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251y0242 12/10/02
ECO251 QBA1
FINAL EXAM
DECEMBER 11, 2002
Name
KEY
Class ________________
Part I. Do all the Following (16 Points) Make Diagrams! Show your work!
x ~ N 4, 6 .


3.50  4 
  3.5  4
z
 P 1.25  z  0.08
P 3.50  x  3.50  P 
6
6 

 P 1.28  z  0  P 0.08  z  0  .3944  .0319  .3625
6.10  4 
 2.10  4
z
 P 0.32  z  0.35
2. P2.10  x  6.10  P 
6 
 6
 P 0.32  z  0  P0  z  0.35  .1255  .1368  .2623
24  4 
7  4
z
 P0.50  z  3.33
3. P7  x  24  P 
6 
 6
 P0  z  3.33  P0  z  0.50  .4996  .1915  .3081
6.01  4 

 Pz  0.335
4. Px  6.01  P  z 
6 

 Pz  0  P0  z  0.33  .5  .1293  .3707 or
 Pz  0  P0  z  0.335  .5  .1312  .3688 or
 Pz  0  P0  z  0.34  .5  .1331  .3669
2.73  4 

 Pz  0.21
5. F 2.73 (Cumulative Probability) Px  2.73  P  z 
6 

 Pz  0  P 0.21  z  0  .5  .0832  .4168
6. x.41 (Find z .41 first). (3 points) We want a point x.41 , so that Px x .41   .41 . Make a diagram for
z , showing zero in the middle, 50% below zero, and the area above zero divided into 9% between z.41 and
zero and 41% above z .43 . From the diagram, P0  z  z.41   .0900. . The closest we can come is
P0  z  0.23  .0910 . So z.41  0.23 , and x    z.41  4  0.236  4  1.38 , or 5.38 .
1.
5.38  4 

Px  5.38  P z 
  Pz  0.23
6 

 Pz  0  P0  z  0.23  .5  .0910`  .4090  .41
To check this note that
7. A symmetrical region around the mean with a probability of 41%. (3 points) We want two points
and
x .705
x.295 , so that Px.705  x  x 295   .4100 . Make a diagram for z , showing zero in the middle, an
area in the middle of 41%, split in two by zero so that 20.5% is above zero and 20.5% is above zero. Since
.5 - .205 = .295, the points we want are z.295 and  z.295 . From the diagram, P0  z  z.295   .2050 .
P0  z  0.54  .2054 . So use z.295  0.54 , and
x    z.285  4  0.546  4  3.24 ,
The closest we can come is
7.24  4 
 0.76  4
z
P0.76  x  8.24  P

6
6


 P 0.54  z  0.54  2P0  z  0.54  2.2054  .4106  .41.
or 0.76 to 7.24. To check this note that
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251y0242 12/10/02
II. (10 points-2 point penalty for not trying part a .) Show your work!
We are investigating the reliability of a machine that fills 16ounce bottles. A sample of six is taken
with the following results (Assume that it is correct to do a confidence interval from such a small sample):
Be careful! There is a lot of chance for rounding error in this problem!
15.68
16.18
16.05
16.14
16.17
16.28
a. Compute the sample standard deviation, s , for the number of ounces. Show your work. (4)
b. Compute a 98% Confidence interval for the mean. (6)
c. (Extra credit) Can you say that the mean that you got is significantly different from 16? You
must give a reason for your answer to be read. (2)
Solution: Since the confidence level is 1    .98, the significance level is   .02 .
a.
x2
x
1
2
3
4
5
6
15.68
16.18
16.05
16.14
16.17
16.28
96.50
245.862
261.792
257.602
260.500
261.469
265.038
1552.263
 x  96.50 ,
x
s x2 
b.

x
sx 
2
 nx
n 1
sx
n

2
2
 1552.263 and n  6 . So x 
 x  96.50  16.0833
n
6
0.2278
1553.263  616.0833

 0.0456 s x  0.0456  0.213 .
5
5
2

0.0456
 0.0872. The degrees of freedom are n  1  6  1  5 .
6
.02
 .01 . From the t-table, tn2 1  t.015   3.365 . But what compelled some of you to decide
2
2
that 0.213 was s in a) but  in b)?

Putting this all together
16.38. More formally,
  x  t n1 s x  16.083  3.3650.0872  16.083  0.293
2
or 15.79 to
P15.79    16.38  .98 .
c. Since 16 is included in this interval, we cannot say that the mean is significantly different from 16. Note
that, because of the small numbers involved, Minitab gets a somewhat different solution.
MTB > TInterval 98.0 c1.
Confidence Intervals
Variable
N
Mean
C1
6
16.0833
StDev
0.2110
SE Mean
0.0861
98.0 % C.I.
( 15.7935, 16.3732)
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251y0242 12/10/02
III. Do at least 4 of the following 6 Problems (at least 12 each) (or do sections adding to at least 48 points Anything extra you do helps, and grades wrap around) . Show your work! Please indicate clearly what
sections of the problem you are answering! If you are following a rule like E ax  aEx  please state it! If
you are using a formula, state it! If you answer a 'yes' or 'no' question, explain why! If you are using the
Poisson or Binomial table, state things like n , p or the mean. Avoid crossing out answers that you think
are inappropriate - you might get partial credit. Choose the problems that you do carefully – most of us are
unlikely to be able to do more than half of the entire possible credit in this section!)
1. 16 employees are randomly drawn from a large corporation and their sick days checked. Note that the
sample size is 16 throughout this problem!
a. The sample mean is 5.4. Assume that the population standard deviation is known to be 1.7 and
create a 99% confidence interval for the average number of sick days. (4)
b. Assume that the facts in part a are correct but that the corporation only has 215 employees.
Create a 99% confidence interval for the average number of sick days again, highlighting what has changed.
(4)
c. Assume that the sample mean is 5.4 and the corporation has 215 employees, but that 1.7 is a
sample standard deviation. Create a 99% confidence interval for the average number of sick days again,
highlighting what has changed. (4)
d. The sample mean is 5.4. Assume that the population standard deviation is known to be 1.7 and
create a 91% confidence interval for the average number of sick days. (3)
e. Explain the effect of the following on the size of a confidence interval (keep the reasons brief)
(3):
(i) A smaller sample size.
(ii) A smaller population size.
(iii) A lower significance level.
Solution: Note that the sample size is 16 throughout this problem. Since the confidence level is
1    .99, the significance level is   .01 .
Repeat after me! " z goes with  (sigma - population variance); t goes with s (sample variance)!"
Most of you seem to have been so hypnotized by the old exams that you had that you never noticed that
most of this question concerned population variances.
  1 .7 ,  x 
x
1.7
 0.425 . From the t-table z 2  z.005  2.576 .
16
n
  x  z 2 x  5.4  2.5760.425  5.4  1.09 or 4.31 to 6.49. More formally
a)

 P4.31    6.49  99% .
b) The population size, N  215 , is less than 20 times the sample size n  16 , so

N n
1.7 215  16
x  x

 0.425 0.92991  0.4098 . From the t-table
16 215  1
n N 1
z 2  z.005  2.576 .
  x  z  x  5.4  2.5760.4098  5.4  1.049
2
or 4.35 to 6.45. More formally
 P4.35    6.45  99% .
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251y0242 12/10/02
s  1.7. The degrees of freedom are n  1  16  1  15 . Since we are given s rather than  , we
n 1
15
must use t. From the t-table, t 
 t.005
 2.947 .
2
c)
The population size,
N  215 , is less than 20 times the sample size n  16 , so s x 
sx
n
N n
N 1
215  16
 0.425 0.92991  0.4098 . Putting this all together
16 215  1
  x  tn1 s x  5.4  2.9470.4098  5.4  1.21 or 4.19 to 6.61. More formally,
2

1.7
P4.19    6.61  .99 .
x
1.7
d)  x 

 0.425 . . Since the confidence level is 1    .91, the significance level is
16
n
  .09 . We need z 2  z.045. Make a diagram for z . It should show 50% above zero, divided into
4.5% above
z.045 and 50% - 4.5% = 45.5% below z.045 . So P0  z  z.045   .4550
P0  z  1.70  .4554 .
  x  z 2 x  5.4  1.700.425  5.4  0.72 or 4.68 to 6.12. More formally
From the Normal table, the closest we can come is
 P4.68    6.12  91% . It’s remarkable how many of you could find something like z .045 on
page one but not here.
e) The basic formula for a confidence interval is
  x  z  x
2
or   x  tn1 s x
2
s
(i) A smaller sample size. If we use the formula   x  tn1 s x and s x  x , a smaller sample size will
n
2
make the standard error larger because the sample size is in the denominator and will also make t
larger, as we can see from the t table.
(ii) A smaller population size. The formula for the standard error is s x 
size grows,
sx
n
 n 1
N n
. As the population
N 1
N n
approaches one from below, making the confidence interval larger. So for a smaller
N 1
population size, the standard deviation and the interval will be smaller.
(iii) A lower significance level. If we look at the
t table, we see the value of t gets larger as the
significance level gets smaller. So a smaller significance level makes the interval larger.
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251y0242 12/10/02
2. A sample of 6 is taken from a large Normal population with a mean of 3 and a standard deviation of 12.
a. What is the probability that an individual item x in the sample lies between 2 and 4? (2)
 
b. What is the probability that the sample mean x  lies between 2 and 4? (2)
c. What is the probability that at least one of the 6 measurements lies between 2 and 4? (2)
d. What is the probability that all the 6 measurements lie between 2 and 4? (2)
e. Do b) again assuming that the sample of 6 is taken from a population of 300 (1)
f. Find the 90th percentile of the distribution of x (2)
g. Find the 90th percentile of the distribution of x (1)
Solution:
x ~ N ,   N 3,12 . Since  x 
x ~ N  ,  x   N 3,4.8990 , n  6.
x
n

12
6
 4.8990,
4  3
2  3
z
 P 0.08  z  0.08
P2  x  4  P 
12 
 12
 2P0  z  0.08  2.0319  .0638 .
43 
 23
z
 P 0.20  z  0.20
b) P2  x  4  P 
4.8990 
 4.8990
 2P0  z  0.20  2.0793  .1586
c) This problem and the next problem are Binomial problems with n  6, p  .0638 and
q  1  .0638  .9362 . Remember Px   C xn p x q n  x .
a)
Px  1  1  P0  1  C06 .9362  1  .6579  .3421
6
d)
Px  6  C66 .0638  .000000067 .
6
e) Because the population size is more than 20 times the sample size, the answer is essentially the same as
b).
f) Since the 90th percentile of z is z.10  1.282 , the 90th percentile of x is x    z.10
 3  1.28212  18.38
g) Since the 90th percentile of z is
z.10  1.282 , the 90th percentile of x is
x    z.05 x  3  1.2824.8990  3  6.281  9.28
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251y0242 12/10/02
3. An eight-sided die is rolled four times. It has sides numbered one through 8, all of which are equally
likely. A one, two or three wins $5.00, so that the probability of winning on each roll is 3/8.
Let x represent the number of wins and y represent the amount won. Find the following:
a. The complete distribution of x and y (5) – note that you do not have to do this part to
answer the remainder of the question.
b. The mean and variance of x . (2)
c. The mean and variance of y .(2)
d. The chance of at least one win (2 points if you did not do a) (1)
e. Assume that the die is rolled 40 times, find the probability of 20 or more wins. (Note that your
binomial tables won’t help and that if you try to do this without using a table you will be here
until Christmas)
(i) Can this problem be done using the Poisson distribution? – Why? (1)
(ii) Can this problem be done using the Normal distribution? – Why? (1)
(iii) Decide which of the two distributions is correct and answer the question. (2 or 2.5)
Solution: This problem is a Binomial problems with n  4,
Px   C xn p x q n  x  C x4 83 

p  3 and q  1  3  5 . Remember
8
8
8
x 5 4 x
.
8
The distribution is as follows:
x
y
0
0
1
5
2
10
3
15
4
20
C04 38 
 8 4  58 4  .15259
1
3
C14  3 8  5 8   4.0915527   .36621
2
2
C 24 3 8  5 8   6.0549316  .32959
3
1
C34 38  5 8   4.0329590  .13184
4
0
C 44 3 8  5 8   .0197754  .01978
0 5
There is a minimal rounding error since these add to 1.00001.
  np  438   1.500
 2  npq  1.5005 8   0.9375
2
c. The prize is y  5 x so, using the formulas Eax   aEx  and Var ax   a Var x  we get
E5x  5Ex  51.5  7.5 and Var5x   5 2 Varx   250.93755  23.4375
d. Px  1  1  P0  1  .15259  .84721 .
e. (i) n  40 , p  3 8 . If we test to see if the Poisson Distribution can be used, we find
b. The mean is
and the variance is
n 40

 106.67. Since this is not above 500, we cannot use the Poisson Distribution.
p 38
(ii)   np  403 8   15  5. nq  40 5 8  40  15  25  5. Since these are both above 5, we can
use the Normal distribution with 
(iii) With the continuity correction
 np  4038   15 and  2  npq  155 8   9.375.

19.5  15 
  Pz  1.47   .5  .4292  .0708
Px  19.5  P z 
9.375 

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251y0242 12/10/02
4. a. I buy two mainframes from a computer manufacturer. One computer has a 22% chance of breaking
down during the first year, the second has a 25% chance of breaking down during the same period. Let A
represent the event that the first computer breaks down in the first year and B represent the event that the
second computer breaks down. Tell how you will note the complement of these events and assume that the
events are independent. Find the following, noting if each is an event like A  B, A  B, or similar
events involving the complement. Assume that events A and B are independent.
(i) The probability that both computers break down in the first year . (2)
(ii) The probability that one computer breaks down in the first year . (2)
(iii) The probability that no computer breaks down in the first year . (2)
(iv) If a breakdown costs you $1000 (and no machine will break down more than once in a year),
what is the mean and variance of your breakdown costs? (4)
b. I use an average of 2 boxes of paper in a day , but it takes 6 days to get a delivery. Given the average
usage over 6 days , do the following:
(i) What is the probability of using at least 5 boxes in the 6 day period? (1)
(ii) What is the probability of using more than 6 boxes? (1)
(iii) If x is the number of boxes used, what is P 4  x  10 ? (2)
(iv) If I want to keep the probability of running out of paper below 5%, what is the minimum
number of boxes I should have on hand when I reorder? (Why?) (2)


A is the complement of A.
A  B  PAPB  .22.25  .055
(ii) PA  B   PA  B   P APB   PA PB   .22.75  .78.25 
 .165  .195  .360 . (This is P A  B  P A  B )
(iii) PA  B   PA PB   .78.75  .585 . These probabilities add to one.
(iv) If x represents the number of breakdowns, y  1000 x is the cost.
x
xPx
Px
x 2 P x 
Solution:
a. (i) P
0
1
2
sum
.585
.360
.055
1.0000
0
.360
.110
.470
0
.360
.220
.580
2
 x  Ex   x  Px   .47  x2  Varx  Ex 2    x2  .580  .47  .3591
Eax  aEx and Varax   a 2Varx  , so E1000x  1000Ex  1000.47  470 and
Var1000 x   1000 2 Varx   359100.
b. (i) Poisson distribution   26  12. Px  5  1  Px  4  1  .00760  .99240
(ii) Px  6  1  Px  6  1  .04582  .95418
(iii) P4  x  10  Px  10  Px  3  .34723  .00229  .34494
(iv) You must have 18 boxes. Px  18  .96258 .This means Px  15 is below 5%.
7
251y0242 12/10/02
5. (Lee et. al.) The following joint probability table measures the satisfaction that customers have with local
food stores. x is the satisfaction level, with 4 representing the highest value. y represents the number of
years that the consumer has lived in the area, with 2 representing long term residents.
x
1
2
3
4
Total
1
.03
.14
.23
.07
.47
2
.08
.17
.23
.05
.53
Total
.11
.31
.46
.12
a. Are x and y independent? Why? (2)
b. How do we know that this is a valid distribution? (1)
c. Compute the mean and standard deviation of y (2)
d. Compute the covariance between x and y and interpret it. Does this mean that long term residents are
more satisfied? (3.5)
e. Compute the correlation. What does this tell us that we couldn’t learn from the covariance? (2.5)
f. Remember that these are joint probabilities. Find the conditional probability that a long-term resident is
y
very dissatisfied
Px  1 y  2 . (2)
g. Use the same method to find the complete conditional probability of x for long term residents, show that
this is a valid distribution and compute the conditional mean for long-term residents. (5.5)
h. If you are ready for some real thinking, find the conditional mean for short-term residents and use it to
compare the satisfaction levels for the two kinds of residents. (To do this correctly we would need variances
as well – have a nice break and we’ll worry about that next semester.)
Solution: a. x and y are not independent. For example
Px  1   y  1  .03  Px  1 P y  1  .11.47  .0517
b. The probabilities add to one and are not negative or above 1.
y
c.
P y
yP y
 
1
2
sum
 
.47
.53
1.00
y 2 P y 
.47
1.06
1.53
0.47
2.12
2.59
 y  E y    yP y   1.53  y2  E y 2    y2  2.59  1.532  0.2491
 y  .2491  .4991.
x
d.
y
1
2
Px
xPx
x 2 P x 
P y 
yP y 
y 2 P y 
0.47
1.06
1.53
0.47
2.12
2.59
1
.03
.08
.11
2
.14
.17
.31
3
.23
.23
.46
4
.07
.05
.12
.47
.53
1.00
.11
.62
1.38
.48
2.59
.11
1.24
4.14
1.92
7.41
 Px  1,   Ex   xPx  2.59 , Ex    x Px  7.41,
 P y   1,   E y    yP y   1.53 and Ey    y P y  2.59
2
To summarize
2
x
2
2
y
E xy 
=
.03(1)(1)
+.14(2)(1)
+.23(3)(1)
+.07(4)(1)
+.08(1)(2)
0.03
+ 0.16
+.17(2)(2)
+ 0.28
+ 0.68
+.23(3)(2)
+ 0.69
+ 1.38
+.05(4)(2)
+ 0.28
+ 0.40
=3.90
251y0242 12/10/02
8
 xy  Covxy  E xy   x  y  3.90  2.591.53  0.0627 .
Negative, so
x and y move
oppositely. This means that long term residents are less satisfied.
e.
 x2  Ex 2    x2  7.41  2.592  0.7019
and
  E y     2.59  1.53  0.2491
 xy
 0.0627
 .06227


 .14995
So that  xy 
 x y
.7019 .2491 0.83780.4991
2
y
2
2
y
2
Since the square of the correlation is between .02 and .03, on a zero-one scale it is quite weak.
f. By the multiplication rule
Px  1 y  2 
Px  1   y  2 .08

 .1509
P y  2 
.53
g. If we divide the whole top row by .47, and the bottom row by .53 we get
y
1
2
1
.064
.151
2
.298
.321
x
3
.489
.434
4
.149
.094
Total
1.00
1.00
For short-term residents .064(1) + .298(2) + .489(3) + .149(4) = .064 + .596 + 1.467 + .596 = 2.723
For long-term residents .151(1) + .321(2) + .434(3) + .094(4) = .151 + .641 + 1.302 + .376 = 2.470.
Anyway long-term residents seem less satisfied.
9
251y0242 12/10/02
6. Answers to a-c can be left in factorial form.
a. A bridge hand consists of 13 cards. A deck is a population of 52 cards, with a total of 4 kings. What is
the probability of 3 kings? What is the distribution you are using? (3)
b. What are the mean and variance of the number of kings in a hand? (2)
c. Now comes the fun. Let’s say we take a load of decks – you do not need to know how many cards we
have, and we deal you 13 cards, what is the probability of 3 kings now? (3)
d. Are the mean and variance of the number of kings in your hand in c) the same as the values in b)?
What changes and why? (1)
e. (Keller, Warrack) The amount of gasoline sold by one of your service stations daily is uniformly
distributed between a minimum of 2500 and a maximum of 5500 gallons.
(i) What is the mean and standard deviation of sales? (1.5)
(ii) What is the probability that sales on a given day will fall between 3000 and 3500 gallons? (2)
(iii)What is the probability that sales will be over 4000 gallons? (1)
(iv)What is the probability of sales between 1800 and 2800 gallons? (1)
(v) If you did not do (iv) using cumulative distributions, do it now (2)
(vi)If you own 9 identical service stations, what is the probability that at least one has sales over
4000 gallons? (3)
Solution:
a. There are 4 kings in a deck of 52. So this is a Hypergeometric distribution with
M  4, N  52, n  13 and x  3 . Px  
M
x
N M
n x
N
n
C C
C
. So
P3 
4
3
48
10
52
13
C C
C
 4!  48! 
 

3!1!  38!10! 


 52! 


 39!13! 
 48  47  46  45  44  43  42  41 40  39 

4
10  9  8  7  6  5  4  3  2 1
439




 52  51 50  49  48  47  46  45  44  43  42  41 40   52  51 50  49 

 

13 12 1110  9  8  7  6  5  4  3  2 1

  13 12 11 
43913 12 11  .04120

52  51 50  49
4
1
 4

 .07692   np  13    1 ,
52 13
 52 
 N  n
 52  13   4  48  39  4  48
2 
 .7647113.07101
npq  
13    13 
 N 1 
 52  1   52  52  51  52  52
b. p 
 .70588 so   0.70588  0.84017 .
c. Binomial P  x   C p q
n
x
x
1312 11  1   12 

   
3 2 1  13   13 
3
10
n x
13!  1 
. So P3  C p q 
 
3!10!  13 
13
3
3
10
3
10
 12 
 
 13 
 286.0004551661.449137  .05847
d. Because the population is now infinite, we remove the .76471 finite population adjustment from the
variance formula. The mean is unchanged.
10
251y0242 12/10/02
e. (Keller, Warrack) The amount of gasoline sold by one of your service stations daily is uniformly
distributed between a minimum of 2500 and a maximum of 5500 gallons.
(i) What is the mean and standard deviation of sales? (1.5)
(ii) What is the probability that sales on a given day will fall between 3000 and 3500 gallons? (2)
(iii)What is the probability that sales will be over 4000 gallons? (1)
(iv)What is the probability of sales between 1800 and 2800 gallons? (1)
(v) If you did not do (iv) using cumulative distributions, do it now (2)
(vi)If you own 9 identical service stations, what is the probability that at least one has sales over
4000 gallons? (3)
This is a continuous uniform distribution with c  2500 and d  5500 .
1
1
1


d  c 5500  2500 3000
2
2


d  c
5500  2500
c  d 2500  5500
2

 4000 ,  

 750000
(i)  
12
12
2
2
so   750000  866.025
(ii) P3000  x  3500  
500
 .1667
3000
5500  4000
 .5000
(iii) P x  4000  
3000
2800  2500
 .1000
(iv) P1800  x  2800  
3000
xc
for c  x  d , F x  0 for x  c and F x  1 for x  c and that
(v) Recall that F  x  
d c
F 500 means Px  500 .
2800  2500
 0  .1000 .
P1800  x  2800  F 2800  F 1800 
3000
n x n x
(vi) This is your basic Binomial problem P  x   C x p q . p  .5 , q  .5 and n  9. You can use
the Binomial table for this one.
Px  1  1  P0  1  .00195  .99805
11
251y0242 12/10/02
6. (ctd)
f. (Keller, Warrack) (Extra credit) The number of hours an alkaline battery lasts is exponentially
distributed with a parameter c of 0.05.
(i) What are the mean and standard deviation of a battery’s life? (2)
(ii) What is the probability that the battery will last between 9 and 14 hours? (2)
(iii)What is the probability that it will last for more than 15 hours (1)
(iv)Are you surprised that a jorcillator has two such batteries and that it works as long as one of
the batteries works. What is the probability that the jorcillator lasts more than 15 hours? (2)
(v) (Difficult) What is the probability that the jorcillator lasts between 9 and 14 hours? (The
answer to this will not be published!) (7)
g. The Muggle detector in front of my tower will identify a Muggle correctly as a Muggle 99% of the time
and will identify a Wizard correctly as a Wizard 94% of the time. Assume that 3% of the population are
Wizards and the rest Muggles. The Muggle detector says that you are Wizard. What is the probability that
you really are a Wizard? I only admit Wizards to my tower. (Hint: To do this you need decent notation or a
good tree – Let Y (yes!) be the event that it says you are a Wizard and N (no!) be the event that it says you
are a Muggle. W is the event that you really are a Wizard and M is the event that you are a Muggle. You
need conditional probability to do this.) (6)

Solution: f) You were told in advance that this section would use the exponential distribution. In the
exponential distribution from the outline
  
1
c
Note that this is only for
F x   1  e  cx , when the mean time to a success is
1
.
c
x  0 . There is no probability below zero.
1
1

 20 . (ii) F 9  Px  9  1  e .059   1  e 0.45  1  .63763
c .05
F 14  Px  14  1  e .0514  1  e .70  1  .49659 . So
P9  x  14  Px  14  Px  9  .63763  .49659  .14104
(i)
  


(iii) Px  15  1  F 15  1  1  e
e
 .47237
(iv) The probability that one component lasts less than 15 hours is 1-.47237 = .52763. The probability that
.0515
0.75
both fail before 20 hours is .52763  .27840 . The complement of this is .72160.
(v) Hah!
2
g. Not a hard problem. You are given
PN M   .99 , PY W   .94 , PM   .97 and
PW   .03 . This implies that PY M   .01 and PN W   .06 You are asked for PW Y . By
Bayes’ Rule
PW Y  
PY W PW 
.
PY 
PY   PY W PW   PY M PM   .94.03  .01.97  .0282  .0097  .0379 So
PW Y  
PY W PW 
PY 
 .0282 

  .7441
 .0379 
Another way to do this is to say that of 10000 people who come to my tower, 9700 will be Muggles and 300
will be Wizards. Of the Muggles, 99% or 9603 will be identified as Muggles. The remaining 1% or 97 will
be wrongly identified as Wizards. Of the 300 Wizards, 94% or 282 will be correctly identified. So 97 + 282
= 379 are identified as Wizards. Of the 379, 282 or 74.4% actually are Wizards.
12