251y0471 11/29/04
ECO251 QBA1
THIRD EXAM
Dec 1, 2004
Name: _____________________
Student Number: _____________________
Class Time (Circle) 10am 11am 1pm 2pm
Part I: 8 points.
z follows the standardized Normal distribution z ~ N 0,1 .
Find the following. Make diagrams! (0.5 extra credit for each useful diagram)
1. P1.57  z  1.47   P1.57  z  0  P0  z  1.47   .4418  .4292  .8710
2. P2.27  z  1.57   P2.27  z  0  P1.57  z  0  .4884  4418  .0466
3. P1.47  z  3.75   P0  z  3.75   P0  z  1.47   .4999  .4292  .0707
4. Pz  1.57   Pz  0  P0  z  1.57   .5  .4418  .9418
Add vertical line at zero to diagrams below.
Normal Curve with Mean 0 and Standard Deviation 1
The Area Between -1.57 and 1.47 is 0.8710
Density
0.4
0.2
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
2
3
4
2
3
4
2
3
4
Data Axis
Normal Curve with Mean 0 and Standard Deviation 1
The Area Between -2.27 and -1.57 is 0.0466
Density
0.4
0.2
0.0
-5
-4
-3
-2
-1
0
1
Data Axis
Normal Curve with Mean 0 and Standard Deviation 1
The Area Between 1.47 and 3.75 is 0.0707
Density
0.4
0.2
0.0
-5
-4
-3
-2
-1
0
1
Data Axis
Normal Curve with Mean 0 and Standard Deviation 1
The Area to the Left of 1.57 is 0.9418
Density
0.4
0.2
0.0
-5
-4
-3
-2
-1
0
Data Axis
1
251y0471 11/29/04
Part II: (22+ 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.
The covariance
a) must be between -1 and +1.
b) must be positive.
c) *can be positive or negative.
d) must be between zero and +1
2.
The portfolio expected return of two investments
a) will be higher when the covariance is zero.
b) will be higher when the covariance is negative.
c) will be higher when the covariance is positive.
d) *does not depend on the covariance
3.
. If n = 10 and p = 0.70, then the standard deviation of the binomial distribution is
a) 0.07
b) *1.45
c) 7.00
d) 14.29
Explanation: q  1  p  1  .7  .3,   np  10.7  7 ,  2  npq  7 .3  2.10 so
  npq  2.10  1.45
4.
On the average, 1.8 customers per minute arrive at any one of the checkout counters of a grocery
store. What type of probability distribution can be used to find out the probability that there will
be no customer arriving at a checkout counter in a given minute?
a) binomial distribution.
b) *Poisson distribution.
c) hypergeometric distribution.
d) none of the above
5.
In a taste test four individuals are asked to taste three cups of soda. One of the cups contains Coke
and the other two cups contain Pepsi. Suppose that none of the four individuals can tell the
difference between the colas so that they just guess which is Coke. What is the probability that
exactly three out of four correctly identify the cup that contains Coke? (Extra credit: Do this last!
What is the probability that at least two identify their cup that contains Coke correctly?) Show
your work! (3)
[11]
3
1
8
 .0988
Solution: a) p  13 , n  3 . P3  C 34 13 23  81

16  48
48
2
 1  2 
 1
 .4074
b) Px  2  1  P0  P1  1     C14     1 
3
3
3
81
81
 
  
4
3
2
251y0471 11/29/04
Questions 6-10 are based on exhibit 1. Show your work if you expect full credit!
Exhibit 1: The table below shows average Fahrenheit temperature and yield in lbs./acre for an industrial crop.
F
Y
65
11
70
15
.
75
17
79
16
80
20
The following calculations are done for you. One more column is needed.
6.
Row
x
x2
y
1
2
3
4
5
65
70
75
79
80
369
4225
4900
5625
6241
6400
27391
11
15
17
16
20
79
y2
121
225
289
256
400
1291
Find the sample standard deviation of Fahrenheit temperatures, x (2)
Solution: This is Problem K2 with the numbers changed. The missing column is at right in
x  369 ,
x 2  27391 ,
y  79 ,
question 7. The summary of our calculations is
 y  1291 and  xy  5904 .
 x  369  73.8 , s   x
x



2
n
2
x
5
2
 nx 2
n 1

27391  573 .82 158 .8

 39 .7 ,
4
4
s x  39 .7  6.30
7.
Find the sample covariance between Fahrenheit temperature and yield.(3)
Solution: We will need y and s 2y in Questions 7 and 8.
y
 y  79  15.8 ,
s 2y 
n
y
5
2
 ny 2
n 1

1291  515 .82 42 .8

 10 .7 ,
4
4
s y  10.7  3.2710
sT2F y  s xy 
8.
xy
715
1050
1275
1264
1600
5904
 xy  nxy  5904  573.815.8  73.8  18.45
n 1
4
4
Find the sample correlation between Fahrenheit temperature and yield. (2) [18]
Solution: rxy  rTF Y 
s xy
sx s y

18 .45
39 .7 10 .77

18 .45 2 
39 .710 .77 
0.8013  .8951
9.
If the conversion formula for Celsius temperature is C = 5/9(F-32), find the covariance between
Celsius temperature and yield. (Hint: You are finding the sample covariance between two random
variables, one of which is w  5 9 x  1609 and the other of which is v  1y  0 . (2)
Solution: The formula relating Celsius and Fahrenheit is TC  5 9 TF  5 9 32  . The Outline says
that if w  ax  b and v  cy  d , Cov(w, v)  Cov(ax  b, cy  d )  acCov( x, y) . So here a  5 9
and c  1, so that Cov(w, v)  Cov( 5 9 x  1609 ,1y  0)  5 9 118.45  10.25.
10. If the conversion formula for Celsius temperature is C = 5/9(F-32), find the correlation between
Celsius temperature and yield. (2) [22]
Solution: The Outline says that if w  ax  b and v  cy  d , Corr w, v   SignacCorr( x, y) ,
where Signac  1 if ac is positive and Signac  1 if ac is negative.
So here rTCY  rwv  signacrxy  sign5 9 1.8951  sign5 9 .8951   1.8951  .8951.
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251y0471 11/29/04
11. We have sixteen units of equipment in a bin of which four are defective. Pull three out at random.
What is the probability that exactly one will be defective if we:
a. sample without replacement? (3)
b. sample with replacement? (2)
c. If we sample with replacement, what is he probability that the first item that we find
that is defective is the third item that we pick? (2) [29]
To get full credit for this problem, identify the distribution that you are using and show your
work! This is a slightly expanded version of Problem L9.
C N M C M
a) For the Hypergeometric distribution we have Px   n  x N x , which gives the probability of
Cn
x successes in a sample of n taken from a population of N in which there are M successes. In
this case N  16, M  4 and n  3 .
12! 4!
 12 11  4 

 
16 4 4
12 4
C
C
C C
10! 2! 3!1!  2 1  1  12 11  4  3 1584
P1  3116 1  2 16 1 



 .4714
16!
16 15 14
16 15 14
3360
C3
C3
3  2 1
13! 3!
b) If we sample with replacement or take our sample of 3 from a population that is larger than 60
(20 times 3), but is still 416  25 % defective, we can use the Binomial distribution with p  .25 ,
1
2
q  1  .25  .75 and n  3 . Px  C xn p x q n x , so P1  C13 .25  .75   3.25 .5625   .4219 .
If we are smart and use the table with p  .25 and n  3 , P1  Px  1  Px  0
 .84375  .42187  .42188 .
c) Since this is a problem about the probability of a first success, it involves the Geometric
distribution where P( x)  q x1 p . So P( x)  q x 1 p  .75 2 .25   .1406
12. If ten people are selected at random from a large (continuous) population what is the
probability that at least one makes more than the median income? To get full credit for this
problem, identify the distribution that you are using and show your work! (2) [31]
Solution: This is Problem L1f. If people are selected from a relatively large population, the
probability that each individual has an income above the median is .5. Since there are 10 people,
x ~ B.5,10  (Binomial with p  .5 and n  10 ) . Px  1  1  P0  1  .00098  .99902
13. If x is a random variable with a mean of 7 and a standard deviation of 6, find
a) Cov7 x  3, 9 x  2 (3)
b) Corr 7 x  3, 9 x  2 (2) [36]
Solution: This is Problem K3. Use the same formulas as in Problems 9 and 10 above, but
substitute x for y .
a) If w  ax  b and v  cx  d , Cov( w, v)  Cov(ax  b, cx  d )  acCov( x, x) . We showed that
Cov( x, x)  Var x  , which is s 2 or  2 . So here a  7 and c  9, so that
Cov( w, v)  Cov(7 x  3 ,9 x  2)  7962  2268 .
b) If w  ax  b and v  cx  d , Corrw, v   SignacCorr( x, x) , where Signac  1 if ac is
positive and Signac  1 if ac is negative. We showed that since
Corr x, y  

Covx, y 
, it must be true that
StdDevx StdDev y 
Corr x, x  
Covx, x 
StdDevxStdDevx
Var x 
 1. So Corr7 x  3,9 x  2  Sign72Corr ( x, x)   11  1.
Var x 
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251y0470t 11/30/04
ECO251 QBA1
THIRD EXAM
Dec 1, 2004
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 (20 Points) Show your work!
1. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that the jorcillator works as long as one
component works. For example, if the Phillinx failed last year and the
flubberall fails this year, the phillinx fails this year. This problem, which is a simplified version of Problem H4 has three time periods,
last year, this year and Ever After (after this year). You have been in Tibet for the last year, so you have no idea on what actually
happened last year.
The probability of the phillinx failing is given by a standardized Normal Distribution, so , if z represents the life of the Phillinx, the
probability of the phillinx failing last year is
it lasting into the Ever After is
Pz  1 .
Pz  0 , the probability of it failing this year is P0  z  1 and the probability of
The probability of the Flubberall failing is given by a Continuous Uniform Distribution, where
c  1 and d  2  f , where f
is the fourth digit of your student number, divided by 10. For example if Seymour’s student number is 012345, his value of
d  2  .3  2.3 . If the fourth digit of your student number is zero, use d  3.1 . If x represents the life of the Flubberall, the
probability of the Flubberall failing last year is
of it lasting into the Ever After is
Px  1 .
Px  0 , the probability of it failing this year is P0  x  1 and the probability
Failure of components is assumed to be independent, so that, if the probability of the Phillinx failing in the first year is .1 and the
probability of the Flubberall failing last year is .7, the probability of both components failing last year is (.1) (.7) =.07 (This is not
necessarily the probability that the jorcillator failed last year!)
a) What is the probability the Flubberall failed last year? This year? After this year? (1.5)
b) What is the probability the Phillinx failed last year? This year? After this year? (1.5)
c) What is the probability that the jorcillator failed last year? (2)
d) What is the probability that the jorcillator will fail this year? (2)
e) What is the probability that the jorcillator will fail in the Ever After.? (2)
f) Using the probabilities that you found in c) – e) , associate the number -0.5 with last year, 0.5 with this year and 2.0 with
Ever after and find the expected value and standard deviation of the life of the jorcillator. (2)
2. (Bowerman and O’Connell, modified) An air traffic control center experiences a week with 3 errors (planes placed too close
together either vertically or horizontally).
On the basis of records of the center, you decide that the center has an average number of errors of d 10 per week, where d is the
number you used in the previous problem.
a) (4) Compute a Poisson table with the following values
P0 , P 1 , P2 , P3 and P4 . The easiest way to do this is to
assume that P5 and any probabilities above it are essentially zero and to assume that after you have computed P0  through
P3 , that P4 is whatever is required to make this a valid distribution.
e 0.40  .67032 . Using that in the Poisson
formula, you would find the following P0  .6703 , P1  .2681 , P2  .0536 and P3  ..0072 . In order to get a
valid distribution, you need P4  .0008 . (If you check this against the Poisson table you will find that the actual value of P4
Example: If
d  4.0,
you would assume that the mean is 0.40 and would compute
is .0007, but the difference won’t affect your results.) You can see if you are on the right track for your numbers by comparing your
results with the Poisson tables for parameters of 0.3 and 0.4. Your results should be close.
b) (3)To test the hypothesis that the mean is still the value of d 10 that you are using, you compute the probability that, if the
hypothesis is true you would get results as extreme or more extreme than you actually got. In this case that would be
Px  3 . If the
probability is less than 1%, you can say that you strongly doubt that the hypothesis is correct and demand an investigation. If the
probability is between 1% and 5% you can say that you somewhat doubt that the hypothesis is correct and suggest that the controllers
clean up their act. If the probability is above 5%, you would have to say that you cannot doubt the hypothesis and you would have to
shut up. In view of the probability you computed, what would you do?
c) (2)Using the table that you have made, repeat the analysis assuming that during the week there were 2 errors instead of 3.
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251y0470t 11/30/04
3. (Extra Credit) (Bowerman and O’Connell, modified) a) A young nurse observes that 8 of the last 30000 patients of a hospital went
into a coma during anesthesia. Upon investigation she finds that nationally 6 out of 100000 patients go into comas after anesthesia
(This is 0.006 per cent!). 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? Find the probability of 8 or more people out of 30000 going into a
coma. Explain whether we have a binomial table that will do this. If not, show that this is an appropriate place to use a Poisson
distribution and use the Poisson tables to find the probability that x  8. Use the approach in the previous problem to see if the
number of comas is consistent with the national rate.
b) How would you change your opinion if the number that went into a coma was 6?
1. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that the
jorcillator works as long as one component works. For example, if the Phillinx failed last year and the
flubberall fails this year, the phillinx fails this year. This problem, which is a simplified version of Problem
H4 has three time periods, last year, this year and Ever After (after this year). You have been in Tibet for
the last year, so you have no idea on what actually happened last year.
The probability of the phillinx failing is given by a standardized Normal Distribution, so , if z represents
the life of the Phillinx, the probability of the phillinx failing last year is Pz  0 , the probability of it
failing this year is P0  z  1 and the probability of it lasting into the Ever After is Pz  1 .
The probability of the Flubberall failing is given by a Continuous Uniform Distribution, where c  1 and
d  2  f , where f is the fourth digit of your student number, divided by 10. For example if Seymour’s
student number is 012345, his value of d  2  .3  2.3 . If the fourth digit of your student number is zero,
use d  3.1 . If x represents the life of the Flubberall, the probability of the Flubberall failing last year is
Px  0 , the probability of it failing this year is P0  x  1 and the probability of it lasting into the Ever
After is Px  1 .
Failure of components is assumed to be independent, so that, if the probability of the Phillinx failing in the
first year is .1 and the probability of the Flubberall failing last year is .7, the probability of both components
failing last year is (.1) (.7) =.07 (This is not necessarily the probability that the jorcillator failed last year!) I
will give the events names when I figure them out.
a) What is the probability the Flubberall failed last year? This year? After this year? (1.5)
Solution: c  1 and for good ol’ Seymour d  2.3 .
Make a diagram. The distribution is represented by a box from -1 to 2.3. The length of
the box is 1 + 2.3 = 3.3, so the height is 1  0.3030 .
3.3
Last year: Shade the box between -1 and zero. Px  0 = .3030. Event F1
This year: Shade the box between 0 and 1. P0  x  1 = .3030. Event F2
Ever after: Shade the box from 1 to 2.3 Px  1  2.3  1
1
 .3939 . Event F3
3.3
Note that, except for a rounding error, these probabilities add to 1.
0  1
1
Or: If we use a cumulative distribution: Px  0  F 0 

 .3030
2.3   1 3.3
1  1
1
2
1
P0  x  1  F 1  F 0 



 .3030
2.3  (1) 3.3 3.3 3.3
Px  1  1  F 1  1 
2
1.3
xc

 .3939 Note: F x  
.
3.3 3.3
d c
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251y0470t 11/30/04
b) What is the probability the Phillinx failed last year? This year? After this year? (1.5)
Solution: z ~ N 0,1
Make a diagram. The distribution is represented by a Normal curve centered at zero.
Last year: Shade the area below zero. Pz  0 = .5. Event Ph1
This year: Shade the area between 0 and 1. P0  z  1  .3413 from the Normal table.
Event Ph2
Ever after: Shade the area above 1. Px  1  Pz  0  P0  z  1  .5  .3413
 .1587 Event Ph3
Note that these probabilities add to 1.
Flubberall probabilities (Second through 4th row). Poisson P0 (5th row)
2.1
2.2
2.3
2.4
2.5
2.6
2.7
d
.3226
.3125
.3030 .2941
.2857 .2778
.2702
Px  0
P0  x  1
2.8
.2632
2.9
.2564
3.1
.2439
.3226
.3125
.3030
.2941
.2857
.2778
.2702
.2632
.2564
.2439
Px  1
.3548
.3750
.3939
.4118
.4286
.4444
.4595
.4737
.4872
.5122
0.1d
.8106
.8025
.7945
.7866
.7788
.7711
.7634
.7558
.7483
.7334
e
c) What is the probability that the jorcillator failed last year? (2)
Solution: From the previous page - PF1  = .3030, PF2  = .3030 and PF3   .3939 ;
PPh1  = .5, PPh2   .3413 and PPh3   .1587 . Let’s make a joint probability table.
F1
(Columns don’t all add because of rounding) F2
F3
Ph1
Ph2
Ph3
.1515
.1515
.1970
.1034
.1034
.1344
.0481
.0481
.0625
.3030
.3030
.3939
.5000 .3413 .1587 1.0000
Now let’s make a table of the joint events and when they down the Jorcillator.
Joint Event
Probability
Downs
Jorcillator in
period
.1515
1
F1 Ph1
.1034
2
F1 Ph 2
F1 Ph3
.0481
3
F2 Ph1
.1515
2
F2 Ph2
.1034
2
F2 Ph3
.0481
3
F3 Ph1
.1970
3
F3 Ph2
.1344
3
F3 Ph3
.0625
3
So the probability for last year (Period 1) is .1515 = PF1  Ph1   PF1 PPh1 
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251y0470t 11/30/04
d) What is the probability that the jorcillator will fail this year? (2)
Solution: The probability for this year (Period 2) is .1034 + .1515 + .1034 =.3583 or
1  PF1  Ph1   PF3  Ph3   1 - .1515 - .4901 = .3584
e) What is the probability that the jorcillator will fail in the Ever After.? (2)
Solution: The probability for Ever After (Period 3) is .0481 + .0481 + .1970 + .1344 + .0625
=.4901 or PF3  Ph3   PF3   PPh3   PF3  Ph3  = .3939 + .1587 - .0625 = .4901
f) Using the probabilities that you found in c) – e) , associate the number -0.5 with last year, 0.5
with this year and 2.0 with Ever after and find the expected value and standard deviation of the
life of the jorcillator. (2)
x
Px 
xPx  x 2 Px 
P x   1
 0.5 .1515  .07575 .037875
xPx   1.09595
0.5 .3584
.17915 .089575 So E x    x 

2.0 .4901
Total 1.0000
.98020
1.09595
1.96040
2.08785

   x
Ex
2
2
Px   2.08785
 
Finally  x2  Var x   E x 2   x2  2.08785  1.09595 2  0.087703098 .
So  x  0.078803098  0.94797 . Of course, there is no reason to carry this many decimal
places, but it seemed like less work than thinking about rounding.
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251y0470t 11/30/04
2. (Bowerman and O’Connell, modified) An air traffic control center experiences a week with 3 errors
(planes placed too close together either vertically or horizontally).
On the basis of records of the center, you decide that the center has an average number of errors of d 10 per
week, where d is the number you used in the previous problem.
a) (4) Compute a Poisson table with the following values P0 , P 1 , P2 , P3 and P4 . The easiest
way to do this is to assume that P5 and any probabilities above it are essentially zero and to assume that
after you have computed P0 through P3 , that P4 is whatever is required to make this a valid
distribution.
Example: If d  4.0, you would assume that the mean is 0.40 and would compute e 0.40  .67032 . Using
that in the Poisson formula, you would find the following P0  .6703 , P1  .2681 , P2  .0536 and
P3  ..0072 . In order to get a valid distribution, you need P4  .0008 . (If you check this against the
Poisson table you will find that the actual value of P4 is .0007, but the difference won’t affect your
results.) You can see if you are on the right track for your numbers by comparing your results with the
Poisson tables for parameters of 0.3 and 0.4. Your results should be close.
Solution: Seymour used d  2.3 . The Poisson formula is Px  
P0 
P1 
P2 
e m m x
so
x!
e m m 0
 e 0.23  0.794534
0!
e  m m1 e 0.23 0.23 

 0.23 P0  .182742
1!
1!
e .23 0.23 2 0.23
e m m 2


P1  .021015
2!
2
2!
e  m m 3 e 0.23 0.23 3 0.23


P2  .001611
3!
3
3!
The sum of these is .999902 and 1 - .999902 = .000098. The actual value of P4 is
P3 
e 0.23 0.23 4 0.23
e m m 4


P3  .000093 , so that I wasn’t way off, but it wouldn’t
4!
4
4!
make much difference for the rest of the problem if I were.
b) (3)To test the hypothesis that the mean is still the value of d 10 that you are using, you compute the
probability that, if the hypothesis is true you would get results as extreme or more extreme than you
actually got. In this case that would be Px  3 . If the probability is less than 1%, you can say that you
strongly doubt that the hypothesis is correct and demand an investigation. If the probability is between 1%
and 5% you can say that you somewhat doubt that the hypothesis is correct and suggest that the controllers
clean up their act. If the probability is above 5%, you would have to say that you cannot doubt the
hypothesis and you would have to shut up. In view of the probability you computed, what would you do?
Solution: Since what we actually computed was Px  4  .000098 , we can say either that Px  3
 1  Px  2  P3  Px  4  .001611  .000098  .001709 . Since this is below 1% we strongly doubt
that, under present conditions, the mean is still 0.23 and we demand an investigation.
P4 
c) (2)Using the table that you have made, repeat the analysis assuming that during the week there were 2
errors instead of 3.
Solution: Px  2  P2  Px  3  .021015  .001709  .022724 . Since this is between 1% and 5% we
somewhat doubt that, under present conditions, the mean is still 0.23 we suggest that the controllers clean
up their act.
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251y0470t 11/30/04
3. (Extra Credit) (Bowerman and O’Connell, modified) a) A young nurse observes that 8 of the last 30000
patients of a hospital went into a coma during anesthesia. Upon investigation she finds that nationally 6 out
of 100000 patients go into comas after anesthesia (This is 0.006 per cent!). 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? Find the probability of 8 or more people out of 30000 going into a
coma. Explain whether we have a binomial table that will do this. If not, show that this is an appropriate
place to use a Poisson distribution and use the Poisson tables to find the probability that x  8. Use the
approach in the previous problem to see if the number of comas is consistent with the national rate.
b) How would you change your opinion if the number that went into a coma was 6?
6
 .00006 and n  30000 .
Solution: a) This is a binomial problem. We want Px  8 when p 
100000
since the lowest value of p on the binomial table is .01, we cannot use a binomial table, but, with a small
probability and a large number of tries, this is classic Poisson territory. Note that
n
30000

 5 10 8 is
6
p
100000
6
 1.8 . According to the Poisson(1.8) table
100000
Px  8  1  Px  7  1  .99944  .00056 . Since this is below 1% we strongly doubt that, under present
conditions, the chance of a coma in this hospital is still 0.00006 and we demand an investigation, just as the
nurse in the movie Coma did.
above 500. The mean is m  np  30000
b) Px  6  1  Px  5  1  .98962  .01038 Since this is between 1% and 5% we somewhat doubt that,
under present conditions, the chance of a coma in this hospital is still 0.00006 we suggest that the hospital
clean up their act.
10