251solngr3-08b 6/25/08
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Graded Assignment 3
Name: Key

 x  16  for the following distributions (Use tables in d, e, f, h and j.):
a. Continuous Uniform with c  1, d  14 (Make a diagram!).
b. Continuous Uniform with c  10 , d  25 (Make a diagram!).
c. Continuous Uniform with c  1, d  10 (Make a diagram!).
d. Binomial Distribution with p  .35, n  25 .
e. Binomial Distribution with p  .65, n  25 .
f. Binomial distribution with p  .03, n  20 (Approximate Solution)
g. Geometric Distribution with p  .35.
1. (Source 03) Find P 10
h. Poisson Distribution with parameter of 9.
i. Show how you would do this for a Hypergeometric Distribution with
p  .35, n  25 ,
N  100 . Remember M  Np .
j. Hypergeometric Distribution with
p  .35, n  25 , N  1000 . (Approximate Solution)
p  .35, n  25 , N  100 , find Px  1 . (Extra credit if you actually
k. For a Hypergeometric Distribution with
evaluate this.)
l. (Extra credit) Find P 10

 x  16  for an exponential distribution with c  .01 .
2. Find the Mean and Standard deviation for the following distributions.
a. Continuous Uniform Distribution with c  10 , d  25 .
p  .65, n  25 .
c. Geometric Distribution with p  .35
b. Binomial Distribution with
d. Poisson Distribution with parameter of 9.
e. Hypergeometric Distribution with p  .35,
n  25 , N  100 .
f. Compare means and standard deviations for a Binomial distribution with
Hypergeometric Distribution with
p  .35, n  25 , and a
p  .35, n  25 , N  1000 .
c  .01
g. (Extra credit) Exponential distribution with
3. Identify the distribution and do the following problems. Use tables where possible.
a. This problem uses all four discrete distributions.
(i) An IRS auditor is about to audit 6 returns from a group of returns for improper deductions.
Assume that these returns come from a group of 80 returns in which 25% have improper deductions, what is the
probability that exactly one of the 6 has improper deductions?
(ii) What is the probability that at least one of the six has improper deductions?
(iii) If two or more of the 6 returns have improper deductions, the entire group of 80 will be audited. What is the chance
that that happens?
(iv) Assume that the 6 returns come from a large group of returns (more than 120) that is 25% improper, and if two or
more are improper, the whole group will be audited, what is the probability that the whole group will be audited?
(Answers that use a specific population size will not be counted.)
(v) Assume that a large number of returns are audited from this group that is 25% improper, what is the chance that the
second return is the first one found with improper deductions?
(vi) What is the chance that the first improper return will be among the first ten audited?
(vii) Assume that only 2% of the returns are improper and a sample of 100 taxpayers is taken, what is the chance that at
least 3 have improper returns?
b. A fast food hamburger must be cooked a minimum of 90 seconds to be safe. At my lunch spot the time that burgers are
on can be represented by a continuous uniform distribution with limits of 80 and 120 seconds. (i) What is the chance that
a hamburger will be safe? (ii) What is the chance that, if I order 2 hamburgers both will be safe? (iii) What is the chance
that if I order 3 hamburgers, at least one will be unsafe?
c. The number of students who come to my office has a Poisson distribution with a parameter (mean) of 2.5 a day. (i)
What is the chance that no students come in in a day? (ii) In a 5-day week? (iii) What is the chance that more than 5 come
in in a day (iv) in a 5-day week?
d. Complaints come in to my website at an average of seven an hour and take an hour to adjust. How many complaint
handlers do I need to keep the probability of a complainer having to wait below 1%? (Hint: Look at a Poisson table with a
mean of 7. What is the probability that x is above 10?)
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e. If 80% of shoppers at a website abandon their shopping carts before quitting and there are currently 20 shoppers on my
site, what is the chance that at least one will buy something:? What is the average number that will buy something? What
is the most likely number that will buy something?
f. According to Bowerman and O’Connell, in the movie Coma a young nurse finds that 10 of the last 30000 patients of a
Boston Hospital went into a coma during anesthesia. Upon investigation she finds that nationally 6 out of 100000 patients
go into comas after anesthesia. Assume that the national probability is correct and that the Binomial distribution is
appropriate, what is the mean number of patients out of 30000 that will go into a coma? In a hypothesis test, a researcher
finds the probability of getting a number as large as or larger than what actually happened. So, find the probability of 10 or
more people out of 30000 going into a coma. Obviously, we don’t have a binomial table that will do this. Show that this is
an appropriate place to use a Poisson distribution and use the Poisson tables to find the probability that x  10 . On the
basis of your result; do you think that something strange is going on? (The answer is shorter than the problem.)
g. (Extra credit) A telemarketer finds that the length of a call has an exponential distribution with a mean of
1 2
c
minutes. Find the probability that the length of a call will be (i) No more than 3 minutes, (ii) Between 1 and 2 minutes,
(iii) More than 5 minutes.
h. A coin is tossed 6 times. Define the following events.
A : Exactly one head
B : Exactly two heads
C : At least one head
(i) Find P  A - Use a table in all three parts of this problem. (ii) Find P B  . (iii) Find P C 
Solution to Question 1
1. Find P10  x  16  for the following distributions (Use tables in d, e, f, h j.):
a. Continuous Uniform with c  1, d  14 (Make a diagram!).
1
1
1


 .07692 . In the diagram below, shade the area between 10 and 14. (There is
d  c 14  1 13
no area between 14 and 16.) The horizontal line has a height of 113 .
0
1
10
14 16
14  10
 .3077 .
14  1
Another way to do a problem of this type is to remember that for any continuous distribution, we
can use differences between cumulative distributions, Pa  x  b  F b  F a  where the
xc
cumulative distribution is F x0   Px  x0  and F x  
for c  x  d ,
d c
F x   0 for x  c and F x   1 for x  d . If we use the cumulative distribution method,
So P10  x  16   P10  x  14  
remember F x   1 for x  d . So P10  x  16   F 16   F 10   1 
10  1
9
 1   .3077 .
14  1
13
b. Continuous Uniform with c  10 , d  25 (Make a diagram!).
1
1
1


 .066667 . In the diagram below, shade the area between 10 and 16.
d  c 25  10 15
The horizontal line has a height of 115 .
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10
16
25
16  10
 .4000 . If we use the cumulative distribution method,
So P10  x  16  
25  10
remember F x   1 for x  d . So P10  x  16   F 16   F 10 

16  10 10  10
6


 .4000 .
25  10 25  10 15
c. Continuous Uniform with c  1, d  10 (Make a diagram!). Find P10  x  16  .
1
1
1

 . In the diagram below, try to shade the area between 10 and 16. (Surprise!
d  c 10  1 9
There is no area between 10 and 16.) The probability is thus zero.
0 1
10
16
So P10  x  16   F 16   F 10   1  1  0 .
d. Binomial Distribution P10  x  16  with p  .35, n  25 .
P10  x  16   Px  16   Px  9  .99921  .63031  .3689
e. Binomial Distribution P10  x  16  with p  .65, n  25.
P10  x  16  can't be done directly with tables that stop at p  .5 , so try to do it with
failures. (The probability of failure is 1 - .65 = .35.) 10 successes correspond to 25 – 10 = 15
failures out of 25 tries. 16 successes correspond to 9 failures. So try 9 to 14 successes when
p  .35 and n  25 . P9  x  15   Px  15   Px  8  .99706  .46682  .53078
f. Binomial distribution P10  x  16  with p  .03, n  20 (Approximate Solution)
You do not have a Binomial table for p  .03 . However since
n 20

 666 .67  500 , you can
p .03
use the Poisson distribution for this problem. The mean is m  np  20.03   0.6 . If you use the
0.6 part of the Poisson table, you get P10  x  16   Px  16   Px  9  1  1  0
g. Geometric Distribution P10  x  16  with p  .35.
Remember that F c  Px  c  1  q c , because success at try c or earlier implies
that there cannot have been failures on the first c tries. q  1  p  1  .1  .9

P10  x  16  Px  16  Px  9  F 16  F 9  1  .6516  1  .659
 .65  .65
9
16

 .02071  .00102  .0197
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h. P10  x  16  for Poisson Distribution with parameter of 9. (Parameter of 9 means
9  m    2)
P10  x  16   Px  16   Px  9  .98889  .58741  .40148
i. Show how you would do this for a Hypergeometric Distribution with p  .35, n  25 ,
N  100 . Remember M  Np .
We have no cumulative tables or cumulative distribution formula for this distribution, so the only
method is to add together probabilities over the range 10 to 16.
65
C 35C 25
C M C N M
x
Px   x Nn x and M  Np  100 .35   35 so Px   x 100
Cn
C 25
and P10  x  16  
65 
 C x35C 25
1
x 
 100
100

C 25
x 10
 C 25
16
 
 C
16
45 65
x C 25 x

x 10
j. P10  x  16  for a Hypergeometric Distribution with p  .35, n  25 , N  1000 .
(Approximate Solution). Since 20 n  2025   500  1000 , we can use the Binomial distribution
with p  .35, n  25 . P10  x  16   Px  16   Px  9  .99921  .63031  .3689
k. For a Hypergeometric Distribution with p  .35, n  25 , N  100 , find Px  1 . (Extra credit
if you actually evaluate this.) Px  1  1  Px  0 , M  Np  100 .35   35 and
Px  
C xM C nNxM
C nN
. So 1  P0  1 
65
C 035 C 25
100
C 25
35!  65! 


35!0!  40! 25! 
 1
100!
75! 25!
65  64  62  61  60  59  58  57  56  55  54  53  52  51  50  49  48  47  46  45  44  43  42  41
100  99  98  97  96  95  94  93  92  91  90  89  88  87  86  85  84  83  82  81  80  79  78  77  76
I get 1  .018512  .981488 . This is pretty likely to be right, since it’s close to the Binomial
probability.
 1
l. (Extra credit) Find P10  x  16  for an exponential distribution with c  .01 .
In ‘Great Distributions I Have Known, we have the following information.
x is usually the
Exponential
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 .
Since this is a continuous distribution, P10  x  16   F 16   F 10 



 1  e 0.16  1  e .10  e .10  e .16  .90484  .85214  .0527
Note: F 10   .09516 and F 16   .14786 .
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Solution to Question 2
2. Find the Mean and Standard deviation for the following distributions.
Some of these averages could be expressed in words – try it! (For example, 'The average number of
successes when …… is ……')
a. Continuous Uniform Distribution with c  10 , d  25 .

d  c  25  10   225  18.75
c  d 10  25

 17 .5 ,  2 
2
2
12
12
12
2
2
So   18.75  4.3301
b. Binomial Distribution with p  .65, n  25 .
(Remember that q  1  p and that both p and q must be between 0 and 1.)
q  1  p  1  .65  .35,   np  25.65   16 .25 ,  2  npq  16.25 .35  5.6875
so   npq  5.6875  2.3848 . The average number of successes in 25 tries,
when the probability of a success in an individual try is 65% is 16.25.
c. Geometric Distribution with p  .35
If q  1  p  1  .35  .65,  
1
1
q
.65
.65

 2.8571 ,  2  2 

 5.3061
2
.
1225
p .35
p
.35
so   5.3061  2.3035 . If we play a game repeatedly and the probability of a success
in an individual try is 35%, on the average our first success will occur on the 2nd or 3rd
try.
d. Poisson Distribution with parameter of 9.
  m  9 ,  2  m  9 , so   m  9  3 .
e. Hypergeometric Distribution with p  .35, n  25 , N  100 .
M
and q  1  p then   np  25.35   8.75 and
N
N n
100  25
75
8.75 .65   .757576 5.6875   4.3087 .

npq 
25.35 .65  
N 1
100  1
99
If p 
2
So   4.3087  2.0757 . If we take a sample of 25 from a population of 100 of which
35% are successes, on the average we will get 8.75 successes.
f. Compare means and standard deviations for a Binomial distribution with p  .35, n  25 and a
Hypergeometric Distribution with p  .35, n  25 and N  1000 .
If p 
M
and q  1  p  1  .35  .65 , then   np  25.35   8.75 for both
N
distributions. For the binomial distribution  2  npq  25.35.65  5.6875
so that   5.6875  2.3848 . For the Hypergeometric distribution,
N n
1000  25
975
5.6875 
2 
npq 
25.35 .65  
N 1
1000  1
999
 0.97598 5.6875   5.5509 . So   0.97598 5.6875
 .987915 2.3848   2.3560 . There is a difference of less than 2% between these two.
Note that the finite population correction factor in the variance formula has
relatively little effect this time, which is why we can justify using the binomial
distribution in place of the hypergeometric distribution here.
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g. (Extra credit) Exponential distribution with c  .01 ,
1
1
1
1
We have   
 100 and   
 100 .
c .01
c .01
Solution to Question 3
3. Identify the distribution and do the following problems.
Remember:
(i) If you are looking for numbers of successes when the number of tries is given and the
probability of success is constant, you want the Binomial distribution.
(ii) If you are looking for the try on which the first success occurs out of many possible
tries when the probability of success is constant, you want the Geometric
distribution.
(iii) If you are looking for numbers of successes when the number of tries is given and
the average number of successes per unit time or space is given, you want the
Poisson distribution.
(iv) If you are looking for numbers of successes when the number of tries is given and
the probability of success is not constant because the total number of successes in the
population is limited, you want the Hypergeometric distribution.
a. This problem uses all four discrete distributions.
(i) An IRS auditor is about to audit 6 returns from a group of returns for improper deductions.
Assume that these returns come from a group of 80 returns in which 25% have improper
deductions, what is the probability that exactly one of the 6 has improper deductions?
20 out of 80 returns are improper. This is Hypergeometric because we want to know the
probability of a number of ‘successes, but the total number in the population is limited.
C M C N M
N  80, n  6, M  Np  80.25   20 . Px   x nn x and
CN
60!
60  59  58  57  56
20
20  60  59  58  57  56  6
55!5! 
5  4  3  2 1
P1 


 .3635 . If we
80
80!
80  79  78  77  76  75
80  79  78  77  76  75
C6
74!6!
6  5  4  3  2 1
wrongly used the Binomial distribution here, we would have found Px  1 when n  6 and
C120 C 560
20
p  .25. The binomial formula says Px   C xn p x q n x , so if p  .25, Px  1  C16 p1q 5
 6.25 .75 5  .3560 , but we would be better off using a binomial table with n  6 and p  .25.
P1  Px  1  Px  0  .53394  .17798  .3560 .
(ii) What is the probability that at least one of the six has improper deductions?
Hypergeometric:
60!
60  59  58  57  56  55
1
C 020 C 660
54
!
6
!
6  5  4  3  2 1
Px  1  1  Px  0  1 
 1
 1
80
80
!
80

79
 78  77  76  75
C6
74!6!
6  5  4  3  2 1
60  59  58  57  56  55
 1
 1  .16660  .8333 . This can be argued more directly if we realize
80  79  78  77  76  75
that on the first try, the probability of not getting an improper return is 60 80 , on the next try the
probability becomes
59
79
and so on, so that on the last try the probability is
55
75
. Since these are
60  59  58  57  56  55
conditional probabilities, we can multiply them together to get P0  
80  79  78  77  76  75
 .16660 . If we had wrongly used the Binomial distribution, we would have found Px  1 when
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n  6 and p  .25. If p  .25, Px  1  1  P0  1  C 06 p 0 q 6  1 .75 6  .8220 , but we
would have been better off using a binomial table with n  6 and p  .25.
Px  1  1  Px  0  1  .17798  .8220 .
(iii) If two or more of the 6 returns have improper deductions, the entire group of 80 will be
audited. What is the chance that that happens?
Hypergeometric: Px  2  1  Px  0  Px  1
 1  .16660  .3635  .8333  .3635  .4698
(iv) Assume that the 6 returns come from a large group of returns (more than 120) that is 25%
improper, and if two or more are improper, the whole group will be audited, what is the
probability that the whole group will be audited? (Answers that use a specific population size will
not be counted.)
Binomial: Px  2 when n  6 and p  .25. If we use the binomial table with n  6
and p  .25, Px  2  1  Px  1  1  .53394  .4661 .
(v) Assume that a large number of returns are audited from this group that is 25% improper,
what is the chance that the second return is the first one found with improper deductions?
Geometric: Px   q x1 p . p  .25, q  1  p  .75 , Px  2  .75 .25   .1875
(vi) What is the chance that the first improper return will be among the first ten audited?
Geometric: F c  Px  c  1  q c , so F 10  Px  10  1  .7510  1  .0531  .9437 .
(vii) Assume that only 2% of the returns are improper and a sample of 100 taxpayers is taken,
what is the chance that at least 3 have improper returns?
Binomial: Px  3 when n  100 and p  .02. But we do not have the appropriate
table. Unless we want to do the problem by hand, call on the Poisson distribution, since
n  100
p
.02  5000, which is safely over 500. Our parameter is m  np  100 .02   2.
Px  3  1  Px  2  1  .67668  .3233 .
b. A fast food hamburger must be cooked a minimum of 90 seconds to be safe. At my lunch spot the time
that burgers are on can be represented by a continuous uniform distribution with limits of 80 and 120
seconds.
(i) What is the chance that a hamburger will be safe?
x c
Continuous Uniform: The cumulative distribution is F x0   Px  x0   0
d c
90  80
10
 1
 .75
for c  x  d , c  80 and d  120 . Px  90   1  Px  90   1 
120  80
40
(ii) What is the chance that, if I order 2 hamburgers both will be safe?
Binomial: Px  2 when n  2 and p  .75. The binomial formula says
Px   C xn p x q n x , so if p  .75, P2  C 22 p 2  .75 2  .5625 . . It’s probably not worthwhile
to use the binomial table, since we don’t have one for p  .75. But, if the probability of success is
.75, the probability of failure is .25 and 2 successes is no failures. Sure enough, for n  2 and
p  .25, P0  .5625 .
(iii) What is the chance that if I order 3 hamburgers, at least one will be unsafe?
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Binomial: Px  1 when n  3 and p  .25. Note that if the probability of a safe
hamburger is .75, the probability of an unsafe one is .25. The binomial formula says
Px   C xn p x q n x , so if p  .25, Px  1  1  P0  1  C 03 p o q 3  1  .75 3
 1  .421875  .5781 . It’s probably not worthwhile to use the Binomial table, since we don’t
have one for p  .75. But, if the probability of success (a safe burger) is .75, the probability of
failure is .25 and we want 1 or more failures. So we have the same thing from the
table
Px  1  1  P0  1  .42187  .5781 .
c. The number of students who come to my office has a Poisson distribution with a parameter (mean) of 2.5
a day.
(i) What is the chance that no students come in during a day?
From the Poisson table with m  2.5 , P0  .08208 .
(ii) What is the chance that no students come in during a 5-day week?
From the Poisson table with m  52.5  12 .5 , P0  .000004 .
(iii) What is the chance that more than 5 come in during a day
From the Poisson table with m  2.5 , Px  5  1  Px  5  1  .95798  .0420 .
(iv) What is the chance that more than 5 come in during a 5-day week?
From the Poisson table with m  12 .5 , Px  5  1  Px  5  1  .01482  .09852 .
d. Complaints come in to my website at an average of seven an hour and take an hour to adjust. How many
complaint handlers do I need to keep the probability of a complainer having to wait below 1%?
Poisson. If we look at the table with parameter of 7, the first cumulative probability above 99% is
Px  14   .99428 . We need to hire 14 people.
e. If 80% of shoppers at a website abandon their shopping carts before buying and there are currently 20
shoppers on my site,
(i) What is the chance that at least one will buy something?
Binomial: p  .20 , n  20 . Px  1  1  P0  1  .01153  .9885
(ii) What is the average number that will buy something?
Binomial: p  .20 , n  20 .   np  20.2  4.
(iii) What is the most likely number that will buy something?
P0  Px  0  .01153
P1  Px  1  Px  0  .06918  .01153  .05765
P2  Px  2  Px  1  .20608  .06918  .13690
P3  Px  3  Px  2  .41145  .20608  .20537
P4  Px  4  Px  3  .62965  .41145  .21815
P5  Px  5  Px  4  .80421  .62965  .17456
If we check the rest of the table we find that no difference between subsequent values of
the cumulative distribution is more than .15, so 4 seems to be the mode. Since the
Binomial distribution is not usually symmetrical for low values of n, we do have to
check individual probabilities. We should be able to do this by calculus since
Px   C xn p x q n  x  C x20 .2 x .8n x should be differentiable, and the derivative could be
set equal to zero, but I can’t remember how to differentiate C xn . One of my manuals says
that the mode is any integer value or values between   q and   p . This seems to
work when the mean is not an integer.
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f. According to Bowerman and O’Connell, in the movie Coma a young nurse finds that 10 of the last 30000
patients of a Boston hospital went into a coma during anesthesia. Upon investigation she finds that
nationally 6 out of 100000 patients go into comas after anesthesia. Assume that the national probability is
correct and that the Binomial distribution is appropriate, what is the mean number of patients out of 30000
that will go into a coma? In a hypothesis test, a researcher finds the probability of getting a number as large
as or larger than what actually happened. So, find the probability of 10 or more people out of 30000 going
into a coma. Obviously, we don’t have a binomial table that will do this. Show that this is an appropriate
place to use a Poisson distribution and use the Poisson tables to find the probability that x  10 . On the
basis of your result, do you think that something strange is going on? (The answer is shorter than the
problem.)
6
This is a binomial problem. We want Px  10  when p 
and n  30000 . Note
100000
6
n
30000
 1.8 . According
that

 5 10 8 is above 500. The mean is m  np  30000
6
100000
p
100000
to the Poisson (1.8) table Px  10   1  Px  9  1  .99998  .00002 . This is an extremely small
probability and should make us suspect that something is wrong.
g. (Extra credit) A telemarketer finds that the length of a call has an exponential distribution with a mean of
1  2 minutes. Find the probability that the length of a call will be
c
(i) No more than 3 minutes,
F x  1  ecx . Since 1 c  2 , c  1 2  .5 calls per minute. If x  3, cx  .53  1.5
and Px  3  F 3  1  e 1.5  1  .22313  .7769
(ii) Between 1 and 2 minutes,
F x  1  ecx If x  1, cx  .51  0.5 , if x  2, cx  .52  1 and
P1  x  2  F 2  F 1  1  e 1  (1  e .0.5 )  e 0.5  e 1  .60653  .36788  .2387 .
(iii) More than 5 minutes.
If x  5, cx  .55  2.5 and Px  5  1  F 5  1  1  e 2.5  e 2.5  .0820.


h. A coin is tossed 6 times. Define the following events.
A : Exactly one head
B : Exactly two heads
C : At least one head
(i) Find P A - Use a table in all three parts of this problem. (ii) Find PB  . (iii) Find PC  .
This is a binomial problem with p  .5 and n  6.
(i) P1  Px  1  Px  0  .10938  .01563  .09375 or C16 p 1 q 5  6.56 
6
 .09375 .
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6!
.56  6  5 1
(ii) P2  Px  2  Px  1  .34375  .10938  .23437 or C 26 p 2 q 4 
4! 2!
2 1 64
 .234375
1
6
 .984375 .
(iii) Px  1  1  Px  0  1  .01563  .98437 or 1  C 06 p 0 q 6  1  1.5  1 
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