251solngr3-071 4/10/07
Graded Assignment 3
Name: Solution
(Open this document in 'Page Layout' view!)
Class days and time:
For full credit assignment must be neat and stapled and class day and time must be clearly indicated.
Show your work! Look at the solved problems for Section L and ‘Great Distributions’ before you
start.
1. Find P12  x  17  for the following distributions (Use tables in c, d, f and h. All probabilities should
show 4 places to the right of the decimal point):
a. Continuous Uniform with c  1, d  14 (Make a diagram!).
b. Continuous Uniform with c  15, d  25 (Make a diagram!).
c. Binomial Distribution with p  .45, n  25 .
d. Binomial Distribution with p  .85, n  25 .
e. Geometric Distribution with p  .15.
f. Poisson Distribution with parameter of 15.
g. Show how you would do this for a Hypergeometric Distribution with p  .45, n  25 ,
N  80. Remember M  Np .
h. (Extra credit) Hypergeometric Distribution with p  .45, n  25 , N  520 .
i. (Extra credit) Exponential distribution with c  .01 .
j. For the hypergeometric distribution in g) find Px  1.
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  11, d  25 .
b. Binomial Distribution with p  .45, n  25 .
c. Geometric Distribution with p  .15
d. Poisson Distribution with parameter of 15.
e. Hypergeometric Distribution with p  .45, n  25 , N  80 .
f. Hypergeometric Distribution with p  .45, n  25 , N  520 .
g. (Extra credit) Exponential distribution with c  .01 .
3. Identify the distribution and do the following problems. Use tables when 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
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251solngr3-071 4/10/07
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 a day? (ii) In a 5-day week?
(iii) What is the chance that more than 5 come 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%?
e. If 80% of shoppers at a website abandon their shopping carts without buying 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 minutes. Find the probability that the length of a call will be (i) No more than 3
c
minutes, (ii) Between 1 and 2 minutes, (iii) More than 5 minutes.
1. Find P12  x  17  for the following distributions (Use tables in c, d, f and h. All probabilities should
show 4 places to the right of the decimal point):
a. Continuous Uniform with c  1, d  14 (Make a diagram!).
1
1
1


 .07692 . In the diagram below, shade the area between 12 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
12
14 17
14  12
 .1538 .
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
for c  x  d ,
cumulative distribution is F x0   Px  x0  and F x  
d c
F x   0 for x  c and F x   1 for x  d . If we use the cumulative distribution method,
So P12  x  17   P12  x  14  
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251solngr3-071 4/10/07
remember F x   1 for x  d . So P12  x  17   F 17   F 12   1 
12  1
11
 1
14  1
13
 1  .8462  .1538 .
b. Continuous Uniform with c  15, d  25 (Make a diagram!).
1
1
1


 .1000 . In the diagram below, shade the area between 15 and 17. There is
d  c 25  15 10
no probability between 12 and 15. The horizontal line has a height of 110 .
0
12 15
17
25
17  15
So P12  x  17  
 .2000 . If we use the cumulative distribution method,
25  15
xc
remember F x   0 for x  d and F x  
for c  x  d , . So
d c
17  15
2

0 
 .2000
25  15
10
Let’s try again!!!
Find P12  x  17  when we have a Continuous Uniform distribution with c  1, d  11 (Make a
diagram!). If you make the box as above, and then shade the area between 12 and 17, you will find
P12  x  17   F 17   F 12   0  0  0.
c. Binomial Distribution with p  .45, n  25 .
P12  x  17   Px  17   Px  11  .99417  .54257  .4516 . Remember that for a discrete
distribution, the usual way to do a problem of this type is to remember that for any discrete
distribution, we can use differences between cumulative distributions,
Pa  x  b  F b  F a  1 where the cumulative distribution is F x0   Px  x0  .
d. Binomial Distribution with p  .85, n  25 .
P12  x  17  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 - .85 = .15.) 12 successes correspond to 25 – 12 = 13
failures out of 25 tries. 17 successes correspond to 8 failures. So try 8 to 13 successes when
p  .15 and n  25 . P8  x  13  Px  13  Px  7  1.00000  .97453  .0255
e. Geometric Distribution with p  .15.
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  .15  .85

P12  x  17  Px  17  Px  11  F 17  F 11  1  .8517  1  .8511
 .85  .85
11
.85  .85
11
17
17

 .16734  .06311  .1042 or


 .8511 1  .856  .167341  .37715  .1042
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251solngr3-071 4/10/07
f. Poisson Distribution with parameter of 15.
P12  x  17   Px  17   Px  11  .74886  .18475  .5641
g. Show how you would do this for a Hypergeometric Distribution with p  .45, n  25 ,
N  80. Remember M  Np .
We have no cumulative tables or cumulative distribution formula for this distribution, so the only
available method is to add together probabilities over the range 10 to 15.
C 36 C 44
C M C N M
Px   x Nn x and M  Np  80.45   36 so Px   x 8025 x
Cn
C 25
17
44 
 C x36 C 25
1
44
x 

C x36 C 25
 x . This is certainly the answer that I
80
80

C 25
x 12
 C 25 x 12
expected, but we could take this further by using the recursive formula at the end of the outline.
M  x 1  n  x 1 
The formula is Px  
 N  M  n  x  Px  1 and with n  25 , N  80 and M  36
x


and P12  x  17  
this becomes Px  
17
 


36  x  1  25  x  1 
24  13 
 19  x  Px  1 . So P13   13  32  P12  ,
x


 
22  11 
 12 
 33  P13 , P15   15  34  P14 , etc. So if we compute
 
 
36! 44!
36 44
C12
C13
24!12! 31!13!
, the rest is fairly easy. On the other hand, as you will see below,
P12  

80
80!
C 25
55!25!
even computing P0 , which is easier than computing P12  is a pain and a good demonstration of
P14  
23
14
why we have computers. The computer says that P12  x  17   Px  17   Px  11 =
0.998852 - 0.549357 = .44950
h. (Extra credit) Hypergeometric Distribution with p  .45, n  25 , N  520 .
Since N  520 is more than 20 times n  25 , we can use the binomial distribution with p  .45
and n  25 . P12  x  17   Px  17   Px  11  .99417  .54257  .4516
Note that if we actually computed this using the hypergeometric distribution, we would get
P12  x  17   Px  17   Px  11  .995126  .543372  .4518
i. (Extra credit) 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12  x  17   F 17   F 12 



 1  e 0.17  1  e .12  e .12  e .17  .88692 .84366  .0433
Note: F 17   .15634 and F 12   .11308 .
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251solngr3-071 4/10/07
j. For the hypergeometric distribution in g) find Px  1.
p  .45, n  25 , N  80. M  Np  80.45   36 . Px  1  1  P0 and
44!
25!19!
. So P0 

P x  
80
80!
C nN
C 25
55! 25!
44  43  42  41  40  39  38  37  36  35  34  33  32  31  30  29  28  27  26
19 18 17 16 15 14 13 12 11 10  9  8  7  6  5  4  3  2 1

80  79  78  77  76  75  74  73  72  71  70  69  68  67  66  65  64  63  62  61  60  59  58  57  56
 25  24  23  22  21  20 19 18 17 16 15 14 13 12 11 10  9  8  7  6  5  4  3  2 1
44  43  42  41  40  39  38  37  36  35  34  33  32  31  30  29  28  27  26
1

80  79  78  77  76  75  74  73  72  71  70  69  68  67  66  65  64  63  62  61  60  59  58  57  56
 25  24  23  22  21  20
44  43  42  41  40  39  38  37  36  35  34  33  32  31  30  29  28  27  26  25  24  23  22  21  20

80  79  78  77  76  75  74  73  72  71  70  69  68  67  66  65  64  63  62  61  60  59  58  57  56
= .000001134
So the answer is .999998866. Anybody who stuck this one out deserves a virtue badge with fronds.
The computer just says the answer is zero.
C xM C nNxM
44
C 036 C 25
1
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  11, d  25 .

d  c  25  11  225  16.33333
c  d 11  25

 18 ,  2 
2
2
12
12
12
2
2
So   16.33333  4.0415
b. Binomial Distribution with p  .45, n  25 .
(Remember that q  1  p and that both p and q must be between 0 and 1.)
q  1  p  1  .45  .55,   np  25.45   11 .25 ,  2  npq  11.25 .55  6.1875 so
  npq  6.1875  2.4875 . The average number of successes in 25 tries, when the
probability of a success in an individual try is 45% is 11.25.
c. Geometric Distribution with p  .15 .
If q  1  p  1  .15  .85,  
1
1
q
.85
.85

 6.6667 ,  2  2 

 37 .77778 so
2
.0225
p .15
p
.15
  37.7778  6.1464 . When the probability of a success in an individual try is 15% and we
play a game repeatedly, on the average our first success will occur between the 6th and 7th try.
d. Poisson Distribution with parameter of 15.
  m  15 ,  2  m  15 , so   m  15  3.8730
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251solngr3-071 4/10/07
e. Hypergeometric Distribution with p  .45, n  25 , N  80 .
M
and q  1  p then   np  25.45   11 .25 and
N
N n
80  25
55
11 .25 .55   .69620 6.1875   4.30774 so

npq 
25 .45 .55  
N 1
80  1
79
If p 
2
  .69620 6.18755  .83438 2.4875   4.30774  2.07551 . If we take a sample of 25 from
a population of 80 of which 45% are successes, on the average we will get 11.25 successes.
f. Hypergeometric Distribution with p  .45, n  25 , N  520 .
M
and q  1  p then   np  25.45   11 .25 and
N
N n
520  25
495
11 .25 .55   .95376 6.1875   5.90139 so

npq 
25.45 .55  
N 1
520  1
519
If p 
2
  .95376 6.1875  .97660 2.4875   5.90139  2.4292 . 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 here.
g. (Extra credit) Exponential distribution with c  .01 .
1
1
1
1
We have   
 100 and   
 100 .
c .01
c .01
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. This distribution can also be used to replace the Hypergeometric
when the population size is large N  20 n  .
(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 average number of
successes per unit time or space is given, you want the Poisson
distribution. This distribution can be used to replace the binomial when the
n

population is large and the probability of success is small   500  .
p

(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.
6
251solngr3-071 4/10/07
N  80, n  6, M  Np  80.25   20 . Px  
C xM C nNxM
C Nn
and
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
80
80!
80  79  78  77  76  75
80  79  78  77  76  75
C6
74!6!
6  5  4  3  2 1
we wrongly used the Binomial distribution here, we would have found Px  1 when
C120 C 560
20
n  6 and 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?
60!
60  59  58  57  56  55
1
C 020 C 660
54
!
6
!
6  5  4  3  2 1

 1
Px  1  1  Px  0  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
80  79  78  77  76  75
realize that on the first try, the probability of not getting an improper return is 60 80 . If we
do not get an improper return on our first try, there are now 59 ‘proper returns left out of
a total remaining of 79, so 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 conditional probabilities, we can
60  59  58  57  56  55
 .16660 . If we had wrongly
80  79  78  77  76  75
used the Binomial distribution, we would have found Px  1 when n  6 and p  .25.
multiply them together to get P0  
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  1  1  Px  0  Px  1 . Fortunately, we have
already computed these. Px  2  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?
This is Geometric because it asks about the probability of a first success. Px   q x1 p .
p  .25, q  1  p  .75 , Px  2  .75 .25   .1875
7
251solngr3-071 4/10/07
(vi) What is the chance that the first improper return will be among the first ten audited?
Because this asks about a first success, it is 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?
I’m assuming that we still are talking about a large population. If this is true the number
of possible successes is essentially unlimited and this is 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
safely over 500. Our parameter is m  np  100 .02   2.
n
p
 100.02  5000, which is
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
for c  x  d , c  80 and d  120 . Px  90   1  Px  90   1 
120  80
90  80
10
 1
 1
 .75 . You could make a continuous uniform distribution diagram,
120  80
40
1
1
1
with a rectangle with height


 .025 . The base of the rectangle goes
d  c 120  80 40
from c  80 to d  120 . Shade the area in the rectangle above 90. Its area will be
120  90 .025   .75.
(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 q 0  .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?
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 .
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251solngr3-071 4/10/07
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 a day?
From the Poisson table with m  2.5 , P0  .08208
(ii) In a 5-day week?
From the Poisson table with m  52.5  12 .5 , P0  .000000
(iii) What is the chance that more than 5 come in a day ?
From the Poisson table with m  2.5 , Px  5  1  Px  5  1  .95798  .04202 .
(iv) In a 5-day week?
From the Poisson table with m  12 .5 , Px  5  1  Px  5  1  .01482  .99852 .
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. In other words, we want a
number, call it a , so that Px  a   .01 . 14 is the smallest number on the table that has
Px  14   1  Px  14   1  .99429  .00572  .01. If we only had 13 complaint
handlers Px  13  1  Px  13  1  .97300  .02700  .01.
e. If 80% of shoppers at a website abandon their shopping carts without buying and there are currently 20
shoppers on my site,
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 . Note that it seems
more natural to define a success as buying something than failing to buy something. The
probability of buying is p  1  .80  .20
What is the average number that will buy something?
Binomial: p  .20 , n  20 .   np  20.2  4.
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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251solngr3-071 4/10/07
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 . We should
100000
test to see if we can use the Poisson distribution before we start. Note that
n
30000
6

 5 10 8 is above 500. The mean is m  np  30000
 1.8 . According to
100000
p  6 


 100000 
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.
Actually the probability of getting something as extreme or more extreme than the number
actually obtained is called a p-value. What we have here is a hypothesis test. Her hypothesis (the
‘normal’ situation), called a null hypothesis is, if the mean is m , written H 0 : m  1.8 . What she is
afraid is true is called an alternative hypothesis and is H 1 : m  1.8 . The null hypothesis, with 
turned into an equality is tested by assuming that it is true and finding a p-value. Usually, if the pvalue is below .05, a statistician will doubt the null hypothesis.
g. (Extra credit) A telemarketer finds that the length of a call has an exponential distribution with a mean of
1  2 minutes.
c
(i) Find the probability that the length of a call will be 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) Find the probability that the call will be 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) Find the probability that the call will be more than 5 minutes.
If x  5, cx  .55  2.5 and Px  5  Px  5  1  F 5  1  1  e 2.5

e
2.5

 .0820 .
10