251y0342 12/08/03
ECO 251 QBA1
FINAL EXAM
DEC 12 2003
Name KEY _
Class ________________
Exam is normed on 75 points. There are actually 128 possible points.
Part I. Do all the Following (14 Points) Make Diagrams! Show your work!
x ~ N 4, 5 .
5.20  4 
  5.20  4
z
 P 1.84  z  0.24 
1. P5.20  x  5.20   P 
5
5 

 P1.84  z  0  P0  z  0.24   .4671  .0948  .5619
For z make a Normal curve centered at zero and shade the area from -1.64 to 024. It will be on both
sides of zero, so we add.
22  4 
1  4
z
 P 0.60  z  3.60 
2. P1  x  22   P 
5 
 5
 P0.60  z  0  P0  z  3.60   .2257  .4998  .7255
For z make a Normal curve centered at zero and shade the area from -0.60 to 3.60 .It will be on both
sides of zero, so we add.
4.85  4 

 Pz  0.17 
3. Px  4.85   P  z 
5 

 Pz  0  P0  z  0.17   .5  .0675  .4325
For z make a Normal curve centered at zero and shade the area above 0.17 .It will be on one side of
zero, so we subtract
5.20  4 

 Pz  0.24 
4. F 5.20  (Cumulative Probability) Px  5.20   P  z 
5 

 Pz  0  P0  z  0.24   .5  .0948  .5948
For z make a Normal curve centered at zero and shade the entire area below 0.24 . It will be on both
sides of zero, so we add.
11 .3  4 
0  4
z
 P 0.80  z  1.46 
5. P0  x  11 .3  P 
5 
 5
 P0.80  z  0  P0  z  1.46   .2881  .4279  .7160
For z make a Normal curve centered at zero and shade the area from -0.80 to 1.46 .It will be on both
sides of zero, so we add.
6. x.055 (Find z .055 first) Make a diagram for z . Show a Normal curve with a mean of zero in its center.
Remember that z .055 is a point with 5.5% above it and 94.5% below it. Since 50% of the distribution is
bellow zero P0  z  z.055   .9450  .5  .4450 . According to the Normal table P0  z  1.60   .4452 ,
which is close enough. So z.055  1.60 and x    z.055  4  1.60 5  4  8.  12.00 .
Check:
12 .00  4 

Px  12 .00   P  z 
  Pz  1.60   Pz  0  P0  z  1.60   .5  .4452  .0548  .055
5


It is still true that a probability cannot be negative or above 1.
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251y0342 12/08/03
7. A symmetrical region around the mean with a probability of 22% Make a diagram for z . Show a
Normal curve with a mean of zero in its center. You are looking for two points with a probability of .2200
between them. Since the area is split in two by zero, we can say P0  z  z .39   .11 . Since the area above
zero is .5000, and there is a probability of 11% between our point and zero, there must be 50% - 11% =
39% above this point, so it is known as z .39 . If we look at the Normal table, we will find that the closest we
can come to 11% is P0  z  0.28   .1103 . This means that z .39 .  0.28 and that
P z .39  z  z.39   .22 . Your diagram for z will show two points  z .39 and z .39 with zero between them.
You will mark the area between z .39 and zero with 11% and the area above z .39 with 39%. Similarly mark
the area between  z .39 and zero with 11% and the area below  z .39 with 11%. To get from z to x , use
the formula x    z . In this case x    z.39  4  0.28 5  4  1.40 or 2.60 to 5.40.
5.40  4 
 2.60  4
z
 P 0.28  z  0.28 
Check: P2.60  x  5.40   P 
5 
 5
 P0.28  z  0  P0  z  0.28   .1103  .1103  .2206  22%
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II. (10 points+-2 point penalty for not trying part a .) Show your work!
The following numbers represent a random sample of 5 countries from The World Guide – 2003/2004 and
give per capita income (PCI) in thousands of US dollars x  and Human Development Index (HDI - based
on life expectancy, access to education and income)  y  .
1
2
3
4
5
Kazakhstan
Greece
Estonia
Samoa
Burundi
x
y
5.9
16.5
10.6
5.0
0.6
0.742
0.881
0.812
0.701
0.309
These sums have been calculated for you.
 y  3.445 s x2  36.697, x  7.72, and
x2
34.81
272.25
112.36
25.00
0.36
 x  38.6 ,  x
2
 444 .78 ,
y  .6890 . Please calculate the following:
a. The sample standard deviation of y (4)
b. The sample covariance between x and y . (3)
c. The sample correlation between x and y . (2)
d. Given the size and sign of the correlation, what conclusion might you draw on the relation
between PCI and HDI if this were the only evidence available? (1)
e. Assume that the HDI in all 5 countries rose by an additive factor of 0.001 and the PCI rose by a
multiplicative factor of 1.1. What would be the new values of x , s x , s xy and rxy . Use only the values you
computed in a-c and rules for functions of x and y to get your results. If you state the results without
explaining why, or change x and recompute the results, you will receive no credit. (4).
Solution: Finish the calculations
1
2
3
4
5
Kazakhstan
Greece
Estonia
Samoa
Burundi
x
y
5.9
16.5
10.6
5.0
0.6
0.742
0.881
0.812
0.701
0.309
x2
y2
xy
34.81 4.3778
272.25 14.5365
112.36 8.8092
25.00 3.5050
0.36 0.1854
31.2119
0.550564
0.776161
0.659344
0.491401
0.095481
2.572951
 xy by multiplying  x by  y or that you can get  x by squaring
 x or that you can get  x  x 2 by taking  x  x and squaring it, you should take the course
2
If you think you can get
again. Remember that a variance cannot be negative.
a) s x  36 .697  6.0578 .
s 2y 
y
2
 ny 2
n 1

b) We found above that
c) rxy
s xy
2.572951  50.6890 2 0.199346

 0.498365
4
4
 xy  2.572951
1.1541


 .8531
s x s y 6.0576 0.2232 
, so s xy 
s y  0.498365  0.2232
 xy  nxy  2.572951  57.72 0.6890   1.1541
n 1
4
This must be between -1 and one.
d) If we square the correlation we get 0. 7284, which in a zero to one scale is fairly high. It would be
reasonable to belive that there is a relationship between higher PCI and higher HDI. However, you will find
out next term that the sample is too small to lead to much of a conclusion.
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251y0342 12/08/03
e) From the syllabus supplement article:
“Let us introduce two new variables, w and v , so that w  ax  b , and v  cy  d , where
a, b, c, and d are constants. From the earlier part of this section we know the following:
 w2  Varw  a 2Varx  a 2 x2
 w  Ew  Eax  b  aEx   b
 v2  Var v   c 2Var  y   c 2 y2
 v  E v   E cy  d   cE y   d
To this we now add a new rule: Covw, v   wv  acCovx, y   ac xy
To find the correlation between w and v , recall that  wv 
and  v2  c 2 y2 , then
 w2  a 2 x2
 wv 
 wv
. But since
 w v
ac xy
a 2 x2 c 2 y2

ac xy
ac  x y

ac  xy
 signac xy .”
ac  x y
Since these rules work for sample statistics too, then w  1.1x  0 , v  1y  0.001 , so a  1.1, b  0, c  1
and d  0.001 . w  1.1x  1.17.72   8.492
sw 
rwv 
1.12 36.697   1.1
acs xy
a 2 s x2 c 2 s 2y

s w2  1.12 s x2  1.12 36.697 , so
36 .697  6.663 . s wv  1.11s xy  1.11.1541  1.2695 .
acs xy
ac s x s y

ac s xy
 signacrxy  .8535
ac s x s y
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251y0342 12/08/03
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. A company manufactures small motors for use in snow blowers. A sample of 16 is taken from a batch of
5000 motors and the sample mean horsepower is found to be 2.85 hp. Find a 98% confidence interval for
the mean horsepower under the following circumstances.
a. The sample standard deviation is found to be 0.09. (4)
b. The sample standard deviation is found to be 0.09, but the sample of 16 is taken from a batch
of only 101motors. (4)
c. The sample of 16 still has a sample mean of 2.85, but it is taken from a batch of 5000 motors
with a population standard deviation that is assumed to be 0.1. (4)
d. Assume that the population mean is actually 2.95 hp, the population standard deviation is
assumed to be 0.1, the sample is of size 16, and the batch is of 5000 motors, what is the
probability that the sample mean will lie between 2.80 and 2.90? (3)
e. For the same distribution as in d), what is the probability that an individual machine has a
horsepower between 2.80 and 2.90? (2)
f. Under the circumstances of d), find an 89% confidence interval for the mean. (Hint: see page
1) Assume x  2.85 (2)
19- 47
Solution: The solution to problem O1 gives the following formulas:
i)   x  z  x and  x 
2
x
when  is known and the sample is small relative to the
n
population. ii)   x  z  x and  x 
2
x
n
N n
when  is known and the sample is large relative to
N 1
s
the population. iii)   x  tn1 s x and s x  x
2
n
when  is unknown and the sample is small relative
N n
when  is unknown and the sample is
N 1
s
to the population and iv)   x  tn1 s x and s x  x
2
n
15
large relative to the population. z.01  2.327 , t .01
 2.602 . Many of you still do not know that s is the
sample standard deviation and that  is the population standard deviation.
s
0.09
 0.0225
a. n  16, N  5000 , x  2.85, s  0.09 ,   .02 , s x  x 
4
n
  x  t n1 s  2.85  2.6020.0225  2.85  0.059 or 2.791 to 2.909.

2
x
b. n  16, N  101, x  2.85, s  0.09 ,   .02 , s x 
sx
N  n 0.09 101  16

 0.02074
N 1
4
101  1
n

n 1
  x  t  s x  2.85  2.6020.02074  2.85  0.054 or 2.796 to 2.904
2
c. n  16, N  5000 , x  2.85,   0.1,   .02 ,  x 
x

0.1
 0.025
4
n
  x  z 2  x  2.85  2.3270.025  2.85  0.058 or 2.792 to 2.908.
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251y0342 12/08/03
Many of you thought that these questions were about confidence intervals. As usual, you can’t answer a
question you haven’t read.

0.1
 0.025 P2.80  x  2.90 
d. n  16, N  5000 ,   2.95,   0.1,   .02 ,  x  x 
4
n
2.90  2.95 
 2.80  2.95
 P
z
  P 6.00  z  2.00   .5  .4772  .0228
0.025 
 0.025
2.90  2.95 
 2.80  2.95
z
e. n  1, N  5000 ,   2.95,   0.1,   .02 P2.80  x  2.90   P

0.1
0.1


 P1.50  z  0.50   .4332  .1915  .2417
f. n  16, N  5000 , x  2.85,   0.1,   .11 ,  x 
z 2  z.055  1.60.
x

0.1
 0.025 . On page 1 you found that
4
n
So   x  z  x  2.85  1.60.025  2.85  0.040
2
or 2.81 to 2.89.
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251y0342 12/08/03
2. The following joint probability table describes x and y.
x
y
1
2
Px 
1
2
3
4
.09 .18 .27 .36 


 .01 .02 .03 .04 
P y 
Let A1 be the event that x  1 and B 1 be the event that that y  1. Find the following:
Solution: Complete the table as below.
We can summarize our results as follows:
x
2
1
2
3
4
P y  yP y  y P y  Px   1 ,
.09 .18 .27 .36 
1
.9
.9
.9  x  E  x  
xPx   3.0
y


2
.1
.2
.4
2
2
 .01 .02 .03 .04 
Ex 
x Px   10 .00
P y   1 ,
Px 
.1 .2 .3 .4
1.0
1.1
1.3
 y  E y  
yP y   1.1 and
xPx 
.1 .4 .9 .16
3.0

  
x Px 
2
.1 .8 2.7 6.4


   y
E y
10
2

2
P y   1.3
a) P A1  B1  (1) Solution: From the table Px  1  P A1   .10 , P y  1  PB1   .9 and
Px  1   y  1  P A1  B1   .09 . By the addition rule P A1  B1   P A1   PB1   P A1  B1 
 .1  .9  .09  .91
b) P A1  B1  (1) Solution: From the table Px  1   y  1  P A1  B1   .09


c) P A1 B1 (1) Solution: By the multiplication rule PA1 B1  


P A1  B1  .09

 .1 . At this point you
PB1 
.9
may notice that P A1 B1  P A1  which is a definition of independence.


d) P A1  B1 (1) Solution: From the table we could just add together those parts of B1 that are not in
A1 . These are P A2  B1  + P A3  B1  + P A4  B1  = .18 + .27 + .48 = .81. But since we said x and

  
y are independent, we could also write P A1  B1  P A1 PB1   1  .1.9  .81
e)  x , the mean or expected value of x (2) ) Solution: From calculations above
 x  E x  
 xPx  3.0
  x
f)  x , the standard deviation of x (2) ) Solution: From above E x 2 
 
2
Px   10 .00 , so
 x2  E x 2   2  10  32  1 . and  x  1  1. Many of you were still trying to compute a sample
variance here. There is no sample here, so that is unreasonable.,
g)  xy , the covariance of x and y (3) ) Solution: We could say
E xy  
 .09 11
 xyPxy    .0112
 .1821 .  27 31  .36 41  .09  .36  .81  1.44 

 3.3
 .02 22  .0332  .04 42  .02  .08  .18
 .32 
and  xy  Covxy  Exy   x  y  3.3  31.10  0 , or we could just note that since x and y are
independent, their covariance must be zero.
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251y0342 12/08/03
h)  xy , the correlation between x and y (1) Solution: We could say that  xy 
 xy
 x y
and that
 xy  Covxy   0 while  x and  y are not zero, so  xy  0 or we could just say that if x and y are
independent, their correlation is zero. Note that 1   xy  1 always!
i) The mean and variance of x  y (2) As in the 3rd graded assignment,
 
Ex  y   Ex  E y    x   y  3  1.1  4.1 and since  y2  E y 2   2  1.3  1.12  0.09
Var x  y 
  x2
   2 xy  Var x   Var  y   2Covx, y   1  0.09  20  1.09 .
2
y
j) The probability that x  y  4. (3) Solution; The table appears below with the totals of x and y after the
probabilities. Only the three probabilities toward the upper left give us numbers below 4. So the probability
is .27 + .36 + .02 + .03 +.04 = .72.
x
y
1
2
Px 
1
2
3
4
.09  2 .18  3 .27  4 .36  5


 .01  3 .02  4 .03  5 .04  6
P y 
17-64
8
251y0342 12/08/03
3. Assume that a population is 10% defective.
a. If the population consists of 90 items and you take a sample of ten, what is the chance of at
least one defective item?(3)
b. If the population consists of many items and you take a sample of 10, what is the chance of at
least one defective item?(2)
c. If the population consists of many items and you take a sample of 90 items, what is the chance
of at least five defective items? Show that you can use a Poisson distribution here and use it. (2)
d. Show that you can use a Normal distribution here and use it. (3)
e. We are starting a Bayes’ Rule problem, but it starts real easy. When a process is out of control
(OC ) its operation is described by a process in which the probability that an item is defective is
p  .1. In other words, if we take a sample of 10 the probability of one defective item is
described by the binomial distribution with p  .1 and n  10 . What is this probability ? (2) Call this

probability P x  1 OC

f. When the process is in control, the probability of exactly one defective item in a sample of 10 is described
by a binomial distribution with p  .01 and n  10 . What is this probability? (1) Call this


probability P x  1 OC
g. Now assume that the (prior) probability of the process being in control is .95 and the
probability of the process being out of control is .05, and that you take a sample of 10 and find
exactly one defective item. Find P x  1 OC and P x  1 OC . Then use Bayes' rule to figure
out the probability that the process is out of control when you take a sample and find exactly one
is defective. (5)
19 -83




Solution: a) Hypergeometric Distribution - N  90 , M  Np  90.10   9 ,
P0 
81
C 09 C10
90
C10
Px  
C xM C nNxM
so
C nN
81!
81!
1
9 81
C C
71 !10!
71 !10!
. Therefore Px  0  1  P0  1  0 9010  1 

90!
90!
C10
80!10!
80!10!
1
81  80  79  78  77  76  75  74  73  72
 1  0.3284  .6719
90  89  88  87  86  85  84  83  82  81
b) According to the binomial table for p  .1 and n  10 , Px  0  1  P0  1  .34868  .6513 .
 1
c) This is a Binomial problem with p  .1 and n  90 , We test to see that n
  np  90.1  9 , so that Px  5  1  Px  4  1  .05496  .94504
p
 90
.1
 900  500 Note that
d) The same problem can be approximated using the Normal distribution. Note that
  np  90.1  9  5 and nq  90.9  81  5 and  2  npq  9.9  8.1 , so that it is approximately true



4.5  9 
that x ~ N 9, 8.1 . Extend the interval by 0.5 and find Px  4.5  P  z 
  Pz  1.58 
8.1 

 P1.58  z  0  Pz  0  .4429  .5  .9429
9
251y0342 12/08/03


e) P x  1 OC for n  10 and p  .1 is Px  1  Px  0  .73610  .34868  .38742


f) P x  1 OC for n  10 and p  .01 is Px  1  Px  0  .99573  .90438  .09135


g) Bayes rule says P OC x  1 

Px  1 OC POC 
Px  1

Px  1 OC POC 

 
Px  1 OC POC   P x  1 OC P OC
.38742 .05 
 .1825
.38742 .05   .09135 .95 
10
251x0342 12/08/03
4. A salesperson makes sales to 60% of her prospects.
number.
Let
x
a. If she calls on 10 people, what is the mean and the variance of the number of sales she makes?(2)
b. If she calls on a large number of people, what is the probability that her first sale is among her first three calls? (2)
c. What is the probability that she makes at least 6 sales in 10 calls? (Don’t give me exactly six – or exactly any other
)(2)
d. What is the probability that she makes at least 54 sales in 90 calls? (3)
e. Assume that she calls on 10 people every day, so that the mean and variance of her daily sales are as you stated in a).
be the sample mean number of sales she makes a day over 100 days. What does the central limit theorem say the expected
value and standard error of x will be over 100 days? (2)
e. What is the probability that x will exceed 6.2? (3) 14 - 97
Solution: a. If she calls on 10 people, what is the mean and the variance of the number of sales she
makes?(2) This is binomial n  10, p  .6.   np  10.6  6  2  npq  6.4  2.4.
b. If she calls on a large number of people, what is the probability that her first sale is among her
first three calls? (2) This is geometric: Px  3  1  q 3  1  .43  1  .064  .936 .
c. What is the probability that she makes at least 6 sales in 10 calls? (Don’t give me exactly six –
or exactly any other number. )(2) We have to do this problem in terms of failures. She makes
between 6 and 10 sales (successes) . This is between zero and four failures. If n  10 and p  .4 ,
P0  x  4  Px  4  .63310 .
d. What is the probability that she makes at least 54 sales in 90 calls? (3) Normal approximation
to the Binomial distribution. If p  .6 and n  90, there are more than 5 successes and failures,
so we can use the Normal distribution.   np  90.6  54 .  2  npq  54.4  21.6.

53 .5  54 
Px  53 .5  P  z 
  Pz  0.10   .5  .0398  .5398 .
21 .6 

e. Assume that she calls on 10 people every day, so that the mean and variance of her daily sales
are as you stated in a). Let x be the sample mean number of sales she makes a day over 100
days. What does the central limit theorem say the expected value and standard error of x will be
over 100 days? (2) Ex     6 ,  x 

n

2.4
 .1549
100
6.2  6 

 Pz  1.29 
e. What is the probability that x will exceed 6.2? (3) Px  6.2   P  z 
.1549 

 .5  .4015  .0985
11
251x0342 12/08/03
5. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that the
particular jorcillator we have works as long as either component works (This has changed!).
The probability of the phillinx failing is given by a continuous uniform distribution with a range between
c  0.5 and d  2.0, so , if x represents the life of the phillinx, the probability of the phillinx failing in the
first year is P0  x  1 , the probability of it failing in the second year is P1  x  2 and the probability
of it lasting beyond the 2nd year is Px  2 .
The probability of the flubberall failing is given by a Normal distribution with a mean of 1.5 and a standard
deviation of 0.25, so , if y represents the life of the phillinx, the probability of the phillinx failing in the
first year is P0  y  1 , etc.
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 in the first year is .7, the probability of both
components failing in the first year is (.1) (.7) =.07 (This is not necessarily the probability that the
jorcillator will fail in the first year!)
a) What is the probability the Phillinx will fail in the first year? The second year? After the
second year? (1.5)
b) What is the probability the Flubberall will fail in the first year? The second year? After the
second year? (1.5)
c) What is the probability that the jorcillator will fail in the first year? (2)
d) What is the probability that the jorcillator will fail in the second year? (2)
e) What is the probability that the jorcillator will last beyond 2 years? (2)
f) If my firm owns two Jorcillators, what is the probability that at least one lasts beyond two
years? (3)
Solution: a) The probability of the phillinx failing is given by a continuous uniform distribution with a
range between c  0.5 and d  2.0, so , if x represents the life of the phillinx, the probability of the phillinx
failing in the first year is P0  x  1 , the probability of it failing in the second year is P1  x  2 and the
probability of it lasting beyond the 2nd year is Px  2 . What is the probability the Phillinx will fail in the
first year? The second year? After the second year?
1
1
1
2


 . Make a diagram of a box with height 2 between 0.5
The density is
3
d  c 2.0  0.5 1.5 3
and 2.0. Name the following events A1 : Phillinx fails in first year; A2 : Phillinx fails in second year and
A3 : Phillinx fails after second year. Wake up! Some of you have been caught on this 3 times. Make a
diagram with a box of height 2 3 between 0.5 and 2.0 and nothing outside of that range.
(i)
Event A1 : Shade the part of the box below 1. It has area P A2  
(ii)
1  0 .5
.5 1


2  0 .5 1 .5 3
Event A2 : Shade the part of the box above 1. It has area P A2  
(iii)
2 1
1
2


2.0  0.5 1.5 3
Event A3 : Shade the part of the box above 2. Surprise! There is no area above 2. It has
P0  x  1 
P1  x  2 
area P A3   Px  2  0.
Note that these probabilities add to one.
12
251x0342 12/08/03
b) The probability of the flubberall failing is given by a Normal distribution with a mean of 1.5 and a
standard deviation of 0.25, so , if y represents the life of the phillinx, the probability of the phillinx failing
in the first year is P0  y  1 , etc. What is the probability the Flubberall will fail in the first year? The
second year? After the second year? (1.5) y ~ N 1.5, 0.25  Name the following events B1 : Flubberall fails
in first year; B 2 : Flubberall fails in second year and B3 : Flubberall fails after second year.
(i)
(ii)
1  1.5 
 0  1.5
z
 P 6.00  z  2.00 
Event B1 : PB1   P0  y  1  P 
0.25 
 0.25
 P6.00  z  0  P2.00  z  0  .5000  .4772  .0228
Event B 2 : PB2   P1  y  2
2  1.5 
1  1.5
 P
z
 P 2.00  z  2.00   .4772  .4772  .9544
0.25 
 0.25
2  1.5 

 Pz  2.00   .5  .4772  .0228
Event B3 : PB3   Px  2  P  z 
0.25 

Note that these probabilities add to one.
Now, note that whichever component fails last downs the jorcillator. The following nine joint events can be
enumerated.
Joint Event
Probability
Jorcillator fails
1 .0228   .0076
In Year 1
A1  B1
3
(iii)
A1  B2
A1  B3
A2  B1
A2  B2
A2  B3
A3  B1
A3  B2
A3  B3
13 .9544   .3181
13 .0228   .0076
 3 .0228   .0152
2
2 3 .9544   .6363
2 3 ..0228   .0152
0.0228   0
0.9544   .0
0.0228   0
In Year 2
After Year 2
In Year 2
In Year 2
After Year 2
After Year 2
After Year 2
After Year 2
If we add together the probabilities of possible joint events, we find the following.
c) Probability of failing in first year = .0076
e) Probability of it lasting beyond second year = .0076 + .0152 = .0228 or
P A3  B3   P A3   PB3   P A3  B3   0  .0228  0  .0228
d) Probability if it failing in second year = 1 - .0076 - .0228 = .9696 or .3181 + .0152 + .6363 = .9696
or P A2  B2   P A2  B3   P A3  B2   [ P A2   PB2   P A2  B2 ]  P A2  B3   P A3  B2 
 [ 2 3  .9544  .6363 ]  .0152  0  .9848  .0152  .9696
f) There are at least 2 ways to do this. The probability of one Jorcillator failing before the end of the 2 nd
year is 1 - .0228 = .9772. The probability that both fail is .9772 2  .9549 . The probability that at least one
does not fail is 1 - .9549 = .0451. You could also do this as .0228  .0228  .0228 2  .0451 .
13
251x0342 12/08/03
6. The weights of players on my university football team have a Normal distribution with a mean of 115
pounds, with a standard deviation of 24 pounds. (We don’t win many games.)
team
weight
chance
value
will
4140
a. The opposing team has weights that also have a Normal distribution with a much higher mean. If Cadwallader is on my
and he is tackled by someone who is on the opposing team, what is the probability that the tackler is above the median
for his team? (1)
b. If Spike, the guy who tackled Cadwallader, weighs 210 pounds and Spike tackles someone on my team, what is the
that Spike tackles a person who weighs half of what he weighs.? (2)
c. Lets say that I take a sample of 36 from my team ( consider the team to have lots of players on it) what is the expected
and standard deviation of the sample mean?(1.5)
d. What is the probability that that the sample mean of this sample exceeds 120 pounds?(2)
e. If I put 36 members of the team on a bus and the bus has a capacity of 4320 pounds, what is the probability that the bus
be overloaded? (1) 8.5
f. If I randomly pick 36 members of the team and weigh the entire group, the expected value of the sum of their weights is
pounds. What is the standard deviation of the sum of their weights? (2)
g. If you did not do e) using the mean and variance in f), do it now. (2) (If you did do e) this way, you got 2 points for it.)
12.5
e
h. If you learned that this sample of 36 was taken from a team with 80 members, how would this affect your answer to c, d,
and f? Do it! (Some of this involves hard thinking.) (6.5) 19
i. If we implement mandatory testing for steroids, even though we believe that 95% of our players do not use them
( P U  .05 ), we will use a drug testing procedure that has a 93% chance of
testing positive for a user. (Use
 
event that someone tests positive and U for the event that someone is a user.) The procedure has a 3% chance
Po for the
of testing positive for a non - user, a 7% chance of testing negative for a user and a 97% chance of testing negative for a non - user.
Cadwallader tests negative. If we use the 95% as a prior probability, what is the chance that
Cadwallader is a
steroid user?
j. Actually, my university football team gets an average of 1.2 goals per game.
(i) If we assume that the Poisson distribution applies, what is the probability that we get
at least one goal in a game?
(ii) What is the probability of at least one goal in the first quarter?
(iii) Why might the Poisson distribution not apply in this case?
Solution: a. The opposing team has weights that also have a Normal distribution with a much higher
mean. If Cadwallader is on my team and he is tackled by someone who is on the opposing
team, what is the probability that the tackler is above the median weight for his team? (1)
Since the median is the middle number, the probability is .5.
b. If Spike, the guy who tackled Cadwallader, weighs 210 pounds and Spike tackles someone on
my team, what is the chance that Spike tackles a person who weighs half of what Spike weighs?
105  115 

(2) x ~ N 115 ,24  P x 
  Pz  0.42   .5  .1628  .3372
24


c. Lets say that I take a sample of 36 from my team ( consider the team to have lots of players on
it) what is the expected value and standard deviation of the sample mean?(1.5)


24 
  N 115 ,4
x 
. So x ~ N 115 ,

n
36 

d. What is the probability that that the sample mean of this sample exceeds 120 pounds?(2)
120  115 

Px  120   P z 
  Pz  1.25 
4


e. If I put 36 members of the team on a bus and the bus has a capacity of 4320 pounds, what is
the probability that the bus will be overloaded? (1) 8.5
f. If I randomly pick 36 members of the team and weigh the entire group, the expected value of
the sum of their weights is 4140 pounds. What is the standard deviation of the sum of their
weights? (2)
Since variances add (assuming independence), the variance will be
Var
 x  n
2
x
 
 36 24 2  20736 . So  x  n  x   20736  144
14
251x0342 12/08/03
g. If you did not do e) using the mean and variance in f), do it now. (2) (If you did do e) this way,
you got 2 points for it.
4320  4140 

P
x  4320  P z 
  Pz  1.25 
144


h. If you learned that this sample of 36 was taken from a team with 80 members, how would this
affect your answer to c, d, e and f? Do it! (Some of this involves hard thinking.) (6.5)

120  115 
N n
44

c’ )  x  x
4
 2.9852 d’) Px  120   P z 
  Pz  1.67 
2.9852 
79

n N 1


Since the answer to d’) is the same as the answer to e’),
f’ )  x  n  x   n
g’) P
x
n
x
x

 x  n
n x
and
N n
N n
44
 n  x 
 624 
 107 .467 and
N 1
N 1
79
 4140 
 x  4320   P z  4320
  Pz  1.67 
107 .467 
i. If we implement mandatory testing for steroids, even though we believe that 95% of our
players do not use them ( PU   .05 ), we will use a drug testing procedure that has a 93%
chance of testing positive for a user. (Use Po for the event that someone tests positive and U for
the event that someone is a user.) The procedure has a 3% chance of testing positive for a non user, a 7% chance of testing negative for a user and a 97% chance of testing negative for a non user. Cadwallader tests negative. If we use the 95% as a prior probability, what is the chance that
Cadwallader is a steroid user?
Bayes’ Rule.

U
.97 .95 
.9215
 PPoPUPoPU   PPo U PPPoU U PPPo


 .9962




.
97
.
95

.
07
.
05
.
9250
U PU 
P U Po 
j. Actually, my university football team gets an average of 1.2 goals per game.
(i) If we assume that the Poisson distribution applies, what is the probability that we get
at least one goal in a game?
(ii) What is the probability of at least one goal in the first quarter?
(iii) Why might the Poisson distribution not apply in this case?
(i) Poisson (1.2) Px  0  1  P0  1  .30119  .69881
(ii) Poisson (.3) Px  0  1  P0  1  .74082  .259118
(iii) Unlikely that the probability holds steady quarter to quarter. Whether the team scored in the
first quarter is likely to affect what happens later.
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