251y0821 4/4/08
ECO251 QBA1
SECOND HOUR EXAM
March 27, 2008
Solution (30 pages)
Name: __KEY______________
Student Number: ____________________
Hour of Class Registered: _____________
Part I: (3 points) 2 point penalty for not trying. (Problem modified from Webster)
1. A professor wishes to see if the number of absences of his students is related to the number of miles the
students live from the classroom. (Is it true that a miss is as good as a mile?) He takes a sample of eight
students and gets the results below. Compute the sample standard deviation of ‘Miles.’ Show your
work! (3)
Row
y
x
Row
1
2
3
4
5
6
7
8
Miles
5
6
2
0
9
12
16
5
Misses
2
2
4
5
4
2
5
2
7
For your convenience the sum of the first seven numbers in x is
x
 50 and the sum of the first
i 1
7
x
seven numbers squared is
2
 546 .
i 1
Solution: If you took advantage of the numbers that were given
x
2
 x  50  5  55 and
 546  52  546  25  571 . Most people did not take advantage of the numbers computed for
them . In fact, many recomputed the variance of y, which was given to you. It doesn’t make much sense to
complain about the length of the exam if you did not use the time savers given to you.
If you wasted time by not using this freebie, the results are as below. You, of course, do not need the y 2 ,
x  x , or x  x 2 columns.
Miles Misses
Row
1
2
3
4
5
6
7
8
x2
y
x
5
6
2
0
9
12
16
5
55
2
2
4
5
4
2
5
2
26
So the mean of x is x 
s x2 
x
2
 nx
n 1
2

y2
25
36
4
0
81
144
256
25
571
4
4
16
25
16
4
25
4
98
n
8
xx
 x  x 2
-1.875
3.5156
-0.875
0.7656
-4.875 23.7656
-6.875 47.2656
2.125
4.5156
5.125 26.2656
9.125 83.2656
-1.875
3.5156
0.000 192.8748
xy
10
12
8
0
36
24
80
10
180
 x  55  6.875 . If you used the computational formula, you got
571  86.875 2 192 .875

 27 .5536 and s x  27 .5536  5.24915 . If you wasted
7
7
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251y0821 4/4/08
more time by using the definitional formula, you got
s x  27 .55354  5.24935 . Minitab gets
2. Compute

s x2
s x2
 x  x 

n 1
2

192 .8748
 27 .55354 and
7
 27.55357 and s x  5.24915.
7
x y . For your convenience the sum of the first seven numbers in xy is
 xy
 170 . (1)
i 1
 x y  170  52  180 . If you wasted
your time by computing the sum as shown on the previous page you still got  x y  180
Solution: If you took advantage of the numbers that were given
Exhibit 0: According to Minitab y  3.250 and s 2y  1.92857 .!!!!!
3. Compute the sample covariance. What does this show about the relationship between Miles and Misses?
(2.5)
Solution: We already know that the mean of x is x  6.875 .
 x  x  y  y    xy  nx y
180  86.875 3.250  1.25


 0.17857 . This implies that there
7
7
n 1
n 1
is some tendency of ‘Misses’ to rise as ‘Miles’ does. We cannot be sure of how strong this relationship is.
s xy 
4. Compute the sample correlation between Misses and Miles. What does this show about the strength of
the relationship between Miles and Misses? (2.5)
Solution: rxy 
s xy
sx s y

0.17857

0.17857 2

27 .55357 1.92857 
.000600  .024496 . The square of
27 .55357 1.92857
this is .000600 on a zero to one scale. This is very weak.
5. The amount of time in minutes t  it takes in minutes to get to campus can be approximated by the
equation t  20  3x , where x is the number of miles. Using only the results that you have so far find:
(i) The sample variance of the time
(ii) The covariance of time with ‘misses’
(iii) The correlation of time with misses.
Please do not waste our time by computing a column with the 8 values that time would take and then
recomputing the values of these three statistics. Many people did this anyway! Most of those who did not
use the method you were taught (as below) got the wrong answers. There is a much easier way.
Solution: We know that if w  ax  b and v  cy  d , Varw  a 2Varx , Covw, v   acCovx, y  , and
Corr w, v   Sign(ac)Corr x, y  . If our w  t  20  3x and v  1y  0 , a  3 and c  1 .
(i) st2 , the sample variance of time.
s x2  27.55357 . So Var t   32 Var x   927 .55357   247 .98213 .
(ii) s ty , the covariance between time and misses.
s xy  0.17857 . So Covt , y   31Covxy   30.17857   0.53571 .
(iii) rty , the correlation between time and misses.
rxy  .024496. So Corrt , y   Sign(31)Corrx, y   10.024496   .024496 .
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Part II: (24.5+ points) Do most of the following: All questions are 2 points each except as marked. Exam
is normed on 50 points including take-home. [Bracketed numbers are a point total.] If you answer ‘None of
the above’ in most questions, you should provide an alternative answer and explain why. You may receive
credit for this even if you are wrong. This section is long and few people will finish. You only need 24.5
points for a perfect score in this section, so look them over! As always any extra points in this section wrap
around to other sections.
1. If both events A and B are possible and P A  B   0 , the following cannot be true.
a) A and B are complements.
Note that complements are mutually exclusive.
b) A and B are mutually exclusive.
c) *A and B are independent
d) All of the above could be true
e) None of the above can be true
f) Not enough information
Exhibit 1: (Webster) Consider the Following table. It divides the 500 employees of a corporation by their
gender and their job classification. The events are: A1 'Male', A2 'Female', B1 'Staff Job', B2 'Line Job' and
B3 'Auxiliary Job'.
Staff Line Aux
Male  120
150
30 


Female  50 140
10 
Hint: Before doing anything on the following four problems, you may want to make these into proportions
or probabilities. Be sure that you look at both sides of the answer you choose. Guessing is very unlikely to
get you anywhere.
Comment: Before doing anything, you may want to total the rows and columns.
Staff Line Aux Total

Male
120
150
30  300
.
You certainly should make these into probabilities by dividing


Female  50 140
10  200
Total
170 290
40
500
B1 B 2 B3 Total
A1 .24 .30 .06 
.60
by 500 and (if necessary) totaling rows and columns.


A2 .10 .28 .02 
.40
Total .34 .58 .08
1.00
From these we see that P A2  B3 
30
200
40
 .02 , P A2  
 .40 P B3 
 .08
500
500
500
30  50  140  10 230
P A2  B3 

 .46 or P A2  B3  .06  .02  .28  .10  .46 or
500
500
P A2  B3  P A2  PB3  P A2  B3  .40  .08  .02  .46 or
P A2  B3 .02
10 P A2  B3 .02
10


 .25 and PB3 A2 


 .05 .
40
PB3
.08
200
P A2
.40
2. There is no question 2. There are some things I’m just not good at.
PA2 B3 
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251y0821 4/4/08


3. Find the probability that a woman is an Auxiliary employee. Comment: This is P B3 A2 . We found


out that P B3 A2  .05 above.
a. PA2  B3  .02
b. PA2  B3  .46
c. PA2  B3  .25
d. PA2  B3  .05
e. PA2  B3  .02
f. PA2  B3  .46
g. PA2  B3  .25
h. PA2  B3  .05


i. P B3 A2  .02


k. PB3 A2  .25
l.* PB3 A2  .05
m. PA2 B3  .02
n. PA2 B3  .46
o. PA2 B3  .25
p. PA2 B3  .05
j. P B3 A2  .46
q. None of the above – fill in the
correct answer
4. Find the probability that an employee is both a woman and an Auxiliary employee. Comment: This is
PA2 B3  . We found out that P A2  B3  .02 above.
a.* PA2  B3  .02
j. P B3 A2  .46
b. PA2  B3  .46
c. PA2  B3  .25
d. PA2  B3  .05
e. PA2  B3  .02
f. PA2  B3  .46
g. PA2  B3  .25
h. PA2  B3  .05


i. P B3 A2  .02


k. PB3 A2  .25
l. PB3 A2  .05
m. PA2 B3  .02
n. PA2 B3  .46
o. PA2 B3  .25
p. PA2 B3  .05
q. None of the above – fill in the correct
answer.
5. Find the probability that an employee is either a woman or an Auxiliary employee (or both). Comment:
This is PA2 B3  . We found out that P A2  B3  .46 above.
a. PA2  B3  .02
b. PA2  B3  .46
c. PA2  B3  .25
d. PA2  B3  .05
e. PA2  B3  .02
f. * PA2  B3  .46
g. PA2  B3  .25
h. PA2  B3  .05


i. P B3 A2  .02


k. PB3 A2  .25
l. PB3 A2  .05
m. PA2 B3  .02
n. PA2 B3  .46
o. PA2 B3  .25
p. PA2 B3  .05
j. P B3 A2  .46
q. None of the above – fill in the correct
answer.
6. On the basis of what you found in the last three questions, A2 and B3 are:
a) complements.
Note: This means P A2  B3  0 and P A2  B3  1
b) mutually exclusive.
c) independent.
[12]
Note: This means P A2  B3  0
Note: This means P A2  B3  P A2 PB3
d) collectively exhaustive. Note: This means P A2  B3  1
e) All of the above could be true.
f) *None of the above can be true.
g) Not enough information
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7. z is a standardized random variable. (What do you know about the mean and variance of a standardized
variable?) If y  13z 100 , the standard deviation of y  y is:
 
a) -113.
b) -1. Note: We know that Varz   1 . We also know that a standard deviation cannot be < 0.
c) -13.
d) *13. Note: We know that if w  ax  b . Varw  a 2Varx . So Var y   132 Varz  .
e) 169.
f) 113
g) None of the above. Supply the correct answer.
Exhibit 2: A deck of cards has 52 cards divided equally into hearts, diamonds, spades and clubs. There are
4 of each kind of card (2s, 3s, 4s, 5s, 6s, 7s, 8s, 9s, 10s, jacks, queens, kings and aces. 26 of the cards are
black, 26 of the cards are red and 12 of the cards are face cards.
8. You are dealt 4 cards. What is the probability that all are aces? You may present this answer as a
decimal, a per cent or a fraction. (3)
[17]
Solution: Remember that the number of ways you can get r cards from n cards if order is not important is
n!
4!
 1 . The number of
C rn 
. The number of ways you can get 4 aces from 4 aces is C 44 
n  r ! r!
0! 4!
ways you can get 4 cards from 52 cards is C 452 
probability is
C 44 C 048
C 452

52  51  50  49 6497400
52!


 270725 . So the
48! 4!
4  3  2 1
24
1
 3.69379 10 6 . Let x be the number of aces. This is Px  4 .
270725
4
. The
52
3
probability of getting an ace on the second draw, assuming that we got an ace on the first draw is
. If we
51
4  3  2 1
24

 3.69379 10 6
follow this logic completely, we get Px  4 
52  51  50  49 6497400
Another way of looking at this is that the probability of getting an ace on the first draw is
9. What is the probability that you will get 2 aces? You may leave this answer as a fraction involving Cs or
Ps if you wish. (2)
4  3 12
48  47
4!
48!


 6 ways to get two aces. There are C 248 

Solution: There are C 24 
2! 2! 2 1 2
46! 2!
2 1
C 4 C 48
6 1128 
2256
 1128 ways to get two other cards that are not aces. So the probability is 2 522 
270725
2
C4
 .025000 . This is Px  2 .
10. What is the probability that you will get at least 1 ace? Please don’t waste our time by giving me the
probability of exactly one ace. You may leave this answer as a fraction involving Cs or Ps if you wish. (3)
[22]
Solution: The event ‘at least one ace’ is the complement of the event ‘no aces.’ So we have Px  1

 1  P x  0   1 

C 04 C 448
C 452
. You don’t really need C 04 
4!
48!
 1 , but you do need C 448 
4! 0!
44! 4!
C 4 C 48
48  47  46  45 4669920
194580
 1  .718737  .281263 .

 194580 . So Px  1  1  0 524  1 
270725
4  3  2 1
24
C4
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251y0821 4/4/08
48
. The
52
probability of not getting an ace on the second draw, assuming that we did not get an ace on the first draw
48  47  46  45
47
is
. If we follow this logic completely, we get Px  0 
 .718737 . So again Px  1
51
52  51  50  49
Another way of looking at this is that the probability of not getting an ace on the first draw is
 1  Px  0  1  .718737  .281263 .
11. Find the probability that you will get 5 Aces. (1)
Solution: Px  5  0 because it’s impossible.
Exhibit 3: As everyone knows a Jorcillator has two components. These are the Flubberall and the Phillinx.
Let us assume that our company owns a Jorcillator that only works as long as both components work. The
life of a Flubberall is described by a continuous uniform distribution where nonzero frequencies are
between c  0.5 and d  2.0 . The life of a Phillinx is described by a continuous uniform distribution
where c  0.5 and d  2.5 . For example, the probability that the Phillinx fails in year 1 is P0  x  1 and
the probability that the Phillinx fails in year 2 is P2  x  3 . Let A1 be the event that the Flubberall fails
in year 1, A2 be the event that the Flubberall fails in year 2, A3 be the event that the Flubberall lasts
beyond year 2, B1 be the event that the Phillinx fails in year 1, B2 be the event that the Phillinx fails in year
2, B3 be the event that the Phillinx lasts beyond year 2. The lives of the two components can be considered
independent.
12. Find the following. (Make diagrams.) (1 each)
[29]
(a) P A 2 , the probability that the Flubberall fails in year 2.
Solution: The life of a Flubberall is described by a continuous uniform distribution where nonzero
2 1
1
2


frequencies are between c  0.5 and d  2.0 . P A 2  P2  x  3 
2  0.5 1.5 3
(b) P A1 , the probability that the Flubberall fails in year 1.
1  0.5
.5 1


2  0.5 1.5 3
(c) P A 3 , the probability that the Flubberall fails after year 2.
Solution: P A1  P0  x  1 
Solution: P A 3  Px  2  
0
 0 . Make a diagram. You need a rectangle with a base from 0.5 to
2  0.5
2. Since its base is 2 – 0.5 = 1.5 and its area must be 1, its height must be 1  2 . (b) P0  x  1 is
1.5
3
represented by the part of the rectangle between 0.5 and 1 because there is no probability below 0.5. The
width of this area is 1 – 0.5 = 0.5 and its height is 2 3 , so if we multiply these together we get 13 . (a)
P2  x  3 is represented by the part of the rectangle between 1 and 2. The width of this area is 2 – 1 = 1
and its height is 2 3 , so if we multiply these together we get 2 3 . (c) Px  2 would be represented by the
part of the rectangle above 2, but there isn’t any.
0
0.5
1.0
2.0
x
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251y0821 4/4/08
(d) PB 1 , the probability that the Phillinx fails in year 1.
Solution: The life of a Phillinx is described by a continuous uniform distribution where c  0.5 and
1  0.5
0.5 1
d  2.5 . PB 1  P0  x  1 


2.5  0.5
2
4
(e) PB 2 , the probability that the Phillinx fails in year 2.
Solution: PB 2  P1  x  2 
1
1 1
 
2.5  0.5 2 2
(f) PB 3 , the probability that the Phillinx fails after year 2.
2.5  2
0.5 1


2.5  0.5
2
4
Make a diagram. You need a rectangle with a base from 0.5 to 2.5. Since its base is 2.5 – 0.5 = 2 and its
area must be 1, its height must be 1 . (a) P0  x  1 is represented by the part of the rectangle between
2
0.5 and 1 because there is no probability below 0.5. The width of this area is 1 – 0.5 = 0.5 and its height is
1 , so if we multiply these together we get 1 . (b) P2  x  3 is represented by the part of the rectangle
2
4
Solution: PB 3  Px  2 
between 1 and 2. The width of this area is 2 – 1 = 1 and its height is
1
2
, so if we multiply these together we
. (c) Px  2 is represented by the part of the rectangle between 2 and 2.5, because there isn’t any
area above 2.5. The width of this area is 2.5 – 2 = 0.5 and its height is 1 2 , so if we multiply these together
get
2
3
we get
1
4
.
0
0.5
1
2
2.5
13. Assuming that your probabilities are correct, find the probability that both components fail in year 2.
(2)
2 1 1
Solution: If we assume independence, P A 2  B2  P A2 PB2    
3 2 3
14. Find the probability that the Jorcillator fails in the first year. (2)
Solution: The joint events listed below can occur. Our company owns a Jorcillator that only works as long
as both components work, so that whichever component fails first will down the machine. This means that
in the first year if either component fails, the jorcillator fails. In the second year this is true also, but only if
the jorcillator did not fail in the first year. Rather than messing with conditional probabilities, I prefer to list
each joint probability (intersection) separately and then to figure out when the machine will fail it the joint
event occurs.
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251y0821 4/4/08
We know P A1 
1
2
1
1
1
, P A 2  , P A 3  0 , PB 1  , PB 2  and PB 3  .
3
3
4
2
4
Joint Event
A1 B1
Probability
11 1
 
3  4  12
Failure Time
A1 B2
11 2
 
3  2  12
Year 1
A1  B3
11 1
 
3  4  12
Year 1
A2  B1
21 2
 
3  4  12
Year 1
A2  B2
21 4
 
3  2  12
Year 2
A2  B3
21 2
 
3  4  12
Year 2
A3  B1
1
0   0
4
Year 1
A3  B2
1
0   0
2
Year 2
A3  B3
1
0   0
4
After Year 2
Sum
12
1
12
Year 1
So the solution for the probability that the jorcillator fails in the first year is
 P A1  B1 . But notice that this doesn’t work in question 15.
1
2 1
2 1
  

12 12 12 12 2
15. Find the probability that the Jorcillator fails in the second year. (2)
Solution:
2 1 4 8  6  4 10
4 2 1


  . Note that this probability is not P A2  B 2    
.
3 2 12
12
12
12 12 2
16. Find the probability that the Jorcillator fails after the second year. (1)
[34]
Solution: 0 because one of the components cannot last beyond the second year. Note that, because the
jorcillator must fail sometime, the probabilities in the three periods must add to 1.
8
251y0821 4/4/08
Exhibit 4: A coin is tossed 5 times. Define the following events.
A : Exactly one head
B : Exactly two heads
C : At least one head
17. Find P A
5
5
1
 .15625 . This is because the probability of any given
Solution: P1  C15   
32
2
5
 1  1  1  1  1   1 
list of 5 items, whether heads or tails is identical to PHTHTH            , and there
 2  2  2  2  2   2 
are 5 different ways to place the one H in a sequence where all the other events are Ts, so that there are 5
mutually exclusive ways to get one head and there probabilities can be added together.
5
18. Find PB 
5! 1 5  4 1 10
1


 .31250
Solution: P2  C 25   
3! 2! 32 2 1 32 32
2
19. Find PC 
1
1
 .96875
Solution: 1  P0  1  C 05    1 
32
2
5
20. Let x be the number of heads. List all the values that x can take and their probabilities. You have done
most of this already if you realize that the probability of 2 heads and the probability of 2 tails are equal.
Show that this is a valid distribution. (4)
Solution: Because the situation is symmetrical, the probability of 1 head must be the same as the
probability of 1 tail (or 4 heads). The probability or 2 heads must be the same as the probability of 2 tails
(or 3 heads). Similarly, the probability of no heads and the probability of no tails (5 heads) must be the
same. This can be done with tree diagrams, but everyone who tried left out large parts of the tree. The
approach I gave you in class was to write down the 5 slots that could have heads in them and to use
combinations to figure out how many ways we can pick the slots that have the desired number of heads.
The table below has been used to record the probabilities
x
P x 
xPx 
x 2 P x 
0
1
2
3
4
5
Sum
1
32
5
32
10
32
10
32
5
32
1
32
0
0
5
32
20
32
30
32
20
32
5
32
5
32
40
32
90
32
80
32
25
32
32
1
32
80
 2.5 
32
 xPx
240
 7.5 
32
x
2
Px 
21. Find the expected value (mean) and standard deviation using the values of x and the probabilities
found in question 20. Do not guess! Show your work. (3)
[47]
xPx   2.5 , E x 2 
x 2 Px   7.5 and
Solution: From the table above,  x  E x  
 

  
 x2  Varx  E x 2   x2  7.5  2.52  1.25 . So the standard deviation is  x  1.25  1.11803 .
9
251y0821 4/4/08
ECO251 QBA1
SECOND EXAM
March 28, 2007
TAKE HOME SECTION
Name: _________________________
Student Number: _________________________
Throughout this exam show your work! Neatness counts! Please indicate clearly what sections of the
problem you are answering and what formulas you are using.
Part III. Do all the Following (12+ Points) Show your work!
1. Seymour Butz’s student number is 976543
Take the following set of numbers: and the probabilities below
x
P x 
10
.16
20
.17
30
.18
42
.19
53
.20
64
__
75
__
and use the second to last digit of your Student Number to provide the first digit of all seven numbers so
that the become 3 digit numbers the last six numbers, so that, for Seymour, the numbers would become
410, 420, 430, 442, 453, 464 and 475. Now compute the last two probabilities by putting the last digit of
your student number divided by 100 in the first blank. For Seymour, this would mean that he will write in
.03. In the second blank, put in whatever you need to make the distribution valid and explain how you got
that number. Compute a population standard deviation using your probabilities. (2) Find the probability that
x is an even number and find the median of the distribution. Again show your work. (1)
Solution: Let’s assume that I am Seymour. My
table is as below.
x
P x 
xPx  x 2 Px 
410
.16
65.60 26896.00
420
.17
71.40 29988.00
430
.18
77.40 33282.00
442
.19
83.98 37119.16
453
.20
90.60 41041.80
464
.03
13.92 6458.88
475
.07
33.25 15793.75
1.00
436.15 190579.59
 Px  1 (a check for a valid
distribution),  xPx   436 .15  E x   
and  x Px   190479 .59  E x  .
We have
2
x
2
 
So Var x   E x 2   x2  190579 .59  436 .15 2  190579.59 190226.8225  352.7675   x2 and
 x  352 .7675  18 .782106 .
The probability that x is an even number is gotten by adding the probabilities of 410, 420, 430, 442 and
464, but it’s easier to take 1  P453   P475   1  .20  .07  .73 . The median is not affected by the last
two probabilities, since they always add to .10. In the table above, the closest we can come to equal
probabilities above and below a number is Px  430   .16  .17  .33 and
Px  430   .07  .03  .20  .19  .49 . All of the other numbers will have either more than 50% above
them or more than 50% below them. The Number on the third line will be the median in all versions of this
problem.
10
251y0821 4/4/08
The probability that x is an even number will change according to the probability on the 464 or 475 line.
Since the actual values of the numbers are irrelevant, we can say the following.
Version
Probability on 6th line of table
Probability of an even number
Version 1
0
1  .20  .10  .70
Version 2
.01
1  .20  .09  .71
Version 3
.02
1  .20  .08  .72
Version 4
.03
1  .20  .07  .73
Version 5
.04
1  .20  .06  .74
Version 6
.05
1  .20  .05  .75
Version 7
.06
1  .20  .04  .76
Version 8
.07
1  .20  .03  .77
Version 9
.08
1  .20  .02  .78
Version 10
.09
1  .20  .01  .79
The Minitab printout for version 4 has
Data Display
Row
1
2
3
4
5
6
7
x_
310
320
330
342
353
364
375
P
0.16
0.17
0.18
0.19
0.20
0.03
0.07
xP
49.60
54.40
59.40
64.98
70.60
10.92
26.25
Sum of P
Sum of P = 1
xsqP
15376.0
17408.0
19602.0
22223.2
24921.8
3974.9
9843.8
Data Display
E(x)
E(xsq)
336.150
113350
Data Display
varx
352.767
Data Display
stdevx
varx
18.7821
352.767
Because this output used 300 instead of 400 in the x column, Seymour must adjust the means in the
printout up by 100.
So, to use the computer printout of the answers if your probability in the sixth row is k 100 , use the printout
for version k  1 as above. The variances will be correct but the means have to be adjusted by the same
constant that would be added or subtracted to the x column to get your x column.
2. Compute a sample standard deviation for your x column. Ignore the probabilities. (2) [5]
Solution: This time Seymour has (Version 5), using the computational formula,
x  3094 and
x
Row
We have found
x2
1
2
3
4
5
6
7
410
420
430
442
453
464
475
3094
168100
176400
184900
195364
205209
215296
225625
1370894
x
x


2
 1370894 . We thus have
3094
 442 and s x2 
7
x
2
 nx 2
n 1
1370894  7442 
3346

 557 .6667 . So
6
6
2
s x  557 .667  23 .61497
Since I have asked you not to waste your time by using the definitional formula, it is not presented here.
You should get the same answer if you know how to use the formula.
11
251y0821 4/4/08
3. Assume that we roll a pair of dice. Use x for the amount on top of the first die, y for the amount on top
of the second die and w for their sum. Let zo be the value of the last digit of your student number plus 2.
Seymour Butz’s student number is 976543. For Seymour this means zo  5 .
Let event A be w  z0  . Let event B be y is an even number . Let event C be w  z0  . Make sure
that I know what z0 is and show your work. Find (1 each): [12]
a) P A
b) PC 
 
c) P B C
d) PB  C 
e) PB  C 
 
f) P C B

g) P B  C

x
1
2
We have this diagram for dice problems:
y 3
4
5
6
of which has a probability of
1
36
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
4 5 6 7
5 6 7  8
. Counting points, each
6 7 8 9
7  8  9  10 
8  9  10  11 
9  10  11  12 
, works fine. PB   P y  2,4 or 6 
18 1

36 2
1
1
1

. b) PC   Pw  2 
.
36
36 36
c) P B C  P y  2,4or6 w  2 =0. d) PB  C   0 e) PB  C   PB   PC   PB  C 
Version 0: z o  2 a) P A  Pw  2  
 


18 1
19

0 
. f)
36 36
36

17
PC B  0 . g) PB  C   P y  1,3or5  w  2  
36
 1  PB  C  .
x
1
2
y 3
4
5
6
2*
3#
4*#
5#
6*#
7#
1
2
3*# 4#* 
4#
5#
5*# 6*#
6#
7#
7*# 8*#
8#
9#
3
4
5
6
5#*  6#*  7*#
6#
7#
8#
There are 17 points that involve an odd y (*) and a
7*# 8*# 9*#
8#
9# 10 #
9*# 10*# 11*#
10 # 11# 12 #
total above 2. (#).
12
251y0821 4/4/08
Version 1: z o  3 a) P A  Pw  3 
2
1 2
3

. b) PC   Pw  3 
.
36
36
36
x
1
2*
3*#
4
5
6
7
1
2
1
c) P B C  P y  2,4or6 w  3  .
3 y 3
4
5
6
 


2
3*
4
5
6
7
8
3
4
5
6
4 5 6 7
5 6 7 8
Out of 3 points (*), 1
6 7 8 9
7  8  9  10 
8  9  10  11 
9  10  11  12 
point involve an even number (#) in y . d) PB  C  
1
. e) PB  C   PB   PC   PB  C 
36
1
16
18 3
1
20




. f) P C B 
. g) P B  C 
 1  PB  C  .
18
36
36 36 36 36
x

 
1
2
y 3
4
5
6
1
2*
3
4*#
5#
6*#
7 #
2
3*
4#
5*#
6#
7*#
8#
3
4*#
5#
6*#
7 #
8*#
9#

4
5
6
5*# 6*# 7*#
6#
7 #
8#
There are 16 points that involve an odd y (*) and a
7*# 8*# 9*#
8#
9# 10 #
9*# 10*# 11*#
10 # 11# 12 #
total above 3. (#).
Version 2: z o  4 a) P A  Pw  4  
3
1 2  3 6

. b) PC   Pw  4 
.
36
36
36
x
1
2*
3*#
4*
5
6
7
1
2
2
c) P B C  P y  2,4or6 w  4  .
y
3
6
4
5
6
 


2
3*
4*#
5
6
7
8
3
4
5
6
4* 5  6  7 
5 6 7 8
Out of 6 points (*),
6 7 8 9
7  8  9  10 
8  9  10  11 
9  10  11  12 
2 points involve an even number (#) in y . d) PB  C  
2
. e) PB  C   PB   PC   PB  C 
36
2
14
18 6
2 22




. f) P C B 
. g) P B  C 
 1  PB  C  .
18
36
36 36 36 36
 


13
251y0821 4/4/08
x
1
2
y 3
4
5
6
1
2*
3
4*
5#
6*#
7 #
2
3*
4
5*#
6#
7*#
8#
3
4*
5#
6*#
7 #
8*#
9#
4
5
6
5*# 6*# 7*#
6#
7 #
8#
There are 14 points that involve an odd y (*) and a
7*# 8*# 9*#
8#
9# 10 #
9*# 10*# 11*#
10 # 11# 12 #
total above 4. (#).
Version 3: z o  5 a) P A  Pw  5 
4
1  2  3  4 10

. b) PC   Pw  5 
36
36
36
x
1
2
4
c) P B C  P y  2,4or6 w  5 
.
y
3
10
4
5
6

 

1
2*
3*#
4*
5*#
6*
7
2
3*
4*#
5*
6
7
8
(*) 4 points involve an even number (#) in y . d) PB  C  
e) PB  C   PB   PC   PB  C  
3
4*
5*#
6
7
8
9
4
5
6
5* 6  7 
6 7 8
Out of 10 points
7 8 9
8  9  10 
9  10  11 
10  11  12 
4
.
36
 
4
18 10 4
24
. f) P C B 
.



18
36 36 36 36
12
g) P B  C  P y  1,3or5  w  6  
 1  PB  C  .
36
x

1
2
y 3
4
5
6

1
2*
3
4*
5
6#* 
7 #
2
3*
4
5*
6#
7*#
8#
3
4*
5
6*#
7#
8*#
9#
4
5
6
5 *  6*# 7*#
6#
7#
8#
There are 12 points that involve an odd y (*) and a
7*# 8*# 9*#
8#
9# 10 #
9*# 10*# 11*#
10 # 11# 12 #
total above 5. (#).
14
251y0821 4/4/08
Version 4: z o  6 a) P A  Pw  6  
5
1  2  3  4  5 15

. b) PC   Pw  6 
36
36
36
x
1
2
6
c) P B C  P y  2,4or6 w  6 
.
15 y 3
4
5
6

 

1
2*
3*#
4*
5*#
6*
7
2
3*
4*#
5*
6*#
7
8
(*) 6 points involve an even number (#)in y . d) PB  C  
3
4*
5*#
6*
7
8
9
4
5*
6*#
7
8
9
10 
5
6
6* 7 
7 8
Out of 15 points
8 9
9  10 
10  11 
11  12 
6
.
36
 
e) PB  C   PB   PC   PB  C  
6
18 15 6
27
. f) P C B 
.



18
36 36 36 36
9
g) P B  C  P y  1,3or5  w  6  
 1  PB  C  .
36
x


1
2
y 3
4
5
6
1
2*
3
4*
5
6*
7 #
2
3*
4
5*
6
7*#
8#
3
4*
5
6*
7 #
8*#
9#
4
5
6
5*
6 *  7*#
6
7 #
8#
There are 9 points that involve an odd y (*) and a
7*# 8*# 9*#
8#
9# 10 #
9*# 10*# 11*#
10 # 11# 12 #
total above 6. (#).
Version 5: z o  7 a) P A  Pw  7  
6
1  2  3  4  5  6 21

. b) PC   Pw  7 
36
36
36
x
1
2
9
c) P B C  P y  2,4or6 w  7 
.
y
3
21
4
5
6
 


1
2*
3*#
4*
5*#
6*
7#* 
2
3*
4*#
5*
6*#
7 *
8
3
4*
5*#
6*
7*#
8
9
points in C (*) 9 points involve an even number (#) in y . d) PB  C  
 PB   PC   PB  C  
4
5*
6*#
7 *
8
9
10 
5
6*
7  *#
8
9
10 
11 
6
7 *
8
Out of 21
9
10 
11 
12 
9
e) PB  C 
36
 
9
18 21 9 30



. f) P C B 
.
18
36 36 36 36
15
251y0821 4/4/08


g) P B  C  P y  1,3or5  w  7  
6
 1  PB  C  .
36
x
1
2
y 3
4
5
6
1
2*
3
4*
5
6*
7
2
3*
4
5*
6
7 *
8#
3
4*
5
6*
7
8*#
9#
4
5
6
5*
6*
7 *
6
7
8#
There are 6 points that involve an odd y (*) and a
7 *  8#*  9#* 
8#
9# 10 #
9#*  10 #*  11#* 
10 #
11# 12 #
total above 7(#).
Version 6: z o  8 a) P A  Pw  8 
5
1  2  3  4  5  6  5 26

. b) PC   Pw  8 
36
36
36
x
1
2
12
c) P B C  P y  2,4or6 w  8 
.
y
3
26
4
5
6

 

1
2*
3*#
4*
5*#
6*
7 #* 
2
3*
4*#
5*
6*#
7 *
8#* 
3
4*
5*#
6*
7*#
8*
9
points (*) 12 points involve an even number (#)in y . d) PB  C  
 PB   PC   PB  C  
4
5*
6*#
7 *
8*#
9
10 
5
6*
7  *#
8*
9
10 
11 
6
7 *
8*#
Out of 26
9
10 
11 
12 
12
e) PB  C 
36
 
12
18 26 12 32



. f) P C B 
.
18
36 36 36 36
4
g) P B  C  P y  1,3or5  w  8 
 1  PB  C  .
36
x


1
2
y 3
4
5
6
1
2*
3
4*
5
6*
7
2
3*
4
5*
6
7 *
8
3
4*
5
6*
7
8*
9#
4
5
6
5*
6*
7 *
6
7
8
There are 4 points that involve an odd y (*) and a
7 *
8 *  9#* 
8
9# 10 #
9#*  10 #*  11#* 
10 # 11# 12 #
total above 8(#).
16
251y0821 4/4/08
Version 7: z o  9 a) P A  Pw  9  
4
1  2  3  4  5  6  5  4 30

. b) PC   Pw  9 
.
36
36
36
x
1
2
14
c) P B C  P y  2,4or6 w  9 
.
30 y 3
4
5
6
 


1
2*
3*#
4*
5*#
6*
7#* 
2
3*
4*#
5*
6*#
7 *
8#* 
3
4*
5*#
6*
7*#
8*
9#* 
4
5*
6*#
7 *
8*#
9*
10 
points (*) in C, 14 points involve an even number (#)in y . d) PB  C  
 PB   PC   PB  C  
14
18 30 14 34
. f) P C B 
. g) P B  C



18
36 36 36 36
x
1
2
2
 P y  1,3or5  w  9  
 1  PB  C  .
y 3
36
4
5
6
1
2*
3
4*
5
6*
7
2
3*
4
5*
6
7 *
8
6
7 *
8*#
Out of 30
9*
10 
11 
12 
14
. e) PB  C 
36

 
5
6*
7  *#
8*
9#* 
10 
11 

3
4
5
6
4* 5*
6* 7 *
5
6
7
8
6* 7 *
8* 9*
7
8
9  10 #
8 *  9 *  10 #*  11*#
9  10 # 11# 12 #
There are 2 points that involve an odd y (*) and a total above 9(#).
Version 8: z o  10 a) P A  Pw  10  
3
1  2  3  4  5  6  5  4  3 33

. b) PC   Pw  10  
.
36
36
36
x
1
2
16
c) P B C  P y  2,4or6 w  10 
.
33 y 3
4
5
6
 


1
2*
3*#
4*
5*#
6*
7 #* 
2
3*
4*#
5*
6*#
7 *
8#* 
3
4
4*
5*
5*# 6*#
6*
7 *
7*#
8*#
8*
9*
9#*  10 #* 
points (*) 16 points involve an even number (#) in y . d) PB  C  
 PB   PC   PB  C  
 
5
6
6*
7 *
7  *# 8*#
Out of 33
8*
9*
9#*  10*#
10 * 
11 
11 
12 
16
. e) PB  C 
36

16
18 33 16 35



. f) P C B 
. g) P B  C
18
36 36 36 36

17
251y0821 4/4/08
x
1
2
1
 P y  1,3or5  w  10  
 1  PB  C  .
y 3
36
4
5
6
1
2*
3
4*
5
6*
7
2
3*
4
5*
6
7 *
8
3
4*
5
6*
7
8*
9
4
5
6
5* 6* 7 *
6
7
8
7 * 8* 9*
8
9
10 
9 *  10 *  11*#
10  11# 12 #
There is only one point that involves an odd y (*) and a total above 10(#).
Version 9: z o  11 a) P A  Pw  11 
2
1  2  3  4  5  6  5  4  3  2 35

. b) PC   Pw  11 
36
36
36
x
1
2
17
c) P B C  P y  2,4or6 w  11 
.
35 y 3
4
5
6

 

1
2*
3*#
4*
5*#
6*
7 #* 
2
3*
4*#
5*
6*#
7 *
8#* 
3
4
5
6
4*
5*
6*
7 *
5*# 6*# 7  *# 8*#
Out of 35
6*
7 *
8*
9*
7*#
8*# 9#*  10*#
8*
9 *  10 *  11 * 
9#*  10 #*  11#* 
12 
points (*) 17 points involve an even number (#)in y . d) PB  C  
e) PB  C   PB   PC   PB  C  


17
.
36
 
17
18 35 17
.


 1 f) P C B 
18
36 36 36
g) P B  C  P y  1,3or5  w  11  0  1  PB  C  .
x
1
2
y 3
4
5
6
1
2*
3
4*
5
6*
7
2
3*
4
5*
6
7 *
8
3
4*
5
6*
7
8*
9
4
5
6
5* 6* 7 *
6
7
8
There are no points that involve an odd y (*) and a total
7 * 8* 9*
8
9  10 
9 *  10 *  11 * 
10  11  12 #
above 11(#).
18
251y0821 4/4/08
4. (Moore et al) 5% of male high school athletes (in basketball, football or baseball) compete in college. Of
those who compete in college, 1.7% go on to major league sports. 40% of the athletes who compete in both
high school and college and enter major leagues last more than three years. The probability that a high
school athlete who does not compete in college goes on to professional play is .0001.
Although the numbers given above are probably correct, modify the probabilities as follows: Seymour
Butz’s student number is 976543. Take the last digit of your student number and add it to 5%, so, for
example, Seymour will say that (5+3) % = 8% of male high school athletes compete in college; then take
the same number and add it to the 1 in .0001, so, for example, Seymour will say that .0001 + .0003 = .0004
is the probability that a high school athlete who does not compete in college goes on to professional play.
Let A be the event that a high school athlete competes in college. Let B be the event that a high-school
athlete competes professionally. Let C be the event that a high-school athlete has a professional career of
more than 3 years. Use these in your answer. (2 points extra credit for each part)
a) Identify your four probabilities as given the problem and modified by you in terms of events A , B and
C . That is to say probabilities like PB  C  , P B  C , P B C or P A  B  C  . (I do not want to

  
imply that any of these appear in the problem)
b) What is the probability that a high school athlete will play in college and then have a major league career
of more than three years?
c) What is the probability that a high school athlete will play professional sports?
d) What proportion of professional athletes competed in college?
Solution: I have used j for the last digit of your number, so that Seymour has j  3 .
a) Identify your four probabilities as given the problem and modified by you in terms of events A , B and
C . Let j be the number that you add.
A is the event that a high school athlete competes in college. P A  5  j % or .05  .0 j (Seymour has
P A  5  3%  8%  .08 or .05  .03  .08 )
 
B is the event that a high-school athlete competes professionally. PB A  1.7% , P B A  .0001  .000 j .
 
(So Seymour has P B A  .0001  .0003  .0004 )


C is the event that a high-school athlete has a professional career of more than 3 years. P C A  B  .40
b) What is the probability that a high school athlete will play in college and then have a major league career
of more than three years? P A  B  C   P C A  B  P B A  P A  .40.017 .05  .0 j 

  
c) What is the probability that a high school athlete will play professional sports?

 

PB  PB  A  P( B  A  PB AP A  P B A P A  .017 .05  .0 j   (.0001  .00  j )1  .05  .0 j 
d) What proportion of professional athletes competed in college?
P B AP A .017 .05  .0 j 
P A B  

P B 
P B 
 
If j  0 , so that P A  .05 and P B A  .0001
 
 


b) P A  B  C   PC A  B PB A P A  .40 .017 .05   .00034
c) PB  PB  A  P(B  A   PB AP A  PB A PA   .017 .05   .0001 1  .05 
a) P A  .05 , P B A  .017 , P C A  B  .40 , P B A  .0001
 .00085  .000095  .000945
PB AP A .017 .05 

 .899471
d) PA B  
P B 
.000945
19
251y0821 4/4/08
Let’s try doing c)-d) as a box. We can divide
1,000,000 high school athletes so that some
compete in college and some don’t. Yup! To
avoid fractions I’m going to need a million high
school athletes.
If j  0 , we have P A  .05 . So out of 1000000
athletes 50000 compete in college and 950000
do not.
B B

A 


A 


Total
1000000
B B


A


A 


Total
50000
950000
1000000
Out of the 50000 athletes who compete in
college, 1.7% or .017 50000   850 go pro. Out
of the 950000 athletes who do not compete in
college .0001(950000) = 95 go pro.
B B


A 850


A  95 


Total
50000
950000
1000000
We can now fill in the missing numbers.
So c) PB  
B
B
Total
A 850 49150 
50000


A  95 949905  950000
945 999055
1000000
945
850
 .000945 d) P A B  
 .899471
945
1000000
So – For a second example, lets take the most extreme variation from the last example.
 
If j  9 , P A  .14 P B A  .0010
 
 


b) P A  B  C   PC A  B PB A P A  .40 .017 .14   .00095
c) PB  PB  A  P(B  A   PB AP A  PB A PA   .017 .14   .0010 1  .14 
a) P A  .14 , P B A  .017 , P C A  B  .40 , P B A  .0010
 .00238  .00086  .00324
P B AP A .017 .14 

 .734568
d) P A B  
P B 
.00324
20
251y0821 4/4/08
Appendix – Solutions to Questions 1 and 2 of the Take-home.
Data Display
version
1.00000
Data Display
Row
1
2
3
4
5
6
7
x_
10
20
30
42
53
64
75
P
0.16
0.17
0.18
0.19
0.20
0.00
0.10
xP
1.60
3.40
5.40
7.98
10.60
0.00
7.50
xsqP
16.00
68.00
162.00
335.16
561.80
0.00
562.50
Sum of P
Sum of P = 1
Data Display
E(x)
E(xsq)
36.4800
1705.46
Data Display
varx
374.670
Data Display
stdevx
19.3564
Data Display
Row
1
2
3
4
5
6
7
x_
10
20
30
42
53
64
75
xsq
100
400
900
1764
2809
4096
5625
Data Display
sum(x)
sum(xsq)
294.000
15694.0
Sum of x_
Sum of x_ = 294
Sum of Squares of x_
Sum of squares (uncorrected) of x_ = 15694
Data Display
svar(x)
557.667
Data Display
mean(x)
sstdev(x)
42.0000
23.6150
Mean of x0
Mean of x0 = 42
Standard Deviation of x0
Standard deviation of x0 = 23.6150
21
251y0821 4/4/08
Data Display
version
2.00000
Data Display
Row
1
2
3
4
5
6
7
x_
110
120
130
142
153
164
175
P
0.16
0.17
0.18
0.19
0.20
0.01
0.09
xP
17.60
20.40
23.40
26.98
30.60
1.64
15.75
xsqP
1936.00
2448.00
3042.00
3831.16
4681.80
268.96
2756.25
Sum of P
Sum of P = 1
Data Display
E(x)
E(xsq)
136.370
18964.2
Data Display
varx
367.393
Data Display
stdevx
19.1675
Data Display
Row
1
2
3
4
5
6
7
x_
110
120
130
142
153
164
175
xsq
12100
14400
16900
20164
23409
26896
30625
Data Display
sum(x)
sum(xsq)
994.000
144494
Sum of x_
Sum of x_ = 994
Sum of Squares of x_
Sum of squares (uncorrected) of x_ = 144494
Data Display
svar(x)
557.667
Data Display
mean(x)
sstdev(x)
142.000
23.6150
Mean of x1
Mean of x1 = 142
Standard Deviation of x1
Standard deviation of x1 = 23.6150
22
251y0821 4/4/08
Data Display
version
3.00000
Data Display
Row
1
2
3
4
5
6
7
x_
210
220
230
242
253
264
275
P
0.16
0.17
0.18
0.19
0.20
0.02
0.08
xP
33.60
37.40
41.40
45.98
50.60
5.28
22.00
xsqP
7056.0
8228.0
9522.0
11127.2
12801.8
1393.9
6050.0
Sum of P
Sum of P = 1
Data Display
E(x)
E(xsq)
236.260
56178.9
Data Display
varx
360.092
Data Display
stdevx
18.9761
Data Display
Row
1
2
3
4
5
6
7
x_
210
220
230
242
253
264
275
xsq
44100
48400
52900
58564
64009
69696
75625
Data Display
sum(x)
sum(xsq)
1694.00
413294
Sum of x_
Sum of x_ = 1694
Sum of Squares of x_
Sum of squares (uncorrected) of x_ = 413294
Data Display
svar(x)
557.667
Data Display
mean(x)
sstdev(x)
242.000
23.6150
Mean of x2
Mean of x2 = 242
Standard Deviation of x2
Standard deviation of x2 = 23.6150
23
251y0821 4/4/08
Data Display
version
4.00000
Data Display
Row
1
2
3
4
5
6
7
x_
310
320
330
342
353
364
375
P
0.16
0.17
0.18
0.19
0.20
0.03
0.07
xP
49.60
54.40
59.40
64.98
70.60
10.92
26.25
xsqP
15376.0
17408.0
19602.0
22223.2
24921.8
3974.9
9843.8
Sum of P
Sum of P = 1
Data Display
E(x)
E(xsq)
336.150
113350
Data Display
varx
352.767
Data Display
stdevx
18.7821
Data Display
Row
1
2
3
4
5
6
7
x_
310
320
330
342
353
364
375
xsq
96100
102400
108900
116964
124609
132496
140625
Data Display
sum(x)
sum(xsq)
2394.00
822094
Sum of x_
Sum of x_ = 2394
Sum of Squares of x_
Sum of squares (uncorrected) of x_ = 822094
Data Display
svar(x)
557.667
Data Display
mean(x)
sstdev(x)
342.000
23.6150
Mean of x3
Mean of x3 = 342
Standard Deviation of x3
Standard deviation of x3 = 23.6150
Data Display
24
251y0821 4/4/08
Data Display
version
5.00000
Data Display
Row
1
2
3
4
5
6
7
x_
410
420
430
442
453
464
475
P
0.16
0.17
0.18
0.19
0.20
0.04
0.06
xP
65.60
71.40
77.40
83.98
90.60
18.56
28.50
xsqP
26896.0
29988.0
33282.0
37119.2
41041.8
8611.8
13537.5
Sum of P
Sum of P = 1
Data Display
E(x)
E(xsq)
436.040
190476
Data Display
varx
345.418
Data Display
stdevx
18.5854
Data Display
Row
1
2
3
4
5
6
7
x_
410
420
430
442
453
464
475
xsq
168100
176400
184900
195364
205209
215296
225625
Data Display
sum(x)
sum(xsq)
3094.00
1370894
Sum of x_
Sum of x_ = 3094
Sum of Squares of x_
Sum of squares (uncorrected) of x_ = 1370894
Data Display
svar(x)
557.667
Data Display
mean(x)
sstdev(x)
442.000
23.6150
Mean of x4
Mean of x4 = 442
Standard Deviation of x4
Standard deviation of x4 = 23.6150
Data Display
25
251y0821 4/4/08
Data Display
version
6.00000
Data Display
Row
1
2
3
4
5
6
7
x_
510
520
530
542
553
564
575
P
0.16
0.17
0.18
0.19
0.20
0.05
0.05
xP
81.60
88.40
95.40
102.98
110.60
28.20
28.75
xsqP
41616.0
45968.0
50562.0
55815.2
61161.8
15904.8
16531.3
Sum of P
Sum of P = 1
Data Display
E(x)
E(xsq)
535.930
287559
Data Display
varx
338.045
Data Display
stdevx
18.3860
Data Display
Row
1
2
3
4
5
6
7
x_
510
520
530
542
553
564
575
xsq
260100
270400
280900
293764
305809
318096
330625
Data Display
sum(x)
sum(xsq)
3794.00
2059694
Sum of x_
Sum of x_ = 3794
Sum of Squares of x_
Sum of squares (uncorrected) of x_ = 2059694
Data Display
svar(x)
557.667
Data Display
mean(x)
sstdev(x)
542.000
23.6150
Mean of x5
Mean of x5 = 542
Standard Deviation of x5
Standard deviation of x5 = 23.6150
Data Display
26
251y0821 4/4/08
Data Display
version
7.00000
Data Display
Row
1
2
3
4
5
6
7
x_
610
620
630
642
653
664
675
P
0.16
0.17
0.18
0.19
0.20
0.06
0.04
xP
97.60
105.40
113.40
121.98
130.60
39.84
27.00
xsqP
59536.0
65348.0
71442.0
78311.2
85281.8
26453.8
18225.0
Sum of P
Sum of P = 1
Data Display
E(x)
E(xsq)
635.820
404598
Data Display
varx
330.648
Data Display
stdevx
18.1837
Data Display
Row
1
2
3
4
5
6
7
x_
610
620
630
642
653
664
675
xsq
372100
384400
396900
412164
426409
440896
455625
Data Display
sum(x)
sum(xsq)
4494.00
2888494
Sum of x_
Sum of x_ = 4494
Sum of Squares of x_
Sum of squares (uncorrected) of x_ = 2888494
Data Display
svar(x)
557.667
Data Display
mean(x)
sstdev(x)
642.000
23.6150
Mean of x6
Mean of x6 = 642
Standard Deviation of x6
Standard deviation of x6 = 23.6150
Data Display
27
251y0821 4/4/08
Data Display
version
8.00000
Data Display
Row
1
2
3
4
5
6
7
x_
710
720
730
742
753
764
775
P
0.16
0.17
0.18
0.19
0.20
0.07
0.03
xP
113.60
122.40
131.40
140.98
150.60
53.48
23.25
xsqP
80656
88128
95922
104607
113402
40859
18019
Sum of P
Sum of P = 1
Data Display
E(x)
E(xsq)
735.710
541592
Data Display
varx
323.226
Data Display
stdevx
17.9785
Data Display
Row
1
2
3
4
5
6
7
x_
710
720
730
742
753
764
775
xsq
504100
518400
532900
550564
567009
583696
600625
Data Display
sum(x)
sum(xsq)
5194.00
3857294
Sum of x_
Sum of x_ = 5194
Sum of Squares of x_
Sum of squares (uncorrected) of x_ = 3857294
Data Display
svar(x)
557.667
Data Display
mean(x)
sstdev(x)
742.000
23.6150
Mean of x7
Mean of x7 = 742
Standard Deviation of x7
Standard deviation of x7 = 23.6150
Data Display
28
251y0821 4/4/08
Data Display
version
9.00000
Data Display
Row
1
2
3
4
5
6
7
x_
810
820
830
842
853
864
875
P
0.16
0.17
0.18
0.19
0.20
0.08
0.02
xP
129.60
139.40
149.40
159.98
170.60
69.12
17.50
xsqP
104976
114308
124002
134703
145522
59720
15313
Sum of P0
Sum of P0 = 1
Data Display
E(x)
E(xsq)
835.600
698543
Data Display
varx
315.780
Data Display
stdevx
17.7702
Data Display
Row
1
2
3
4
5
6
7
x_
810
820
830
842
853
864
875
xsq
656100
672400
688900
708964
727609
746496
765625
Data Display
sum(x)
sum(xsq)
5894.00
4966094
Sum of x_
Sum of x_ = 5894
Sum of Squares of x_
Sum of squares (uncorrected) of x_ = 4966094
Data Display
svar(x)
557.667
Data Display
mean(x)
sstdev(x)
842.000
23.6150
Mean of x8
Mean of x8 = 842
Standard Deviation of x8
Standard deviation of x8 = 23.6150
Data Display
29
251y0821 4/4/08
Data Display
version
10.0000
Data Display
Row
1
2
3
4
5
6
7
x_
910
920
930
942
953
964
975
P
0.16
0.17
0.18
0.19
0.20
0.09
0.01
xP
145.60
156.40
167.40
178.98
190.60
86.76
9.75
xsqP
132496
143888
155682
168599
181642
83637
9506
Sum of P
Sum of P = 1
Data Display
E(x)
E(xsq)
935.490
875450
Data Display
varx
308.310
Data Display
stdevx
17.5588
Data Display
Row
1
2
3
4
5
6
7
x_
910
920
930
942
953
964
975
xsq
828100
846400
864900
887364
908209
929296
950625
Data Display
sum(x)
sum(xsq)
6594.00
6214894
Sum of x_
Sum of x_ = 6594
Sum of Squares of x_
Sum of squares (uncorrected) of x_ = 6214894
Data Display
svar(x)
557.667
Data Display
mean(x)
sstdev(x)
942.000
23.6150
Mean of x9
Mean of x9 = 942
Standard Deviation of x9
Standard deviation of x9 = 23.6150
30