251y0731 4/30/07
ECO251 QBA1
THIRD EXAM
Apr 18, 2007
Name: __KEY_______________
Student Number: _____________________
Part I. (16 points) Do all the following (2 points each unless noted otherwise). Make Diagrams! Show
your work!
z has the standardized Normal distribution z ~ N 0, 1 for the first four problems.
Material in italics below is a description of the diagrams you were asked to make or a general explanation
and will not be part of your written solution. The x and z diagrams should look similar. If you know what
you are doing, you only need one diagram for each problem. General comment - I can't give you much
credit for an answer with a negative probability or a probability above one, because there is no such
 x 
thing!!! In all these problems we must find values of z  
 corresponding to our values of x before
  
we do anything else. A diagram for z will help us because, if areas on both sides of zero are shaded, we
will add probabilities, while, if an area on one side of zero is shaded and it does not begin at zero, we will
subtract probabilities. Note: All the graphs shown here are missing a vertical line. They are also to
scale. A hand drawn graph should exaggerate the distances of the points from the mean. Note also
that, because of the rounding error necessary to use a conventional normal table, the results for x will
disagree with results for z, especially for small values of z.
1. Pz  1.43   Pz  0  P1.43  z  0  .5  .4236  .0764
This is how we usually find a p-value for a left-sided test when the distribution of the test ratio is Normal.
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the entire area below -1.43. Because this is entirely on the left side of zero, we must subtract the area
between -1.43 and zero from the area below zero.
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2. P0  z  3.21  .4993
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the area between zero and 3.21. Because this is entirely on the right side of zero and starts at zero, it
is precisely the sort of probability in Table 17. Because 3.21 is larger than the numbers on a conventional
Normal table, there is a section below the usual table that says ‘If z 0 is between 3.18 and 3.21
P0  z  z 0   .4993 .
3. P1.45  z  1.45   P1.45  z  0  P0  z  1.45   2 P0  z  1.45  =2(.4265) =.8530
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the entire area between -1.45 and 1.45. Because this is on both sides of zero, we must add the area
between -1.45 and zero to the area between zero and 1.45.
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4. z .14 Solution: z .14 is, by definition, the value of z with a probability of 14% above it. Make a
diagram. The diagram for z will show an area with a probability of 86% below z .14 . 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% - 50% = 36%. The upper tail of the distribution above z .14 has a probability of 14%, so that the
entire area above 0 adds to 36% + 14% = 50%. From the diagram, we want one point z .14 so that
Pz  z .14   .86 or P0  z  z.14   .3600 . If we try to find this point on the Normal table, the closest we
can come is P0  z  1.08   .3599 . So we will use z.14  1.08 , though 1.09 might be acceptable.
x ~ N 3, 7.2 for problems 5 through 8.
 1.43  3 

 Pz  0.62   Pz  0  P0.62  z  0  .5  .2324  .2676
5. Px  1.43   P  z 
7.2 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the entire area below -1.43. Because this is entirely on the left side of zero, we must subtract the area
between -1.43 and zero from the area below zero. If you wish, make a completely separate diagram for x .
Draw a Normal curve with a mean at 3. Indicate the mean by a vertical line! Shade the area below -1.43.
This area is completely to the left of the mean (3), so we subtract the smaller area between -1.43 and the
mean from the larger area below the mean.
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3.21  3 
0  3
z
 P 0.42  z  0.03   P0.42  z  0  P0  z  0.03
6. P0  x  3.21  P 
7.2 
 7.2
 .1628  .0120  .1748
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the entire area between -0.42 and 0.03. Because this area is on both sides of zero, we must add the
area between -0.42 and zero to the area from zero to 0.03. If you wish, make a completely separate diagram
for x . Draw a Normal curve with a mean at 3. Indicate the mean by a vertical line! Shade the area from
0 to 3.21. This area is on both sides of the mean (3), so we add the area between 0 and the mean to the area
between the mean and 3.21.
1.45  3 
  1.45  3
z
 P 0.62  z  0.22 
7. P1.45  x  1.45   P 
7.2 
 7.2
 P0.62  z  0  P0.22  z  0  .2324  .0871  .1453
For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the entire area between -0.62 and -0.22. Because this area is on the left side of zero, we must subtract
the area between -0.22 and zero from the larger area between -0.62 and zero. If you wish, make a
completely separate diagram for x . Draw a Normal curve with a mean at 3. Indicate the mean by a
vertical line! Shade the area from -1.45 to 1.45. This area is on the left side of the mean (3), so we subtract
the area between 1.45 and the mean from the larger area between -1.45 and the mean.
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8. x.14 Solution: Since x ~ N 3, 7.2 , the diagram for x would show 86% probability below x.14 split in
two regions on either side of 3 with probabilities of 50% below 3 and 36% above 3 and below x.14 , and
with 14% above x.14 . On the first page we found z.14  1.08 , so the value of x can then be written
x.14    z.14  3  1.087.2  3  7.776  10.776 .
Check:
10 .776  3 

 Pz  1.08   Pz  0  P0  z  1.08   .5  .3599  .1401  14%
Px  10.776   P  z 
7.2 

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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. Which of the following are correct for a discrete distribution? (You may circle more than one.)
a) P7  x  20   Px  20   Px  7
b) Px  0  1  Px  0 Sorry about this! This should have read Px  1  1  Px  0 in which
case it would be correct for discrete distributions but not continuous distributions. Because of my error, if
you say that this is correct, you will be given credit for the answer.
c) Px  2  1  Px  2
But note that Px  2  Px  3
d) * Px  2  1  Px  2
e) * Px  6  Px  5
f) None of the above is correct. (1.5 for 2, 1 for 1, 0.5 off for each one wrong.)
Explanation: The rule that you were given for a discrete distribution said
Pa  x  b  Px  b  Px  a  1 , so P7  x  20   Px  20   Px  6 . Px  2  1  Px  1
2. Which of the following are correct for a continuous distribution? (You may circle more than one.)
a) * P7  x  20   Px  20   Px  7
b) Px  0  1  Px  0
Note that Px  0 and Px  2 are not defined for a
continuous distribution.
c) Px  2  1  Px  2
d) * Px  2  1  Px  2
But note that Px  2  Px  2
e) Px  6  Px  5
f) None of the above is correct. (1 for 1, 0.5 off for each one wrong.)
Explanation: The rule that you were given for a continuous distribution said
Pa  x  b  Px  b  Px  a  . So P7  x  20   Px  20   Px  7 . Px  0 and Px  2 do not
exist in a continuous distribution. where all probabilities are defined over intervals. Px  6  Px  6 .
Both of the above questions will have points taken off for wrong choices, though no question will be
given a score below zero.
3. Which of the following is not a requirement of a discrete probability distribution?
a) * Equally likely probability of a success.
b) Sum of the possible outcomes is 1.00.
c) The outcomes are mutually exclusive.
d) The probability of each outcome is between 0 and 1.
e) All of the above are required.
f) None of the above is required.
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4. A company that sells annuities must base the annual payout on the probability distribution of the length
of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants
is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. What
proportion of the plan recipients would receive payments beyond age 75? (2.5) [8.5]
Solution:
75  68 

 Pz  2.00   Pz  0  P0  z  2.00   .5  .4772  .0228
Px  75   P  z 
3.5 

For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line!
Shade the entire area above 2.00. Because this is entirely on the right side of zero, we must subtract the area
between zero and 2.00 from the larger area above zero.
5. What type of probability distribution will the consulting firm most likely employ to analyze the
insurance claims in the following problem?
An insurance company has called a consulting firm to determine if the company has an unusually
high number of false insurance claims. It is known that the industry proportion for false claims is
3%. The consulting firm has decided to randomly and independently sample 100 of the company’s
insurance claims. They believe the number of these 100 that are false will yield the information the
company desires.
a) *Binomial distribution.
b) Poisson distribution.
c) Normal distribution.
d) Hypergeometric distribution.
e) Geometric distribution
6. If n = 10 and p = 0.70, then the standard deviation of the binomial distribution of x is
a) 0.021
b) 0.07
c) 0.145
d) *1.45
e) 2.10
f) 7.00
g) 14.29
Explanation: The variance of the binomial distribution is  2  npq , where q  1  p . Here n = 10, p =
0.70 and q  .30 .  2  10.70.30  2.10 . So   2.10  1.4491  1.45
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7. If n = 10 and p = 0.70, then the standard deviation of the binomial distribution of p 
a)
b)
c)
d)
e)
f)
g)
x
is
n
[14.5]
0.021
0.07
* 0.145
1.45
2.10
7.00
14.29
Explanation: The table in ‘Great Distributions I have Known’ says the following.
Distribution
Uses
Formula
Mean
Variance
Binomial
Gives
See above. Since
probability of
pq
E p   p
x
Var p  
p as proportion p  n or x  pn
n
of successes in
n tries
.70 .30 
 0.021 . So   0.021  0.14491  0.145
Here n = 10, p = 0.70 and q  .30 .  2 
10
8. What type of probability distribution will most likely be used to analyze warranty repair needs on new
cars in the following problem?
The service manager for a new automobile dealership reviewed dealership records of the past 20
sales of new cars to determine the number of warranty repairs he will be called on to perform in the
next 90 days. Corporate reports indicate that the probability any one of their new cars needs a
warranty repair in the first 90 days is 0.05. The manager assumes that calls for warranty repair are
independent of one another and is interested in predicting the number of warranty repairs he will be
called on to perform in the next 90 days for this batch of 20 new cars sold.
[16.5]
a) *Binomial distribution.
b) Poisson distribution.
c) Normal distribution.
d) Hypergeometric distribution.
e) Geometric distribution
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9. On the average, 3 customers per minute arrive at any one of the checkout counters of a convenience
store. Checkout time takes 2 minutes. How many clerks are needed to be sure that the chance of someone
having to wait is no more than 10%? Your answer should briefly explain what distribution you are using
and how you got the number you give. (2.5)
[19]
Solution: The average for the 2 minute checkout period is 6. Since we are looking at numbers of successes
when the mean is known, we are using the Poisson distribution. Here is the relevant part of the Poisson
table.
Poisson 6.0
k
0
1
2
3
4
5
6
7
8
9
10
11
12
P(x=k)
0.002479
0.014873
0.044618
0.089235
0.133853
0.160623
0.160623
0.137677
0.103258
0.068838
0.041303
0.022529
0.011264
Poisson 6.5
P(xk)
0.00248
0.01735
0.06197
0.15120
0.28506
0.44568
0.60630
0.74398
0.84724
0.91608
0.95738
0.97991
0.99117
k
0
1
2
3
4
5
6
7
8
9
10
11
12
P(x=k)
0.001503
0.009772
0.031760
0.068814
0.111822
0.145369
0.157483
0.146234
0.118815
0.085811
0.055777
0.032959
0.017853
Poisson 7.0
P(xk)
0.00150
0.01128
0.04304
0.11185
0.22367
0.36904
0.52652
0.67276
0.79157
0.87738
0.93316
0.96612
0.98397
k
0
1
2
3
4
5
6
7
8
9
10
11
12
P(x=k)
0.000912
0.006383
0.022341
0.052129
0.091226
0.127717
0.149003
0.149003
0.130377
0.101405
0.070983
0.045171
0.026350
P(xk)
0.00091
0.00730
0.02964
0.08177
0.17299
0.30071
0.44971
0.59871
0.72909
0.83050
0.90148
0.94665
0.97300
So Px  8  .84724 and Px  9  .91608 . This means Px  8  1  Px  8  1  .84724  .1528 and
Px  9  1  Px  9  1  .91608  .0839 . So if we have 9 clerks the probability is below 10%.
10. 13 of a shipment of 15 hard disks are defective. 4 disks are inspected. What is the chance of finding at
least one defective?
Solution: This is Exercise 5.50 in the text. p  13 , n  4, N  15, M  np  15 13   5
 10! 
 1


  1   6! 4!   1  10  9  8  7  1  .1538461  .84615
 15! 

15 14 13 12



 11! 4! 
11. (Extra Credit) A catalog company that receives the majority of its orders by telephone conducted a
study to determine how long customers were willing to wait on hold before ordering a product. The length
of time was found to be a random variable best approximated by an exponential distribution with a mean
equal to 3 minutes. What proportion of customers will not be on the line after a wait of 4.5 minutes?
a) 0.22313
b) 0.48658
c) 0.51342
d) *0.77687
e) The answer is nowhere near any of the above. Here is my answer.
1
1
Solution: F x 1  e cx when x  0 and the mean time to a success is . Let c  .
c
3
 C 5 C 10
Px  1  1  P0  1   0 154
 C
4

Px  4.5  F 4.5
1
 4.5
 1 e 3
 1  e 1.5  1  .2231  .7769
12. Assume that there is an average of 6 chocolate chips per cookie. Now assume that an inspector will
discard any cookies with fewer than 4 chips. About how many will be discarded from a batch of 100.[23]
Solution: This is Exercise 5.35 in the text. The parameter is m  6.0. We discard cookies with fewer than 4
chips. Px  4  Px  3  .15120 . This represents the proportion that have fewer than 4 chips, so the
quantity is 100  0.15120   15.120 . Seems like it’s 15 or 16.
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ECO251 QBA1
THIRD EXAM
Apr 18, 2005
TAKE HOME SECTION
Name: ____KEY____________________
Student Number: _________________________
Throughout this exam show your work! Please indicate clearly what sections of the problem you are
answering and what formulas you are using.
Part III. Do all the Following (19+ Points) Show your work! Neatness counts!
1. (Webster) 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.
a) The average number of calls that come into a switchboard is 1 each minute.
Let g be the last digit of your student number. Assume that the operator is away from her desk
for 10  g 
minutes (so that the time that you use will be between 5 and 9.5 minutes). What is
2
the average number of calls that will come into the switchboard in that time? (0)
(i) What is the probability that at least one call is missed while the operator is away? (1)
(ii) What is the probability that between 10 and 20 (inclusive) calls are missed while the
operator is away? (1)
b) You purchase parts from a supplier who has an average defect rate of 1  g  units per lot of
100. (Your average for a lot of 100 will be between 1 and 10.)
(i) You order 150 units (1.5 lots) , but say that you will not buy from this supplier again if
more than 4 units are defective. What is the probability that you will not buy from this
supplier again? (1) (Note: If g  9, you order 200 units.)
(ii) The next time you order (if there is a next time), you resolve to take a sample of ten
items from the thousand in this order. You will return the whole mess if more than two
items are defective. If an average of 1  g  units per lot of 100 means that the probability
of an item being defective is 1  g % , what is the probability that you will return the
order? (2) Note: To keep us on a level playing field, you must work out this answer by
hand. Though it is strongly recommended that you check your answer against Table 14
and/or Table 15, no answer using tables will be accepted!
[5]
c) Lawn fertilizer comes in bags whose weight follows the continuous uniform distribution with a
range between 23.8 and 26 0.2g pounds. You need 24.8 pounds to cover your lawn and buy only
one bag. (Your upper number will be between 26 and 28.9 pounds.)
(i) What is the probability that you will not have enough fertilizer in the bag? (1)
(ii) If you buy two bags, what is the mean and standard deviation of their total weight?
(1)
[7]
(iii) If the bags are distributed with a mean of 25.5 pounds and a range of 2  0.1g 
pounds, what is the probability that a bag will be between 20 and 24 pounds? (Your
range will be between 2 and 2.9 pounds) (1)
d) There are only 10  2g officers in your firm that are eligible to be on the Compensation
Committee. Of these half make over $500,000.00 a year.
(i) If you pick 4 officers at random, what is the probability that exactly three make over
$500,000? (2)
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(ii) The shareholders will be very unhappy if they see that more than half of the members
of the Compensation Committee make over $500,000. If you pick 4 officers at random,
what is the probability that more than half make over $500,000? (Note that the answer is
not the same as the answer to (i).) (1)
[11]
e) Only 65% of the many workers in a high security scientific installation seem to be capable of
remembering to carry their badges. If they are among the 35% that forget they will be sent home
to get them!
(i) If the installation opens at 6am, what is the probability that the second employee to
arrive is the first one sent home? (1)
(ii) On the average, if we number the employees by their arrival order, what is the
average number of the first employee to be sent home? (The third? The 4th? The 3.74th?)
(1)
(iii) If a group of 6  g  employees arrive at 9am, what is the probability that at least one
will not be sent home? (1)
(iv) In the same group, what is the probability that at least one will be sent home? (1)
(v) In the same group, what is the probability that more than half will not be sent home?
(1)
[16]
(vi) (Extra Credit) By late morning, the employees are trickling in at only 12 per hour.
The guard decides that if the probability of someone arriving in the next 5 minutes after
you come in is less than 50%, he will hide behind the warning sign and take a few nips
from his bottle. He asks you whether the probability is less than 50%. Do not use the
Poisson distribution for this problem. (2)
2. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that the
jorcillator only works as long as both components work (so that it fails in the first month if either
component fails). Note change.
The probability of the phillinx failing is given by a Normal distribution with   2  .1h and
  0.5  .01h , where h is the second-to-last digit of your student number. For example, if the life of
the phillinx is represented by x1 , the chance of the phillinx failing in the 3rd month is P2  x1  3
and the probability of it failing after the third month is Px1  3 . For example: Ima Badrisk has the
number 375292, so her distribution has a mean of   2  .1h  2.9 and a standard deviation of
  0.5  .019  0.59 . Your mean will be between 2.0 and 2.9 and your standard deviation will be
between 0.50 and 0.59.
The probability of the flubberall failing is: in the first month 40%; in the second month 30%; in the
third month 20% and after the third month 10%. , We will divide time into four periods with the last
period being ‘beyond the third month.’ The jorcillator is guaranteed to fail in one of the four periods.
Failure of components is assumed to be independent, so if the probability of the phillinx failing in the
first month is .1, and the probability of the flubberall failing in the first month is .4, the probability of
both components failing in the first month is (.1) (.4) =.04 (This is not the necessarily the probability
that the jorcillator will fail in the first month!)
In order to maintain my sanity, use the following events.
Failure of the phillinx in period 1, 2, 3, 4 are events A1, A2 , A3 , and A4 .
Failure of the flubberall in period 1, 2, 3, 4 are events B1, B2 , B3 , and B4 .
Failure of the jorcillator in period 1, 2, 3, 4 are events C1, C2 , C3 , and C 4 .
a) What is the probability that the phillinx will fail in month 1? Month 2? Month 3? After Month
3? (2)
b) What is the probability that the jorcillator will fail in the first month? (2)
c) What is the probability that the jorcillator will fail in the second month? (1)
d) What is the probability that the jorcillator will fail in the third month? (1)
11
251y0731 4/30/07
e) What is the probability that the jorcillator will last beyond 3 months? (1)
[21]
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.
f) (Extra Credit) Find the probability that the jorcillator and the Phillinx both fail in the third
month (1)
g) (Extra Credit) Find the probability that the phillinx fails in the third month, given that the
jorcillator fails in the third month i.e. P A3 C3 (1)


h) (Extra Credit) Demonstrate Bayes’ rule by finding showing how to get the probability that the
jorcillator fails in the third month, given that the phillinx fails in the third month, two ways. The first
would be directly by working with joint probabilities that both the jorcillator and the phillinx fail. The
second would be by using Bayes’ rule and your result in g). (1)
3. (Extra Credit) Do the following Binomial problems. If you substitute another distribution for the
Binomial, justify the substitution by showing that a substitution condition is satisfied. (6)
a) 45% of our many management employees have MBAs. If we select 150 at random, what is the
probability that exactly 72 have college degrees?
b) 45% of our many management employees have MBAs. If we select 150 at random, what is the
probability that more than half have college degrees?
c) 5% of our many management employees have MBAs. If we select 90 at random, what is the
probability that more than 6 have MBAs?
12
251y0731 4/30/07
1. (Webster) 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.
a) The average number of calls that come into a switchboard is 1 each minute.
Let g be the last digit of your student number. Assume that the operator is away from her desk
for 10  g 
minutes. What is the average number of calls that will come into the switchboard in
2
that time? (0)
(i) What is the probability that at least one call is missed while the operator is away? (1)
(ii) What is the probability that between 10 and 20 (inclusive) calls are missed while the
operator is away? (1)
Solution: The number of minutes she will be away is between 5 and 8.5. The average number of
calls that come in while she is away will be between 5 and 8.5. This is a Poisson problem because
we are trying to track the number of successes when the mean is given,
(i) Px  1  1  P0
If m  5 , Px  1  1  .00674  .9933
If m  8.5 , Px  1  1  .00020  .9998
(ii) P10  x  20   Px  20   Px  9
If m  5 , P10  x  20   1.00000  .96817  .0318
If m  8.5 , P10  x  20   .99979  .65297  .3468
b) You purchase parts from a supplier who has an average defect rate of 1  g  units per lot of
100.
(i) You order 150 units (1.5 lots), but say that you will not buy from this supplier again if
more than 4 units are defective. What is the probability that you will not buy from this
supplier again? (1) (Note: If g  9, you order 200 units.)
(ii) The next time you order (if there is a next time), you resolve to take a sample of ten
items from the thousand in this order. You will return the whole mess if more than two
items are defective. If an average of 1  g  units per lot of 100 means that the probability
of an item being defective is 1  g % , what is the probability that you will return the
order? (2) Note: To keep us on a level playing field, you must work out this answer by
hand. Though it is strongly recommended that you check your answer against Table 14
and/or Table 15, no answer using tables will be accepted!
[5]
Solution: (i) The average defects per lot are between 1 and 9 per lot if you buy 1.5 lots. This
means that the mean is between 1.5 and 13.5. There is an error here – the 200 should
have been for 1  g  9 . Using a Poisson distribution Px  4  1  Px  4
The average defects per lot are 10 per lot if you buy 2 lots. This means that the mean is
20.
If m  1.5 , Px  4  1  .98142  .01858
If m  12 .5 , Px  4  1  .00535  .9465
If m  20 ,
Px  4  1  .00002  .99998
(ii) p will be between 1% and 10%. You will reject the shipment if x, the number of
defective items is above 2. Px  2  1  Px  2  1  P0  P1  P2 . This is a
binomial problem where n  10 , Px  C xn p x q n x . P0  C010 p 0 q10  q10 ,
P1  C110 p1q 9  10 p1q 9 and P2  C 210 p 2 q 8 
10  9 2 8
p q  45 p 2 q 8 . Computations
2 1
follow for all values of p .
13
251y0731 4/30/07
p
.01
.02
.03
.04
.05
.06
.07
.08
.09
.10
q
0.99
0.98
0.97
0.96
0.95
0.94
0.93
0.92
0.91
0.90
P0
0.904382
0.817073
0.737424
0.664833
0.598737
0.538615
0.483982
0.434388
0.389416
0.348678
P 1
P2
0.091352
0.166750
0.228069
0.277014
0.315125
0.343797
0.364288
0.377729
0.385137
0.387420
0.004152
0.015314
0.031742
0.051940
0.074635
0.098750
0.123388
0.147807
0.171407
0.193710
Px  2
0.999886
0.999136
0.997235
0.993786
0.988496
0.981162
0.971658
0.959925
0.945960
0.929809
Px  2
0.0001138
0.0008639
0.0027649
0.0062137
0.0115036
0.0188378
0.0283421
0.0400754
0.0540400
0.0701908
c) Lawn fertilizer comes in bags whose weight follows the continuous uniform distribution with a
range between 23.8 and 26 0.2g pounds. You need 24.8 pounds to cover your lawn and buy only
one bag.
(i) What is the probability that you will not have enough fertilizer in the bag? (1)
Solution: If g  0, the distribution goes from 23.8 to 26 pounds.
24 .8  23 .8
 .45
26  23 .8
If g  9, the distribution goes from 23.8 to 27.8 pounds.
Px  24 .8 
24 .8  23 .8
 .25
27 .8  23 .8
Make a diagram! You should show a box between 23.8 and a number that is between 26
and 27.8 pounds. If this number is 26 + .2g, the height of the box will be
1
1

. Shade the area in the box between 23.8 and 24.8. Since the
26  .2 g  23 .8 2.2  .2 g
1
length of this area is 1, the shaded area is
2 .2  .2 g
Px  24 .8 
(ii) If you buy two bags, what is the mean and standard deviation of their total weight?
(1)
d  c  . For two bags, if x and x are independent
cd
and  2 
1
2
2
12
and identical. Ex1  x 2   Ex1   Ex 2   2 and Var x1  x 2 
Solution:  
2
 Varx1   Varx 2   2 2 . So StDevx1  x 2   2 2
If g  0, the distribution goes from 23.8 to 26 pounds.
26  23.8  0.403333
23 .8  26
 24 .9 and  2 
2
12
Ex1  x 2   224.9  49.8 and StDevx1  x 2   20.403333   0.8981

2
If g  9, the distribution goes from 23.8 to 27.8 pounds.
27.8  23.82  1.333333
23 .8  27 .8
 25 .8 and  2 
2
12
Ex1  x 2   225.8  51.6 and StDevx1  x 2   21.33333   1.6330

[7]
(iii) If the bags are distributed with a mean of 25.5 pounds and a range of 2  0.1g 
pounds, what is the probability that a bag will be between 20 and 24 pounds? (1)
Solution: If g  0, 2  0.1g  2 and half the range is 1, the distribution goes from
25.5 – 1 = 24.5 to 25.5 + 1 = 26.5 pounds. Make a diagram! The box for this
14
251y0731 4/30/07
distribution will be between 24.5 and 26.5 pounds. This means that there will be no area
to shade between 20 and 24 pounds. P20  x  24   0 .
If g  9, 2  g  2.9 and half the range is 1.45, the distribution goes from
25.5 – 1.45 = 24.05 to 25.5 + 1.45 = 26.95 pounds. Make a diagram! The box for this
distribution will be between 24.05 and 26.95 pounds. This means that there will be no
area to shade between 20 and 24 pounds. P20  x  24   0 .
d) There are only 10  2g officers in your firm that are eligible to be on the Compensation
Committee. Of these half make over $500,000.00 a year.
(i) If you pick 4 officers at random, what is the probability that exactly three make over
$500,000? (2)
5!
5
C 35 C15 3!2!

Solution: If g  0, 10  2g  10 and half of this is 5. P 3 
10!
C 410
6! 4!
5 43
5
60  5  4
 3  2 1

 .2381
10  9  8  7
5040
4  3  2 1
If g  9, 10  2g  28 and half of this is 14. P3 
C 314 C114
C 428
14!
14
11! 3!

28!
22!4!
14 13 12
14
2184 14  4 122304
 3  2 1


 .2489
28  27  26  25
491400
491400
4  3  2 1
(ii) The shareholders will be very unhappy if the see that more than half of the members
of the Compensation Committee make over $500,000. If you pick 4 officers at random,
what is the probability that more than half make over $500,000? (Note that the answer is
not the same as the answer to (i).) (1)
[11]
5!
C 45 C 05
4
 !1!
Solution: If g  0, 10  2g  10 and half of this is 5. P4  
10
10!
C4
6! 4!
5 4 3 2
120
 4  3  2 1 
 .0238 . P3  P4  .2381  .0238  .2689
10  9  8  7 5040
4  3  2 1
If g  9, 10  2g  28 and half of this is 14. P4 
C 414 C 014
C 428
14!
1
10! 4!

28!
24!4!
14 13 12 11
4  3  2 1  24024  .0489
P3  P4  .2489  .0489  .2978

28  27  26  25 491400
4  3  2 1
Why does the probability rise as the group gets larger? If there are only 10 eligible
officers, the probability that the first pick yields someone with earnings above $100,000
15
251y0731 4/30/07
5 1
 . If we get someone in the high-earnings group, the chance that the second pick
10 2
4
yields one too is  .4444 . ? If there are 28 eligible officers, the probability that the first
9
14 1
 . If we get someone in the
pick yields someone with earnings above $100,000 is
28 2
13
 .4815 , so we
high-earnings group, the chance that the second pick yields one too is
27
can see that the larger the pool of high earnings people, the more slowly it will get
depleted.
is
e) Only 65% of the many workers in a high security scientific installation seem to be capable of
remembering to carry their badges. If they are among the 35% that forget they will be sent home
to get them!
(i) If the installation opens at 6am, what is the probability that the second employee to
arrive is the first one sent home? (1)
Geometric: P2  .65.35   .2275
(ii) On the average, if we number the employees by their arrival order, what is the
average number of the first employee to be sent home? (The third? The 4 th? The 3.74th?)
1
1
 2.857
(1)
Geometric:   
p .35
(iii) If a group of 6  g  employees arrive at 9am, what is the probability that at least one
will not be sent home? (1)
You need 1  Pall   1  .65n . If the probability of forgetting is .65, the
probability of all forgetting is Pall   .65n .
If n  6 , 1  .65 6  1  .075419  .9246 . If we want to use the binomial table
with n  6 and p  .35 , what we want is the probability that at least one remembers.
This is 1  P0  1  .07542  .9246
If n  15 , 1  .6515  1  .001562  .9984 . You could also get this by taking
1  P0 using the Binomial table when n  15 and p  .35 .
(iv) In the same group, what is the probability that at least one will be sent home? (1)
Binomial: n is between 6 and 15.
You need 1  Pall   1  .35n . If the probability of remembering is .35, the
probability of all remembering is Pall   .35n .
If n  6 , 1  .35 6  1  .001838  .9982 .Note that if there are 6 individuals, at least one
will be sent home if 5 or fewer remember. According to the binomial table with n  6
and p  .35 , Px  5  .99816 .
1  .3515  1  .00000  1.0000 . Note that you could also get this by taking
Px  14  using the Binomial table when n  15 and p  .35 .
(v) In the same group, what is the probability that more than half will not be sent home?
(1)
For a group of 6, the probability that any given person is sent home is .35 and a
majority is 4. Px  4  1  Px  3  1  .80015  .19985 according to the binomial table
with n  6 and p  .35 .
For a group of 15, the probability that any given person is sent home is .35 and a
majority is 8. Px  8  1  Px  7  1  .88677  .11323 using the Binomial table when
n  15 and p  .35
16
251y0731 4/30/07
(vi) (Extra Credit) By late morning, the employees are trickling is at only 12 per hour.
The guard decides that if the probability of someone arriving in the next 5 minutes after
you come in is less than 50%, he will hide behind the warning sign and take a few nips
from his bottle. He asks you whether the probability is less than 50%. Do not use the
Poisson distribution for this problem. (2)
Solution: For the exponential distribution, the probability that the waiting time is less
1
than or equal to x is F x  1  ecx when x  0 and the mean time to a success is . If
c
1
12 people arrive in an hour, the average time to a success is 5 minutes. If  5, c  .2
c
.25
1
and F 5  1  e
 1  e  1  .3679  .6321. No drink!
2. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that the
jorcillator only works as long as both components work (so that it fails in the first month if either
component fails). Note change.
The probability of the phillinx failing is given by a Normal distribution with   2  .1h and
  0.5  .01h , where h is the second-to-last digit of your student number. For example, if the life of
the phillinx is represented by x1 , the chance of the phillinx failing in the 3rd month is P2  x1  3
and the probability of it failing after the third month is Px1  3 . For example: Ima Badrisk has the
number 375292, so her distribution has a mean of   2  .1h  2.9 and a standard deviation of
  0.5  .019  0.59 . Your mean will be between 2.0 and 2.9 and your standard deviation will be
between 0.50 and 0.59.
The probability of the flubberall failing is: in the first month 40%; in the second month 30%; in the
third month 20% and in the last month 10%. We will divide time into four periods with the last period
being ‘beyond the third month.’ The jorcillator is guaranteed to fail in one of the four periods. Failure
of components is assumed to be independent, so if the probability of the phillinx failing in the first
month is .1, and the probability of the flubberall failing in the first month is .4, the probability of both
components failing in the first month is (.1) (.4) =.04 (This is not the necessarily the probability that
the jorcillator will fail in the first month!)
In order to maintain my sanity, use the following events.
Failure of the phillinx in period 1, 2, 3, 4 are events A1, A2 , A3 , and A4 .
Failure of the flubberall in period 1, 2, 3, 4 are events B1, B2 , B3 , and B4 .
Failure of the jorcillator in period 1, 2, 3, 4 are events C1, C2 , C3 , and C 4 .
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.
17
251y0731 4/30/07
a) What is the probability that the phillinx will fail in month 1? Month 2? Month 3? After Month
3? (2)
We have   2  .1h ,   .5  .01h
1 2 
02
z
If h  0 , P0  x  1  P
  P 4.00  z  2.00 
.
5
.5 

 P4.00  z  0  P2.00  z  0 = .5000 - .4772  .0228  P A1 
22
 1 2
P1  x  2  P
z
  P 2.00  z  0 = .4772  P A2 
.5 
 .5
3 2 
 22
P2  x  3  P
z
  P0  z  2.00  = .4772  P A3 
.5 
 .5
3 2

P  x  3  P  z 
 Pz  2   Pz  0  P0  z  2.00  =.5 - .4772 = .0228  P A4 
.5 

Check: .0228 + .4772 + .4772 + .0228 = 1.0000
1  2.5 
 0  2.5
z
If h  5 , P0  x  1  P
  P 4.54  z  2.72 
.55 
 .55
 P4.54  z  0  P2.72  z  0 = .5000 - .4967 = .0033  P A1 
2  2.5 
 1  2.5
P1  x  2  P
z
  P 2.73  z  0.91
.
55
.55 

 P2.72  z  0  P0.91  z  0 = .4967 - .3186 = .1781  P A2 
3  2.5 
 2  2.5
P2  x  3  P
z
  P 0.91  z  0.91
.55 
 .55
 P0.91  z  0  P0  z  0.91  2.3186  = .6372  P A3 
3  2.5 

Px  3  P  z 
 Pz  .91  Pz  0  P0  z  0.91 = .5 - .3186 = .1814  P A4 
.55 

Check: .0033 + .1781 + .6372 + .1814 = 1.0000
1  2.9 
 0  2.9
x
If h  9 , P0  x  1  P
  P 4.91  z  3.22 
.59 
 .59
 P4.91  z  0  P3.22  z  0 = .5000 - .4994 = .0006  P A1 
2  2.9 
 1  2.9
P1  x  2  P
x
  P 3.22  z  1.53 
.59 
 .59
 P3.22  z  0  P1.53  z  0 = .4994 - .4370 = .0624  P A2 
3  2.9 
 2  2.9
P2  x  3  P
x
  P 1.53  z  0.17 
.59 
 .59
 P1.53  z  0  P0  z  0.17  = .4370 + .0675 = .5045  P A3 
3  2.9 

Px  3  P  z 
 Pz  0.17   Pz  0  P0  z  0.17   .5  .0625  P A4 
.59 

Check: .0006 + .0624 + .5045 + .4325 = 1.0000
b) What is the probability that the jorcillator will fail in the first month? (2)
c) What is the probability that the jorcillator will fail in the second month? (1)
d) What is the probability that the jorcillator will fail in the third month? (1)
18
251y0731 4/30/07
e) What is the probability that the jorcillator will last beyond 3 months? (1)
Assume that h  0 . The solution for parts b) – e) follows.
For the Flubberall PB1   .4 , PB2   .3 , PB3   .2 and PB3   .1
For the Phillinx, if h  0 we have found P A1   0228, P A2   .4772, P A3   .4772 and
P A4   .0228
The way this was done in the last exam was as below. Remember that our new failure rule says whichever
component fails first downs the machine.
Joint Event
Probability
When Fails
Period 1
.0228  .4 = .00912
A1 B1
A1
A1
A1
A2
A2
A2
A2
A3
A3
A3
A3
A4
A4
A4
A4
.0228  .3 = .00684
.0228  .2 = .00456
.0228  .1 = .00228
.4772  .4 = .19088
.4772  .3 = .14316
.4772  .2 = .09544
.4772  .1 = .04772
.4772  .4 = .19088
.4772  .3 = .14316
.4772  .2 = .09544
.4772  .1 = .04772
.0228  .4 = .00912
.0228  .3 = .00684
.0228  .2 = .00456
.0228  .1 = .00228
B2
B3
B4
B1
B2
B3
B4
B1
B2
B3
B4
B1
B2
B3
B4
Period 1
Period 1
Period 1
Period 1
Period 2
Period 2
Period 2
Period 1
Period 2
Period 3
Period 3
Period 1
Period 2
Period 3
Period 4
I had suggested using a joint probability table for the joint probabilities and a similar table for failure times.
This sure saves space. You can also see from the failure time table that the probability of failure in the first
period is P A1  B1  . However, if you are going to try something similar for the next period, note that you
have to exclude failure in the first period. The result is a nightmare and some difficult logic.
P  A2  B2   A1  B1  P  A2  B2   A1  B1  P  A2   A1  B1  B2  A1  B1



 
 







 
 P A2  B1  A1  B2  P A2  B1   P A1  B2  P A2  B2 
 .30000  .00684   .4772  .19088   .14316  .43632 .
Here is the joint probability table and the table relating joint events to failure times.
A1
A2
A3
A4
B1
B2
B3
B4
.00912
.19088
.19088
.00912
.40000
.00684
.14316
.14316
.00684
.30000
.00456
.09544
.09544
.00456
.20000
.00228
.04772
.04772
.00228
.10000
B1
.0228
.4772
.4772
.0228
1.0000
A1
A2
A3
A4
1
1
1
1
B2
B3
B4
1
2
2
2
1
2
3
3
1
2
3
4


b) What is the probability that the jorcillator will fail in the first month? (2)
PC1   P A1  B1    A1  B2    A1  B3    A1  B4    A2  B1    A3  B1    A4  B1 
19
251y0731 4/30/07
 P A1  B1  = .02280 + .40000 - .00912 = .41368
c) What is the probability that the jorcillator will fail in the second month? (1)
PC 2   P A2  B2    A2  B3    A2  B4    A3  B2    A4  B2  = .14316 + .09544
+ .04772 + .14316 + .00684 = .43632
d) What is the probability that the jorcillator will fail in the third month? (1)
PC 3   P A3  B3    A3  B4    A4  B3   .09544  .04772  .00456  .14772
e) What is the probability that the jorcillator will last beyond 3 months? (1)
PC 4   P A4  B4   .00228 Note that these probabilities add to 1.
Assume that h  0 . The solution for parts f) – h) follows.
f) (Extra Credit) Find the probability that the jorcillator and the Phillinx both fail in the third
month (1)
The problem asks for PC 3  A3  . C 3 is the union of three mutually exclusive events, two of
which involve A3 . So PC 3  A3   P A3  B3    A3  B4   .09544  .04772  .14316 . Note
that because C 3 depends on A3 , PC 3  A3   PC3 P A
g) (Extra Credit) Find the probability that the phillinx fails in the third month, given that the
jorcillator fails in the 3rd month i.e. P A3 C3 (1)

PA3 C3  
P A3  C 3  .14316

 .96913
PC 3 
.14772

h) Demonstrate Bayes’ rule by finding showing how to get the probability that the jorcillator fails
in the third month, given that the phillinx fails in the third month, two ways. The first would be
directly by working with joint probabilities that both the jorcillator and the phillinx fail. The
second would be by using Bayes’ rule and your result in g). (1)
P A3  C 3  .14316

 .3
P C3 A3 
P A3 
..4772


PC3 A3  
PA3 C3 PC3 
P A3 

.96913 .14772 
 .3
..4772
Assume that h  5 . The solution for parts b) – e) follows.
For the Flubberall PB1   .4 , PB2   .3 , PB3   .2 and PB3   .1
For the Phillinx, if h  5 we have found that P A1   .0033, P A2   .1781, P A3   .6372 and
P A4   .1814.
Here is the joint probability table and the table relating joint events to failure times.
B1
B2
B3
B4
B1 B 2 B3 B 4
A1
A2
A3
A4
.00132
.07124
.25488
.07256
.40000
.00099
.05343
.19116
.05442
.30000
.00666
.03562
.12744
.03628
.20000
.00033
.01781
.06372
.01814
.10000
.0033
.1781
.6372
.1814
1.0000
A1
A2
A3
A4
1
1
1
1
1
2
2
2
1
2
3
3
1
2
3
4
b) What is the probability that the jorcillator will fail in the first month? (2)
20
251y0731 4/30/07
PC1   P A1  B1    A1  B2    A1  B3    A1  B4    A2  B1    A3  B1    A4  B1 
 P A1  B1  = .00330 + .40000 - .00132 = .40198.
c) What is the probability that the jorcillator will fail in the second month? (1)
PC 2   P A2  B2    A2  B3    A2  B4    A3  B2    A4  B2  = .05343 + .03562
+ .01781 + .19116 + .05442 = .35244.
d) What is the probability that the jorcillator will fail in the third month? (1)
PC 3   P A3  B3    A3  B4    A4  B3   .12744  .06372  .03628  .22744 .
e) What is the probability that the jorcillator will last beyond 3 months? (1)
PC 4   P A4  B4   .01814 Note that these probabilities add to 1.
Assume that h  9 . The solution for parts b) – e) follows.
For the Flubberall PB1   .4 , PB2   .3 , PB3   .2 and PB3   .1
For the Phillinx, if h  9 we have found that For the Phillinx P A1   .0006, P A2   .0624,
P A3   .5045 and P A4   .4325.
Here is the joint probability table and the table relating joint events to failure times.
B1
B2
B3
B4
B1 B 2 B3 B 4
A1
A2
A3
A4
A1
.00024 .00018 .00012 .00006
.0006
1
1
1
1
A2
.02496 .01872 .01248 .00624
.0624
1
2
2
2
A3
.20180 .15135 .10090 .05045
.5045
1
2
3
3
A4
.17300 .12975 .08650 .04325
.4325
1
2
3
4
.40000 .30000 .20000 .10000 1.0000
b) What is the probability that the jorcillator will fail in the first month? (2)
PC1   P A1  B1    A1  B2    A1  B3    A1  B4    A2  B1    A3  B1    A4  B1 
 P A1  B1  = .0006 + .40000 - .00024 = .40036
c) What is the probability that the jorcillator will fail in the second month? (1)
PC 2   P A2  B2    A2  B3    A2  B4    A3  B2    A4  B2  = .01872 + .01248 +
.00624 + .15135 + .12975 = .31854
d) What is the probability that the jorcillator will fail in the third month? (1)
PC 3   P A3  B3    A3  B4    A4  B3   .10090  .05045  .08650  .23785
e) What is the probability that the jorcillator will last beyond 3 months? (1)
PC 4   P A4  B4   .04325 Note that these probabilities add to 1.
Assume that h  9 . The solution for parts f) – h) follows.
f) (Extra Credit) Find the probability that the jorcillator and the Phillinx both fail in the third
month (1)
The problem asks for PC 3  A3  . C 3 is the union of three mutually exclusive events, two of
which involve A3 . So PC 3  A3   P A3  B3    A3  B4   .10090  .05045  .15135
g) (Extra Credit) Find the probability that the phillinx fails in the third month, given that the
jorcillator fails in the 3rd month i.e. P A3 C3 (1)


21
251y0731 4/30/07
PA3 C3  
P A3  C 3  .15135

 .63633
PC 3 
.23785
h) Demonstrate Bayes’ rule by finding showing how to get the probability that the jorcillator fails
in the third month, given that the phillinx fails in the third month, two ways. The first would be
directly by working with joint probabilities that both the jorcillator and the phillinx fail. The
second would be by using Bayes’ rule and your result in g). (1)
P A3  C 3  .15135

 .3
P C3 A3 
P A3 
.5045


PC3 A3  
PA3 C3 PC3 
P A3 

.63633 .23785 
 .3
.5045
3. (Extra Credit) Do the following Binomial problems. If you substitute another distribution for the
Binomial, justify the substitution by showing that a substitution condition is satisfied. (6)
All problems below are done first using a continuity correction. It is obviously necessary in a).
a) 45% of our many management employees have MBAs. If we select 150 at random, what is the
probability that exactly 72 have college degrees?
We need the binomial probability of 72 successes when the expected number of successes is
  np  150 .45   67.5 , the expected number of failures is 150 – 67.5 = 82.5 and
  npq 
67 .5.55  
37 .125  6.09303 . We cannot use the Poisson distribution here
because the probability is large. Since the expected numbers of successes and failures are both
above 5, we can use the Normal distribution.
72.5  67.5 
 71.5  67.5
PB 72   PN 71 .5  x  72 .5  P 
z
 P0.66  z  0.82 
6
.
09303
6.09303 

150!
150 72 78
 .2939  .2454  .0485 . This can also be done as C72
.45 72 .55 78 , but this
p q 
78!72!
looks like much too much work.
b) 45% of our many management employees have MBAs. If we select 150 at random, what is the
probability that more than half have college degrees?
75 .5  67 .5 

PB x  76   PN x  75 .5  P  z 
 Pz  1.31  .5  .4049  .0951
6.09303 

Without the continuity correction
76  67 .5 

PB x  76   PN x  76   P  z 
 Pz  1.40   .5  .4192  .0808
6.09303 

Or if p 
x
, E  p   p  .45 and  p 
n
pq
.45.55 

 .00165  .04062
n
150
.5  .45 

PN  p  .5  P  z 
 Pz  1.23   .5  .3907  .1093
0.04062 

c) 5% of our many management employees have MBAs. If we select 90 at random, what is the
probability that more than 6 have MBAs?
Note that the expected number of successes,   np  90.05   4.5 , is below 5 so that we should
not use the Normal distribution. However, the criterion for use of the Poisson distribution
n
90
 1800 . If we use the Poisson table with a parameter of 4.5
 500 is met since
.05
p
Px  6  1  Px  6  1  .83105  .16895 .
22