing seems to be falling as the number of questions rises.
If we try to use a Normal approximation, we see that we have the probability that z is greater than a
positive number that rises as n rises, but much more slowly. With both of these, when the sample size rises
above 100, the probability will be smaller than Pz  4 , and this probability is less than .0001.





.7  .5 
.2 n 
P  p  .7 p  .5  P  z 
 P z 
 P z  .4000 n


.5 
.5.5 



n 









.5  .25 
.25 
P  p  .5 p  .25   P  z 
 P z 
 P z  .5774 n

.4330 
.25 .75  




n 
n







8
251y0831 4/18/08
d. (Dummeldinger) The diameter of ball bearings has a continuous uniform distribution over the interval
3 .1g millimeters to 5  .1g millimeters.
(i) If ball bearings are only acceptable if they are between 4 and 5 millimeters, what proportion of
those produced is defective?
So c  3 .1g and d  5  .1g . Make a diagram. The base of the box goes from c  3 .1g to
d  5  .1g . Shade the area between 4 and d  5  .1g .
0
3+.1g 4
5-.1g 5
1
1
1


The height of the box is
and the base of the shaded area is
d  c 5  .1g  3  .1g  2  .2 g
5  .1g  4  1  .1g . So the area of the box is P4  x  5 
1  .1g
 .5 .
2  .2 g
If we use the cumulative distribution method, remember F x   1 for x  d and
F x  
4  (3  .1g ) 1  .1g
xc
for c  x  d . So P4  x  5  F 5  F 4  1 

 .5 .
d c
2  .2 g
2  .2 g
So let g  1. Just to check my work above, make a diagram. c  3  .1g  3.1 , d  5  .1g  4.9 .
0
3.1
4
4.9 5
1
1
1


The height of the box is
and the base of the shaded area is
d  c 4.9  3.1 1.8
1  .1g
.9
 .5 . Thus P4  x  5 
4.9  4  0.9 . So the area of the box is P 4  x  5 
1 .8
2  .2 g
1  0.1
9

 .5 For g  9. c  3  .1g  3.9 and d  5  .1g  4.1 . The height of the box is
2  0.2 1.8
1
1
1


and the base of the shaded area is 4.1  4  .1 . So the area of the box is
d  c 4.1  3.9 0.2
.0.1
P4  x  5 
 .5 .
0.2
(ii) If I take a sample of three ball bearings from the assembly line.
What is the chance that none of the ball bearings are defective?
Solution: From the table with n  3 and p  .5 P0  .1250
(iii) What is the chance that exactly one is defective?
Solution: From the table with n  3 and p  .5 P1  Px  1  Px  0  .5  .1250  .3750
(iv) What is the chance that at least one is defective?
Solution: From the table with n  3 and p  .5 Px  1  1  Px  0  1  .1250  .8750
(v) What is the chance that the third ball bearing that I inspect is the first one that is defective?
Solution: From the geometric distribution formula. P3  q 2 p  .53  .1250
[12]

9
251y0831 4/18/08
e. Supposedly 60% of drivers wear their seatbelts. 1000 drivers are stopped.
(i) What is the approximate probability that less than half were wearing seatbelts? (1)
We can approximate the binomial distribution by the Normal distribution because
np  1000 (.6)  600  5 and nq  1000  600  400  5 . The standard deviation of p is

pq
n
.6.4 
.24

 .00024  .015492 . So we can say that p ~ N .6, .015492 . Without a
1000
1000


.5  .6 

 Pz  .6.46   0 . To use a continuity
continuity correction P p  .5  P  z 
.01549 

correction we would make the interval 0  x  499 wider by adding .5 at both ends, so that it
would be 0.5  x  499 .5 . The bottom number is, frankly, silly. So we could say
.4995  .6 

P p  .5995   P  z 
 Pz  .6.49   0
.01549 

(ii) What is the approximate probability that less than 200  g  were wearing seatbelts? (1)[14]


x ~ N np, npq , np  1000 .6  600 ,
npq  600 .4  240  15 .4919
200  600 

For g  1 , without a continuity correction we get Px  201   Px  200   P  z 
15 .4919 

 Pz  25 .82   0 . Pretty obviously, we will get essentially the same result for any other value
of g and a continuity correction won’t affect things.
f. If each member of the State Highway patrol writes an average of 10 tickets a day,
(i) What is the chance that a given policeman will write more than 10  g  tickets in a day? (1)
Use the Poisson distribution with a parameter of 10.
If g  1, Px  11  1  Px  11  1  .69678 = .3032
If g  2, Px  12   1  Px  12   1  .79156  .2984
If g  3, Px  13  1  Px  13  1  .86446  .1355
If g  4, Px  14   1  Px  14   1  .91654  .0835
If g  5, Px  15   1  Px  15   1  .95126  .0487
If g  6, Px  16   1  Px  16   1  .97296  .0294
If g  7, Px  17   1  Px  17   1  98572  .0143
If g  8, Px  18   1  Px  18   1  .99281  .0072
If g  9, Px  19   1  Px  19   1  .99655 = .0035
(ii) Assuming that a policeman averages 50 tickets during a work week, what is the chance that a
given policeman will write more than 510  g  tickets in a week? (I strongly doubt that the answer
is the same as the answer to (i)) (2)

55 .5  50 
If g  1, with the continuity correction Px  55   Px  55 .5  P  z 

50 

5.5 

 P z 
  Pz  0.78   .5  .2823  .2177 . Without the continuity correction
7
.
0711 


56  50 
6 

Px  55   Px  56   P  z 
  Pz  0.85   .5  .3023  .1977 .
  P z 
7
.
0711


50 

10
251y0831 4/18/08

60 .5  50 
If g  2, with the continuity correction Px  60   Px  60 .5  P  z 

50 

11 .5 

 P z 
  Pz  1.48   .5  .4306  .0694 . Without the continuity correction
7
.0711 


61  50 
11 

Px  60   Px  61  P  z 
  Pz  1.56   .5  .4406  .0594 .
  P z 
7
.
0711


50 


65 .5  50 
If g  3, with the continuity correction Px  65   Px  65 .5  P  z 

50 

15 .5 

 P z 
  Pz  2.19   .5  .4857  .0143 . Without the continuity correction
7
.0711 


66  50 
16 

Px  65   Px  66   P  z 
  Pz  2.26   .5  .4881  .0119 .
  P z 
7
.
0711


50 


70 .5  50 
If g  4, with the continuity correction Px  70   Px  70 .5  P  z 

50 

20 .5 

 P z 
  Pz  2.90   .5  .4981  .0019 . Without the continuity correction
7.0711 


71  50 
21 

Px  71  Px  71  P  z 
  Pz  2.96   .5  .4985  .0015 .
  P z 
7.0711 

50 


75 .5  50 
If g  5, with the continuity correction Px  75   Px  75 .5  P  z 

50 

25 .5 

 P z 
  Pz  3.61  .5  .4998  .0002 . Without the continuity correction
7.0711 


76  50 
26 

Px  75   Px  76   P  z 
  Pz  3.68   .5  .4999  .0001 .
  P z 
7.0711 

50 


80 .5  50 
If g  6, with the continuity correction Px  80   Px  80 .5  P  z 

50 

30 .5 

 P z 
  Pz  4.31  .5  .5  0 . Without the continuity correction
7.0711 


81  50 
31 

Px  80   Px  81  P  z 
  Pz  4.38   .5  .5  0 .
  P z 
7
.
0711


50 


85 .5  50 
If g  7, with the continuity correction Px  85   Px  85 .5  P  z 

50 

35 .5 

 P z 
  Pz  5.02   .5  .5  0 . Without the continuity correction
7
.0711 


81  50 
31 

Px  80   Px  81  P  z 
  Pz  4.38   .5  .5  0 .
  P z 
7
.
0711


50 

11
251y0831 4/18/08

90 .5  50 
If g  8, with the continuity correction Px  90   Px  90 .5  P  z 

50 

40.5 

 P z 
  Pz  5.73   .5  .5  0 . Without the continuity correction
7.0711 


91  50 
41 

Px  90   Px  91  P  z 
  Pz  5.79   .5  .5  0 .
  P z 
7.0711 
50 



95 .5  50 
If g  9, with the continuity correction Px  95   Px  95 .5  P  z 

50 

45 .5 

 P z 
  Pz  6.43   .5  .5  0 . Without the continuity correction
7.0711 


96  50 
46 

Px  95   Px  96   P  z 
  Pz  6.51  .5  .5  0 .
  P z 
7.0711 

50 

g. A ‘psychic’ is shown 5 cards and then is expected to be able to guess which one of the cards is randomly
drawn. If the ‘psychic’ is a fake and is merely randomly guessing what card is picked and the experiment is
repeated 20 times, what is the chance that the psychic will pick the right card 4  g  or more times? (1)
Note that this can be interpreted as a hypothesis test too. If we assume that the probability of one success is
at most 20% then the probability of 8 or more successes is below 5% and we should doubt the correctness
of the 5% probability. If there are 10 or more success the probability falls to 1% and we strongly doubt the
null hypothesis.
If g  1 , n  20 , p  .2 and Px  5  1  Px  4  1  .62965  .37035
If g  2, n  20 , p  .2 and Px  6  1  Px  5  1  .80421  .1958
If g  3, n  20 , p  .2 and Px  7  1  Px  6  1  .91331  .0867
If g  4, n  20 , p  .2 and Px  8  1  Px  7  1  .96786  .0321
If g  5, n  20 , p  .2
If g  6, n  20 , p  .2
If g  7, n  20 , p  .2
If g  8, n  20 , p  .2
and Px  9  1  Px  8  1  .99002  .0100
and Px  10   1  Px  9  1  .99741  .0026
and Px  11  1  Px  10   1  .99944  .0008
and Px  12   1  Px  11  1  .99990  .0001
If g  9 , n  20 , p  .2 and Px  13  1  Px  12   1  .99998  .00002
h. The average number of small business bankruptcies in Fredonia are 13  .1g  a month. What is the
probability of at least one bankruptcy in a month? What is the probability of at least 3 in a month? What is
the probability of more than 180 in a year? (3)
This is a Poisson distribution with Px  
P2 
e  m m 2 mP 1
.

2!
2
e m m x
e m m
. This means P0  e m , P1 
 mP0 and
x!
1!
If g  1 , m  13.1, P0  e 13.1  .000002, P1  13.1.000002   .000027 and P2 
13 .1
P1
2
 .000175 So Px  1  1  P0  .999998 and Px  3  1  P0  P1  P2  .999796

181  157 .2 
  Pz  1.90   .5  .4713 = .0287
Px  181 m  13 .112   157 .2  P z 

157 .2 

12
251y0831 4/18/08
If g  2, m  13.2, P0  e 13.2  .000002 , P1  13.2.000002   .000024 and P2 
13 .2
P1
2
 .000161 So Px  1  1  P0  .999998 and Px  3  1  P0  P1  P2  .999806

181  158 .4 
  Pz  1.80   .5  .4641 = .0359
Px  181 m  13 .212   158 .4  P z 

158 .4 

If g  3, m  13.3, P0  e 13.3  .000002 , P1  13.3.000002   .000022 and P2 
13 .3
P1
2
 .000148 So Px  1  1  P0  .999998 and Px  3  1  P0  P1  P2  .999828

181  159 .6 
  Pz  1.69   .5  .4545 = .0455
Px  181 m  13 .312   159 .6  P z 

159 .6 

If g  4, m  13.4, P0  e 13.4  .000002 , P1  13.4.000002   .000020 and P2 
13 .4
P1
2
 .000272 So Px  1  1  P0  .999998 and Px  3  1  P0  P1  P2  .999706

181  160 .8 
  Pz  1.59   .5  .4441 = .0559
Px  181 m  13 .412   160 .8  P z 

160 .8 

If g  5, m  13.5, P0  e 13.5  .000001, P1  13.5.000002   .000019 and P2 
13 .5
P1
2
 .000125 So Px  1  1  P0  .999999 and Px  3  1  P0  P1  P2  .999855

181  162 .0 
  Pz  1.49   .5  .4319 = .0681
Px  181 m  13 .512   162 .0  P z 

162 .0 

If g  6, m  13.6, P0  e 13.6  .000001, P1  13.6.000001   .000017 and P2 
13 .6
P1
2
 .000115 So Px  1  1  P0  .999999 and Px  3  1  P0  P1  P2  .999862

181  163 .2 
  Pz  1.39   .5  .4177 = .0823
Px  181 m  13 .612   163 .2  P z 

163 .2 

If g  7, m  13.7, P0  e 13.7  .000001, P1  13.7.000001   .000015 and P2 
13 .7
P1
2
 .000105 So Px  1  1  P0  .999999 and Px  3  1  P0  P1  P2  .999879

181  164 .4 
  Pz  1.29   .5  .4015 = .0985
Px  181 m  13 .712   164 .4  P z 

164 .4 

If g  8, m  13.8, P0  e 13.8  .000001, P1  13.8.000001   .000014 and P2 
13 .8
P1
2
 .000097 So Px  1  1  P0  .999999 and Px  3  1  P0  P1  P2  .999796

181  165 .6 
  Pz  1.20   .5  .3849 = 1151
Px  181 m  13 .812   165 .6  P z 

165 .6 

If g  9 , m  13.9, P0  e 13.9  .000001, P1  13.9.000001   .000013 and P2 
13 .9
P1
2
 .0000898 So Px  1  1  P0  .999999 and Px  3  1  P0  P1  P2  .999796

181  166 .8 
  Pz  1.10   .5  .3643=.1357
Px  181 m  13 .912   166 .8  P z 

166 .8 

[21]
13
251y0831 4/18/08
i. There are exactly 14 firms in the US that produce jorcillators and 1  g  produce computers as well. If
you pick a random sample of 3 firms to examine, what is the chance that at least one will be a computer
producer . What is the chance that all the firms in your sample are computer producers? Assume that the
same proportion of 70 companies produce computers what would the probabilities be that at least one and
all three in a 3-firm sample be computer producers? (4)
[25]
If g  1, 2 out of 14 produce computers and the remaining 12 do not. n  3 . So
12 1110
1
C 02 C 312
3 21
12 11 10
(i) Px  1  1  P0  1 
 1
 1
 1  .6044  .3956 (ii) This is impossible.
14
14 13 12
14 13 12
C3
3  21
2
 .14286 , q  1  p  .8571 . Since the sample is less than 5% of
14
the population, use the binomial distribution Px  C xn p x q n x .
There are only two available. (iii) p 
Px  1  1  P0  1  C 03 p 0 q 3  1  .8571 3  1  .6297  .3703 . (iv) P3  C 33 p 3 q 0  .14286 3  .0061
If g  2, 3 out of 14 produce computers and the remaining 11 do not. n  3 . So
11 10 9
1
3 11
C C
3 21
11 10  9
(i) Px  1  1  P0  1  0 143  1 
 1
 1  .4533  .5467
14

13

12
14
13 12
C3
3  21
(ii) P3 
C 33 C 011
C 314

3
1
 .0027 (iii) p 
 .2143 , q  1  p  .7857 . Since the sample is less
14 13 12
14
3  2 1
than 5% of the population, use the binomial distribution Px  C xn p x q n x .
Px  1  1  P0  1  C 03 p 0 q 3  1  .7857 3  1  .4850  .5150 . (iv) P3  C33 p 3 q 0  .2143 3  .0098
If g  3, 4 out of 14 produce computers and the remaining 10 do not. n  3 . So
10 98
1
4 10
C C
3 21
10  9  8
(i) Px  1  1  P0  1  0 143  1 
 1
 1  .3297  .6703
14

13

12
14
13 12
C3
3  21
(ii) P3 
C 34 C 010
C 314

4
4
 .0109 (iii) p 
 .2857 , q  1  p  .7243 . Since the sample is less
14 13 12
14
3  2 1
than 5% of the population, use the binomial distribution Px  C xn p x q n x .
Px  1  1  P0  1  C 03 p 0 q 3  1  .7243 3  1  .3800  .6200 . P3  C33 p 3 q 0  .2857 3  .0233
14
251y0831 4/18/08
If g  4, 5 out of 14 produce computers and the remaining 9 do not. n  3 . So
9 87
1
C 05 C 39
3 21
9 8 7
(i) Px  1  1  P0  1  14  1 
 1
 1  .2308  .7692
14 13 12
14 13 12
C3
3  21
5 4
C 35 C 09
5
2 1
 .3571 , q  1  p  .6429 . Since the sample is less
(ii) P3 

 .0275 (iii) p 
14
14

13

12
14
C3
3  2 1
than 5% of the population, use the binomial distribution Px  C xn p x q n x .
Px  1  1  P0  1  C 03 p 0 q 3  1  .6429 3  1  .2657  .7343 . (iv) P3  C33 p 3 q 0  .3571 3  .0455
If g  5, 6 out of 14 produce computers and the remaining 8 do not. n  3 . So
8 76
1
6 8
C C
3 21
87 6
(i) Px  1  1  P0  1  0 143  1 
 1
 1  .1538  .8462
14

13

12
14 13 12
C3
3  21
65 4
C 36 C 08
6
 .4286 , q  1  p  .5714 . Since the sample is less
(ii) P3 
 3  2 1  .0549 (iii) p 
14
14

13

12
14
C3
3  2 1
than 5% of the population, use the binomial distribution Px  C xn p x q n x .
Px  1  1  P0  1  C 03 p 0 q 3  1  .5714 3  1  .1866  .8134 . (iv) P3  C33 p 3 q 0  .4286 3  .0787
If g  6, 7 out of 14 produce computers and the remaining 7 do not. n  3 . So
7 6  5
1
7 7
C C
3 21
765
(i) Px  1  1  P0  1  0 143  1 
 1
 1  .0962  .9038
14

13

12
14
13 12
C3
3  21
765
C 37 C 07
7
 .5000 , q  1  p  .5000 . Since the sample is less
(ii) P3 
 3  2 1  .0962 . (iii) p 
14
14

13

12
14
C3
3  2 1
than 5% of the population, use the binomial distribution Px  C xn p x q n x .
Px  1  1  P0  1  C 03 p 0 q 3  1  .5000 3  1  .1250  .8750 . (iv) P3  C33 p 3 q 0  .5000 3  .1250
If g  7, 8 out of 14 produce computers and the remaining 6 do not. n  3 . So
(i) Px  1  1  P0  1 
C 08 C 36
C 314
65 4
3 21
65 4
 1
 1
 1  .0549  .9451
14 13 12
14 13 12
3  21
1
87 6
8
 .5714 , q  1  p  .4286 . Since the sample is less
(ii) P3 
 3  2 1  .1538 (iii) p 
14
14 13 12
14
C3
3  2 1
than 5% of the population, use the binomial distribution Px  C xn p x q n x .
C 38 C 06
Px  1  1  P0  1  C 03 p 0 q 3  1  .4286 3  1  .0787  .9213 . (iv) P3  C33 p 3 q 0  .5714 3  .1866
15
251y0831 4/18/08
If g  8, 9 out of 14 produce computers and the remaining 5 do not. n  3 . So
(i) Px  1  1  P0  1 
5 43
3 21
5 43
 1
 1
 1  .0274  .9725
14 13 12
14 13 12
3  21
1
C 09 C 35
C 314
9 8 7
9
 .6429 , q  1  p  .3571 . Since the sample is less
(ii) P3 
 3  2 1  .2308 . (iii) p 
14
14

13

12
14
C3
3  2 1
than 5% of the population, use the binomial distribution Px  C xn p x q n x .
C 39 C 05
Px  1  1  P0  1  C 03 p 0 q 3  1  .3571 3  1  .0456  .9544 . (iv) P3  C33 p 3 q 0  .6429 3  .2657 .
If g  9, 10 out of 14 produce computers and the remaining 4 do not. n  3 . So
(i) Px  1  1  P0  1 
C 010 C 34
C 314
4 3 2
3 21
4 3 2
 1
 1
 1  .0110  .9890
14 13 12
14 13 12
3  21
1
10  9  8
10
 .7143 , q  1  p  .2857 . Since the sample is less
(ii) P3 
 3  2 1  .3297 (iii) p 
14
14

13

12
14
C3
3  2 1
than 5% of the population, use the binomial distribution Px  C xn p x q n x .
C 310 C 04
Px  1  1  P0  1  C 03 p 0 q 3  1  .2857 3  1  .0233  .9767 . (iv) P3  C33 p 3 q 0  .7143 3  .3645
j. (Extra credit) The time between landings at an airport follows an exponential distribution with a mean of
30 seconds. What is the probability that there will be a gap of between 15  g  and 45  g  seconds before
the next plane lands? (2) The following comes from ‘Great Distributions I have Known.’
Distribution
Uses
Formula
Mean
Variance
Continuous
Distributions
Exponential
x is usually the
f x   ce cx and
amount of time
1
1
cx


you have to wait F x  1  e
c
c
when x  0 and
until a success.
the mean time to a
1
success is . Both
c
are zero if x  0 .
 45 g 

30 

1
1
So 30  and c 
. P15  g  x  45  g   1  e 
c
30
If g  1, P16  x  46   e
 16 
 
 30 
 46 
 
 e  30 
 24 
 
 54 
 
 15 g 


 1  e  30 

 15 g 


e  30 
 45 g 


 e  30 
 e 0.5333  e 1.5333  .58665  .21582  .3708
If g  9, P24  x  54   e  30   e  30   e 0.8  e 1.8  .44933  .16530  .2840
All other answers should be between these two.
16
251y0831 4/18/08
2. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that the
jorcillator only works as long as either component works (so that it fails in the first period only if both
components fail).
The probability of the phillinx failing is given by a Normal distribution with  18  g and   3. , For
example, if the life of the phillinx is represented by x1 , the chance of the phillinx failing in the first ten
years is P0  x1  10  and the probability of it failing in the second ten years is P10  x  20  For
example: Ima Badrisk has the number 375292, so her distribution has a mean of   18  .g  20 . The
flubberall has exactly the same distribution.
In order to maintain my sanity, use the following events. Period 1 is the first ten years, period 2 is the
second ten years and period three is happily ever after.
Failure of the phillinx in period 1, 2, 3 are events A1, A2 and A3
Failure of the flubberall in period 1, 2, 3 are events B1, B 2 and B 3
Failure of the jorcillator in period 1, 2, 3 are events C1, C 2 and C 3 .
a) What is the probability that the phillinx will fail in period 1? Period 2? Period 3? (2)
Note that, because of the distance zero is from the mean, P0  x  10   Px  10  . I was wrong to say
that zero should not be in this interval, but this did not affect any grades.
10  18 

If g  0 ,   18 . P A1  can be written as Px  10   P z 
  Pz  2.67   .5  .4962
3 

20  18 
 10  18
z
= .0038. P A2   P10  x  20   P
  P 2.67  z  0.67  = .4962 + .2486
3 
 3
20  18 

= .7448. P A3   Px  20   P z 
  Pz  0.67  = .5 - .2486 = .2514. These add to 1.
3 

10  27 

If g  9 ,   27 . P A1  can be written as Px  10   P z 
  Pz  5.67   .5  .5  0
3 

20  27 
 10  27
z
P A2   P10  x  20   P
  P 5.67  z  2.33  = .5 - .4901 = .0099.
3
3 

20  27 

= .7448. P A3   Px  20   P z 
  Pz  2.33  = .4901 +.5 = .9901. These add to 1.
3 

For other values of g, probabilities fall between the numbers above. They always add to 1.
Joint events are below. Joint event
Probability if g  0
downs machine
in period
1
.0038(.0038) = .00001
P A1  B1 
P A1  B2 
P A1  B3 
P A2  B1 
P A2  B2 
P A2  B3 
P A3  B1 
P A3  B2 
P A3  B3 
Probability if g  9
0(0) = 0
2
.0038(.7448) = .00283
0(.0099) = 0
3
.0038(.2514) = .00096
0(.9901) = 0
2
.7448(.0038) = .00283
.0099(0) = 0
2
.7448(.7448) = .55473
.0099(.0099) = .00010
3
.7448(.2514) = .18724
.0099(.9901) = .00980
3
.2514(.0038) = .00096
.9901(0) = 0
3
.2514(.7448) = .18724
.9901(.0099) = .00980
3
.2514(.2514) = .06320
.9901(.9901) = .98030
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251y0831 4/18/08
If we add probabilities for each period we get the table below.
Probability if g  0
PC1  = .00001
PC 2  = .56039
PC3  = .43960
Probability if g  9
PC1  = 0
PC 2  = .00010
PC3  = .99990
For b), c) and d) see above
b) What is the probability that the jorcillator will fail in the first period? (1)
c) What is the probability that the jorcillator will fail in the second period 2? (1)
d) What is the probability that the jorcillator will fail in the third period? (1)
If you haven’t figured it out already, one of the easiest ways to do this is to make a joint probability
table. Put the A events across the top. Put the B events down the side. Figure out what the
probability of the joint events must be if they are independent. Now make a similar table. This time,
instead of probabilities, fill in the period in which the jorcillator fails.
e) (Extra Credit) Find the probability that the jorcillator and the Phillinx both fail in the third
period (1). This is PC 3  A3  = P A3  B1  + P A3  B2  + P A3  B3  = P A3 
f) (Extra Credit) Find the probability that the jorcillator fails in the third period, given that the
PC 3  A3  P A3 

1
phillinx fails in the third month i.e. P C3 A3 (1). This is
P A3 
P A3 




g) Demonstrate Bayes’ rule by finding P A3 C3 and showing that Bayes’ rule explains the
relationship between the conditional probabilities that you have found. (2) [34]
P A3  C 3 
P A3 

. Bayes’ Rule says
P A3 C3 
PC 3 
P A3   PB3   P A3  B3 


PA3 C3  
PC3 A3 P A3 
PC3 

1P A3 
. Just fill in your numbers.
PC3 
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251y0831 4/18/08
3. Assume that you have a variable with a Poisson distribution with a mean of 1.5  .1g  . (My value of
g is 2, so I have a mean of 1.5  .12  1.5  .2  1.3 . Write down the values of P0 , P 1 etc for all
values of x that have probabilities of .0001or larger by subtracting values from the Poisson table.
Compute the expected value and standard deviation using the probabilities and values of x and show
that these are the same as stated in our write-ups on the Poisson distribution. (4) [37]
If g  1, m  1.4 . x
P x 
xPx 
x 2 P x 
1
2
3
4
5
6
7
8
9
10
11
0
1
2
3
4
5
6
7
8
9
10
0.246597
0.345236
0.241665
0.112777
0.039472
0.011052
0.002579
0.000516
0.000090
0.000014
0.000002
1.00000
0.000000
0.345236
0.483330
0.338331
0.157888
0.055260
0.015474
0.003612
0.000720
0.000126
0.000020
1.40000
0.00000
0.34524
0.96666
1.01499
0.63155
0.27630
0.09284
0.02528
0.00576
0.00113
0.00020
3.3600
 Px  1 (This shows that we have a valid distribution),    xPx  1.40000
and E x    x Px   3.36000 . This means that   E x    3.36000  1.4000   1.4000
So
2
2
2
If g  9, m  0.6 . x
1
2
3
4
5
6
7
8
0
1
2
3
4
5
6
7
P x 
0.548812
0.329287
0.098786
0.019757
0.002964
0.000356
0.000036
0.000003
1.00000
xPx 
0.000000
0.329287
0.197572
0.059271
0.011856
0.001780
0.000216
0.000021
0.60000
2
2
2
x 2 P x 
0.000000
0.329287
0.395144
0.177813
0.047424
0.008900
0.001296
0.000147
0.9600
 Px  1 (This shows that we have a valid distribution),    xPx  0.60000
and E x    x Px   0.9600 . This means that   E x    0.9600  0.6000   0.6000 .
So
2
2
2
2
2
2
Rounding error may knock these totals slightly off.
xPx   1.3 and E x 2  2.99. So  2  E x 2   2  1.30
If g  2, m  1.3 .  
 
 

If g  3, m  1.2 .    xPx   1.2 and Ex 2   2.64. So  2  E x 2    2  1.20
If g  4, m  1.1 .    xPx   1.1 and Ex 2   2.31. So  2  E x 2    2  1.10
If g  5, m  1.0 .    xPx   1.0 and Ex 2   2.00. So  2  E x 2    2  1.00
If g  6, m  0.9 .    xPx   0.9 and Ex 2   1.71. So  2  E x 2    2  0.90
If g  7, m  0.8 .    xPx   0.8 and Ex 2   1.44. So  2  E x 2    2  0.80
If g  8, m  0.7 .    xPx   0.7 and Ex 2   1.19. So  2  E x 2    2  0.70
19
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