"Version 750"
Key
The 1st Hourly
Math 1107
Fall Semester 2006
Protocol
You will use only the following resources:
Your individual calculator, individual tool-sheet (single 8.5 by 11 inch
sheet), writing utensils, Blank Paper (provided by me) and this copy of the
hourly. Share these resources with no-one else. Show complete detail and work
for full credit. Follow case study solutions and sample hourly keys in
presenting your solutions. Work all four
cases. Using only one side of the blank sheets provided, present your work.
Do not write on both sides of the sheets provided, and present your work only
on these sheets. Do not share information with any other students during this
hourly.
Sign and
Acknowledge: I agree to follow this protocol.
________________________________________________________________________
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(PRINTED)
Signature Date
Case One
Long Run Argument
Perfect Samples
Xyrkztin’s Syndrome
Xyrkztin's Syndrome (XS) is a rare,
fictitious disease that I
invented for this test. Symptoms include:
Severity of cases
of Xyrkztin's Syndrome (XS) is noted
as: (F)atal, (S)evere, (Mo)derate, or
(Mi)ld. Suppose that severity probabilities for the population of XS patients
are given below:
|
XS Severity |
Probability |
|
Mild |
.15 |
|
Moderate |
.025 |
|
Severe |
.175 |
|
Fatal |
.65 |
|
Total |
1.00 |
1.a) Interpret each probability using the
Long Run Argument. Be specific and complete for full credit.
In long runs of sampling with replacement
from the population of cases of Xyrkztin's Syndrome, approximately 15% of
sampled cases will show mild severity, approximately 2.5% will show moderate
severity, approximately 17.5% will show severe severity and approximately 65%
will show fatal severity.
1.b) Compute the perfect sample of n=750 XS
cases, and describe the relationship of this perfect sample to real samples of XS
cases.
ExpectedMild, 750=750*PMild
= 750*.15 = 112.5
ExpectedModerate, 750=750*PModerate
= 750*.025 = 18.75
ExpectedSevere, 750=750*PSevere
= 750*.175 = 131.5
ExpectedFatal, 750=750*PFatal
= 750*.65 = 487.5
In random samples of 750 cases of
Xyrkztin's Syndrome, approximately 112 or 113 sampled cases will show mild
severity, approximately 18 or 19 will show moderate severity, approximately 131
or 132 will show severe severity and approximately 487 or 488 will show fatal
severity.
Case Two
Color Slot Machine
Computation of Conditional Probabilities
Here is our slot
machine – on each trial, it produces a 10-color sequence, using the table
below:
|
Sequence* |
Probability |
|
RRBBRRYRRB |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYRGYGBR |
.15 |
|
GRRGRGBRGB |
.10 |
|
BGYGYRYGYY |
.25 |
|
RRGYGRRBBB |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
1.00 |
*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1st
to 6th , from left to right: (1st 2nd 3rd
4th 5th6th7th 8th 9th
10th )
Compute the following conditional
probabilities:
2.1) Pr{ R Shows 4th | Yellow
Shows}
|
Sequence* |
Probability |
|
RRBBRRYRRB |
.10 |
|
BBYYRGYGBR |
.15 |
|
BGYGYRYGYY |
.25 |
|
RRGYGRRBBB |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
1.00 |
Pr{Yellow Shows} = Pr{One of RRBBRRYRRB ,
BBYYRGYGBR, BGYGYRYGYY, RRGYGRRBBB or YYGBYYBGRR Shows} = Pr{RRBBRRYRRB}+
Pr{BBYYRGYGBR}+
Pr{ BGYGYRYGYY}+ Pr{RRGYGRRBBB}+
Pr{YYGBYYBGRR} = .1+.15+.25+.1+.2 = .80
Pr{ R Shows 4th and Yellow
Shows}= 0
Pr{ R Shows 4th | Yellow Shows}
= Pr{ R Shows 4th and Yellow Shows}/ Pr{Yellow Shows} = 0/.8 = 0
2.2) Pr{ Green Shows | “BR” Shows }
|
Sequence* |
Probability |
|
RRBBRRYRRB |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYRGYGBR |
.15 |
|
GRRGRGBRGB |
.10 |
|
Total |
1.00 |
Pr{ “BR” Shows } = Pr{One of RRBBRRYRRB,
RRGGRGBRRB, BBYYRGYGBR, GRRGRGBRGB Shows} = Pr{RRBBRRYRRB}+
Pr{RRGGRGBRRB}+ Pr{BBYYRGYGBR}+
Pr{GRRGRGBRGB} = .1+.1+.15+.1 = .45
Pr{ Green Shows | “BR” Shows } = Pr{One of RRGGRGBRRB, BBYYRGYGBR, GRRGRGBRGB
Shows} = Pr{RRGGRGBRRB}+ Pr{BBYYRGYGBR}+
Pr{GRRGRGBRGB} =.1+.15+.1 = .35
Pr{ Green Shows | “BR” Shows }= Pr{ Green Shows and “BR” Shows }/ Pr{ “BR” Shows
} = .35/.45 = 7/9 » .7777
2.3) Pr{ Yellow Shows | Red Shows}
|
Sequence* |
Probability |
|
RRBBRRYRRB |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYRGYGBR |
.15 |
|
GRRGRGBRGB |
.10 |
|
BGYGYRYGYY |
.25 |
|
RRGYGRRBBB |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
1.00 |
Pr{Red Shows} = Pr{One of RRBBRRYRRB,
RRGGRGBRRB, BBYYRGYGBR, GRRGRGBRGB, BGYGYRYGYY, RRGYGRRBBB or YYGBYYBGRR Shows}
= Pr{RRBBRRYRRB}+ Pr{RRGGRGBRRB}+ Pr{ BBYYRGYGBR}+ Pr{GRRGRGBRGB}+
Pr{BGYGYRYGYY}+ Pr{RRGYGRRBBB}+ Pr{ YYGBYYBGRR} = .1+.1+.15 +.1 +.25+.1+.2 = 1
Pr{Yellow and Red Show} = Pr{One of
RRBBRRYRRB, BBYYRGYGBR, BGYGYRYGYY, RRGYGRRBBB or YYGBYYBGRR Shows} =
Pr{RRBBRRYRRB} + Pr{ BBYYRGYGBR}+ Pr{BGYGYRYGYY}+ Pr{RRGYGRRBBB}+ Pr{
YYGBYYBGRR} = .1+.15 +.25+.1+.2 = .80
Pr{ Yellow Shows | Red Shows}= Pr{ Yellow
and Red Show}/ Pr{Red Shows} = .80/1 = .80
Case Three
Random Variable
Pair of Fair Dice
We have a pair of
independently operating, fair, dice: the first die with face values 1,2,3,4 and
the second die with face values 1,2. Each trial of our experiment involves
tossing the pair of dice, and noting the pair of face values. We compute a
random variable from the pair of face values as follows: First, compute the sum of the faces. Then:
if sum of faces is even,
compute RandomVariable = HighTie
if sum of faces is odd,
compute RandomVariable = LowTie
Where HIGHTIE = higher of the two face
values or common value if tied and LOWTIE = lower of the two face values or
common value if tied.
3.1)
List all
the possible pairs of face values, and compute the probability for each pair,
showing full detail.
|
|
1 |
2 |
3 |
4 |
|
1 |
(1,1) |
(2,1) |
(3,1) |
(4,1) |
|
2 |
(1,2) |
(2,2) |
(3,2) |
(4,2) |
Pr{(a,b)} = Pr{a from 1st
die}*Pr{b from 2nd die} = (1/4)*(1/4) = 1/16
(1,1): Pr{(1,1)} = Pr{1 from 1st
d4}*Pr{1 from 2nd d4} = (1/4)*(1/2) = 1/8
(1,2): Pr{(1,2)} = Pr{1 from 1st
d4}*Pr{2 from 2nd d4} = (1/4)*(1/2) = 1/8
(2,1): Pr{(2,1)} = Pr{2 from 1st
d4}*Pr{1 from 2nd d4} = (1/4)*(1/2) = 1/8
(2,2): Pr{(2,2)} = Pr{2 from 1st
d4}*Pr{2 from 2nd d4} = (1/4)*(1/2) = 1/8
(3,1): Pr{(3,1)} = Pr{3 from 1st
d4}*Pr{1 from 2nd d4} = (1/4)*(1/2) = 1/8
(3,2): Pr{(3,2)} = Pr{3 from 1st
d4}*Pr{2 from 2nd d4} = (1/4)*(1/2) = 1/8
(4,1): Pr{(4,1)} = Pr{4 from 1st
d4}*Pr{1 from 2nd d4} = (1/4)*(1/2) = 1/8
(4,2): Pr{(4,2)} = Pr{4 from 1st
d4}*Pr{2 from 2nd d4} = (1/4)*(1/2) = 1/8
3.2)
Compute
the values of the random variable, showing in detail how these values are
computed from each pair of face values.
if sum of faces is even, compute
RandomVariable = HighTie
if sum of faces is odd, compute
RandomVariable = LowTie
|
Pair Product Even/Odd RV |
1 |
2 |
3 |
4 |
|
1 |
(1,1) 2 Even HT=1 |
(2,1) 3 Odd LT=1 |
(3,1) 4 Even HT=3 |
(4,1) 5 Odd LT=1 |
|
2 |
(1,2) 3 Odd LT=1 |
(2,2) 4 Even HT=2 |
(3,2) 5 Odd LT=2 |
(4,2) 6 Even HT=4 |
(1,1) Þ 1+1=2Þ Even
Þ
HighTie=1
(1,2) Þ 1+2=3Þ Odd
Þ
LowTie=1
(2,1) Þ 2+1=3Þ Odd
Þ
LowTie=1
(2,2) Þ 2+2=4Þ
Even Þ HighTie=2
(3,1) Þ 3+1=4Þ Even
Þ
HighTie=3
(3,2) Þ 3+2=5Þ Odd
Þ
LowTie=2
(4,1) Þ 4+1=5Þ Odd
Þ
LowTie=1
(4,2) Þ 4+2=6Þ
Even Þ HighTie=4
3.3)
Compute
the probability for each value of the random variable in 3.2, showing full
detail.
Pr{Random Variable = 1} = Pr{One of (1,1),
(1,2), (2,1) or (4,1) Shows}= Pr{(1,1)} + Pr{(1,2)}+ Pr{(2,1)}+ Pr{(4,1)} =
(1/8) + (1/8) + (1/8) + (1/8) = 4/8
Pr{Random Variable = 2} = Pr{One of (2,2), (3,2) Shows}= Pr{(2,2)} + Pr{(3,2)} =
(1/8) + (1/8) = 2/8
Pr{Random Variable = 3} = Pr{(3,1)} = 1/8
Pr{Random Variable = 4} = Pr{(4,2)} = 1/8
Case Four
Color Slot Machine
Probability Rules
Here is our slot
machine – on each trial, it produces a 10-color sequence, using the table below:
|
Sequence* |
Probability |
|
RRBBRRYRRB |
.10 |
|
RRGGRGBRRB |
.10 |
|
BBYYRGYGBR |
.15 |
|
GRRGRGBRGB |
.10 |
|
BGYGYRYGYY |
.25 |
|
RRGYGRRBBB |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
1.00 |
*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1st
to 6th , from left to right: (1st 2nd 3rd
4th 5th6th7th 8th 9th
10th )
Compute the
following probabilities. In each of the following, show your intermediate steps
and work. If a rule is specified, you must use that rule for your
computation.
4.1) Pr{“RG” Shows }
|
Sequence* |
Probability |
|
RRGGRGBRRB |
.10 |
|
BBYYRGYGBR |
.15 |
|
GRRGRGBRGB |
.10 |
|
RRGYGRRBBB |
.10 |
|
Total |
1.00 |
Pr{“RG” Shows}=Pr{One of RRGGRGBRRB,
BBYYRGYGBR, GRRGRGBRGB or RRGYGRRBBB Shows}=Pr{ RRGGRGBRRB
}+ Pr{ BBYYRGYGBR }+ Pr{ GRRGRGBRGB }+ Pr{ RRGYGRRBBB } =
.1+.15+.1+.1 = .45
4.2) Pr{ Blue Shows 4th }
|
Sequence* |
Probability |
|
RRBBRRYRRB |
.10 |
|
YYGBYYBGRR |
.20 |
|
Total |
1.00 |
Pr{ Blue Shows 4th } = Pr{One of
RRBBRRYRRB, YYGBYYBGRR
Shows}= Pr{One of RRBBRRYRRB,
YYGBYYBGRR Shows}= Pr{RRBBRRYRRB} + Pr{YYGBYYBGRR} =
.1+.2=.30
4.3) Pr{ Yellow Shows } – Use the
Complementary Rule
Other Event = “Yellow does not Show”
|
Sequence* |
Probability |
|
RRGGRGBRRB |
.10 |
|
GRRGRGBRGB |
.10 |
|
Total |
1.00 |
Pr{ Yellow Does Not Show} = Pr{One of RRGGRGBRRB, GRRGRGBRGB Shows}= Pr{
RRGGRGBRRB } + Pr{ GRRGRGBRGB } = .1+.1 = .2
Pr{ Yellow Shows} = 1 -
Pr{ Yellow Does Not Show} = 1 -
.20 = .80
Work all four cases. Show full detail and work for full credit.