Key | 1st
Hourly | Math 1107 | Fall Semester 2010
You will use only the
following resources: Your individual calculator; Your
individual tool-sheet (one (1) 8.5 by 11 inch sheet); Your writing utensils; Blank
Paper (provided by me) and this copy of the hourly.
Do not share these resources
with anyone 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. When you are finished:
Prepare a
Cover Sheet: Print only your name on an otherwise blank sheet of
paper. Then stack your stuff as follows: Cover Sheet (Top), Your Work Sheets, The Test Papers and Your Toolsheet.
Then hand all of this in to me.
Sign and
Acknowledge: I agree to follow this protocol: (initial)
Sign___________________________Name______________________________Date________________
Show
full work and detail for full credit. Be sure that you have worked all four
cases.
Case One, Long Run Argument and Perfect Samples
Consider the population of resident
live births in the United States in year 2007 – suppose that the following
probability model accurately describes the distribution of year 2007 United States
resident live births by maternal age ( at delivery) category:
|
Maternal
Age (Years) |
Probability |
|
Strictly
Younger than 20 |
0.11 |
|
20-24 |
0.25 |
|
25-34 |
0.50 |
|
35
or Older |
0.14 |
|
Total |
1.00 |
a) Interpret each probability
using the Long Run Argument.
In long runs of random sampling from year
2007 US resident live births, approximately 11% of sampled births are to
mothers aged strictly less than 20 years of age.
In long runs of random sampling from year
2007 US resident live births, approximately 25% of sampled births are to
mothers aged between 20 and 24 years of age.
In long runs of random sampling from year
2007 US resident live births, approximately 50% of sampled births are to
mothers aged between 25 and 34 years of age.
In long runs of random sampling from year
2007 US resident live births, approximately 14% of sampled births are to
mothers aged 35 or more years of age.
b) Compute and discuss
Perfect Samples for n=5000.
|
Maternal
Age (Years) |
Probability |
Expectedn=5000 |
|
Strictly
Younger than 20 |
0.11 |
5000*.11 = 550 |
|
20-24 |
0.25 |
5000*0.25 = 1250 |
|
25-34 |
0.50 |
5000*0.50 = 2500 |
|
35
or Older |
0.14 |
5000*0.14 = 700 |
|
Total |
1.00 |
5000*1.00 = 5000 |
Expectedn=5000, <20 = 5000*0.11 = 550
In random samples of 5,000 year 2007 US
resident live births, approximately 550 of 5,000 sampled births are to mothers
aged strictly less than 20 years of age.
Expectedn=5000, 20-24 = 5000*0.25 =1250
In random samples of 5,000 year 2007 US
resident live births, approximately 1,250 of 5,000 sampled births are to
mothers aged between 20 and 24 years of age.
Expectedn=5000, 25-34 = 5000*0.50 = 2500
In random samples of 5,000 year 2007 US
resident live births, approximately 2,500 of 5,000 sampled births are to
mothers aged between 25 and 34 years of age.
Expectedn=5000, 35+ = 5000*0.14 = 700
In random samples of 5,000 year 2007 US
resident live births, approximately 700 of 5,000 sampled births are to mothers
aged 35 or more years of age.
Show all work and full detail for full
credit. Provide complete discussion for full credit.
Case Two: Random Variables, Pair of Dice
We have a pair of
independently operating dice, whose
probability models are given below:
|
1st
Face Value |
Probability |
|
2nd
Face Value |
Probability |
|
1 |
1/10 |
|
3 |
9/20 |
|
2 |
4/10 |
|
4 |
2/20 |
|
6 |
4/10 |
|
5 |
9/20 |
|
7 |
1/10 |
|
Total |
20/20 |
|
Total |
10/10 |
|
|
|
Each trial of our
experiment involves tossing the pair
of dice: (1st Face Value, 2nd
Face Value), and noting the pair of face values. We compute random variables
from the pair of face values as follows: Define
DIFF = (2nd
Face Value) – (1st Face Value). Compute the values and probabilities
for DIFF.
a) List all the
possible pairs of face values, and compute the probability for each pair,
showing full detail.
|
2nd ß
1st Ţ |
1(1/10) |
2(4/10) |
6(4/10) |
7(1/10) |
|
3(9/20) |
(1,3) |
(2,3) |
(6,3) |
(7,3) |
|
4(2/20) |
(1,4) |
(2,4) |
(6,4) |
(7,4) |
|
5(9/20) |
(1,5) |
(2,5) |
(6,5) |
(7,5) |
(1,3),(2,3),(6,3),(7,3),(1,4),(2,4),(6,4),(7,4),(1,5),(2,5),(6,5),(7,5)
Pr{(1,3)} = Pr{1from 1st}*Pr{3 from
2nd } = (1/10)*(9/20) =
9/200
Pr{(2,3)} = Pr{2from 1st}*Pr{3 from
2nd } = (4/10)*(9/20) = 36/200
Pr{(6,3)} = Pr{6from 1st}*Pr{3 from
2nd } = (4/10)*(9/20) = 36/200
Pr{(7,3)} = Pr{7from 1st}*Pr{3 from
2nd } = (1/10)*(9/20) = 9/200
Pr{(1,4)} = Pr{1from 1st}*Pr{4 from
2nd } = (1/10)*(2/20) = 2/200
Pr{(2,4)} = Pr{2from 1st}*Pr{4 from
2nd } = (4/10)*(2/20) = 8/200
Pr{(6,4)} = Pr{6from 1st}*Pr{4 from
2nd } = (4/10)*(2/20) = 8/200
Pr{(7,4)} = Pr{7from 1st}*Pr{4 from
2nd } = (1/10)*(2/20) = 2/200
Pr{(1,5)} = Pr{1from 1st}*Pr{5 from
2nd } = (1/10)*(9/20) =
9/200
Pr{(2,5)} = Pr{2from 1st}*Pr{5 from
2nd } = (4/10)*(9/20) = 36/200
Pr{(6,5)} = Pr{6from 1st}*Pr{5 from
2nd } = (4/10)*(9/20) = 36/200
Pr{(7,5)} = Pr{7from 1st}*Pr{5 from
2nd } = (1/10)*(9/20) = 9/200
b) Compute the values and probabilities
for DIFF, showing full detail.
DIFF{(1st
Face Value, 2nd Face Value)}
= (2nd Face Value) – (1st
Face Value).
Diff{(1,3)} = 3 – 1 = 2 (@9/200)
Diff{(2,3)} = 3 – 2 = 1 (@36/200)
Diff{(6,3)} = 3 – 6 = -3 (@36/200)
Diff{(7,3)} = 3 – 7 = -4 (@9/200)
Diff{(1,4)} = 4 – 1 = 3(@2/200)
Diff{(2,4)} = 4 – 2 = 2 (@8/200)
Diff{(6,4)} = 4 – 6 = -2 (@8/200)
Diff{(7,4)} = 4 – 7 = -3 (@2/200)
Diff{(1,5)} = 5 – 1 = 4 (@9/200)
Diff{(2,5)} = 5 – 2 = 3 (@36/200)
Diff{(6,5)} = 5 – 6 = -1 (@36/200)
Diff{(7,5)} = 5 – 7 = -2 (@9/200)
Pr{Diff = 4} = Pr{(1,5)} = 9/200
Pr{Diff = 3} = Pr{One of (2,5),(1,4) Shows} =
Pr{(2,5)}+ Pr{(1,4)} = (36/200) + (2/200) = 38/200
Pr{Diff = 2} = Pr{One of (1,3),(2,4) Shows} =
Pr{(1,3)}+ Pr{(2,4)} = (9/200) + (8/200) = 17/200
Pr{Diff = 1} = Pr{(2,3) Shows} = 36/200
Pr{Diff = -1} = Pr{(6,5) Shows} = 36/200
Pr{Diff = -2} = Pr{One of (6,4),(7,5) Shows} =
Pr{(6,4)}+ Pr{(7,5)} = (8/200) + (9/200) = 17/200
Pr{Diff = -3} = Pr{One of (6,3),(7,4) Shows} =
Pr{(6,3)}+ Pr{(7,4)} = (36/200) + (2/200) = 38/200
Pr{Diff = -4} = Pr{(7,3) Shows} = 9/200
Case Three |
Color Slot Machine | Probability Rules
Here is our slot machine – on
each trial, it produces a color sequence, using the table below:
|
Sequence* |
Probability |
Sequence* |
Probability |
|
BBRRYRRR |
.10 |
BGYRYGYY |
.25 |
|
GGRGBRRB |
.10 |
YYGRRBBY |
.10 |
|
BBYYGGBR |
.15 |
YBYYBGRR |
.20 |
|
GRRYBRGG |
.10 |
|
|
*B-Blue, G-Green, R-Red, Y-Yellow
Compute the following conditional
probabilities. Show full work and detail for full credit.
Pr{ Yellow Shows Exactly Twice }
|
Sequence* |
Probability |
Sequence* |
Probability |
|
BBRRYRRR |
.10 |
BGYRYGYY |
.25 |
|
GGRGBRRB |
.10 |
YYGRRBBY |
.10 |
|
BBYYGGBR |
.15 |
YBYYBGRR |
.20 |
|
GRRYBRGG |
.10 |
|
|
Pr{ Yellow Shows Exactly Twice
} = Pr{ BBYYGGBR } = 0.15
Pr{ “RB” Shows }
|
Sequence* |
Probability |
Sequence* |
Probability |
|
BBRRYRRR |
.10 |
BGYRYGYY |
.25 |
|
GGRGBRRB |
.10 |
YYGRRBBY |
.10 |
|
BBYYGGBR |
.15 |
YBYYBGRR |
.20 |
|
GRRYBRGG |
.10 |
|
|
Pr{ “RB” Shows } = Pr{One
of GGRGBRRB , YYGRRBBY
Shows} = Pr{GGRGBRRB} + Pr{YYGRRBBY Shows} = 0.10 +
0.10 = 0.20
Pr{ Yellow Shows } – Use the Complementary Rule
Other Event = OE = “Yellow Does Not Show”
|
Sequence* |
Probability |
Sequence* |
Probability |
|
BBRRYRRR |
.10 |
BGYRYGYY |
.25 |
|
GGRGBRRB |
.10 |
YYGRRBBY |
.10 |
|
BBYYGGBR |
.15 |
YBYYBGRR |
.20 |
|
GRRYBRGG |
.10 |
|
|
Pr{Other Event} = Pr{OE} =
Pr{“Yellow Does Not Show”} = Pr{GGRGBRRB} = 0.10
Pr{ Yellow Shows } = 1 –
Pr{“Yellow Does Not Show”} = 1 – 0.10 = 0.90
Case Four |
Conditional Probability | Color Slot Machine
Here is our slot
machine – on each trial, it produces an 8-color sequence, using the table
below:
|
Sequence* |
Probability |
Sequence* |
Probability |
|
RRBBRRYR |
.21 |
YYYRYGYY |
.20 |
|
BYGGYGBR |
.14 |
RYGRRBBY |
.35 |
|
GRGYBRGG |
.10 |
|
|
*B-Blue, G-Green, R-Red, Y-Yellow
Compute the following conditional probabilities. Show full work and detail for full credit.
a) Pr{“RRB” Shows | Green Shows}
|
Sequence* |
Probability |
Sequence* |
Probability |
|
YYYRYGYY |
.20 |
||
|
BYGGYGBR |
.14 |
RYGRRBBY |
.35 |
|
GRGYBRGG |
.10 |
|
|
Pr{Green
Shows} =
Pr{One
of YYYRYGYY, BYGGYGBR, RYGRRBBY or GRGYBRGG Shows} =
Pr{YYYRYGYY}
+ Pr{BYGGYGBR} + Pr{RYGRRBBY} + Pr{GRGYBRGG} = .20+.14+.35+.10 = 0.79
|
Sequence* |
Probability |
Sequence* |
Probability |
|
|
|
||
|
RYGRRBBY |
.35 |
||
|
|
|
Pr{“RRB”
and Green Shows} = Pr{RYGRRBBY} = 0.35
Pr{“RRB”
Shows | Green Shows} = Pr{“RRB” and Green Shows}/ Pr{Green Shows} = 0.35/0.79
b) Pr{ “BR” Shows | Green Shows }
|
Sequence* |
Probability |
Sequence* |
Probability |
|
YYYRYGYY |
.20 |
||
|
BYGGYGBR |
.14 |
RYGRRBBY |
.35 |
|
GRGYBRGG |
.10 |
|
|
Pr{Green
Shows} =
Pr{One
of YYYRYGYY, BYGGYGBR, RYGRRBBY or GRGYBRGG Shows} =
Pr{YYYRYGYY}
+ Pr{BYGGYGBR} + Pr{RYGRRBBY} + Pr{GRGYBRGG} = .20+.14+.35+.10 = 0.79
|
Sequence* |
Probability |
Sequence* |
Probability |
|
|
|
||
|
BYGGYGBR |
.14 |
|
|
|
GRGYBRGG |
.10 |
|
|
Pr{“BR”
and Green Shows} = Pr{One of BYGGYGBR
or GRGYBRGG Shows} =
Pr{BYGGYGBR}
+ Pr{GRGYBRGG} = 0.14 + 0.10 = 0.24
Pr{ “BR”
Shows | Green Shows } = Pr{ “BR”
and Green Shows }/ Pr{ “BR” Shows } =
0.24/0.79
c) Pr{ Blue Shows | Red Shows}
|
Sequence* |
Probability |
Sequence* |
Probability |
|
RRBBRRYR |
.21 |
YYYRYGYY |
.20 |
|
BYGGYGBR |
.14 |
RYGRRBBY |
.35 |
|
GRGYBRGG |
.10 |
|
|
Pr{Red
Shows} =
Pr{One
of RRBBRRYR, YYYRYGYY, BYGGYGBR,
RYGRRBBY or GRGYBRGG Shows} = Pr{RRBBRRYR} + Pr{YYYRYGYY} + Pr{BYGGYGBR} +
Pr{RYGRRBBY} + Pr{GRGYBRGG} = 0.21+0.20 +0.14+0.35+0.10 = 1
|
Sequence* |
Probability |
Sequence* |
Probability |
|
RRBBRRYR |
.21 |
|
|
|
BYGGYGBR |
.14 |
RYGRRBBY |
.35 |
|
GRGYBRGG |
.10 |
|
|
Pr{Blue
and Red Show} =
Pr{One
of RRBBRRYR, BYGGYGBR, RYGRRBBY or
GRGYBRGG Shows} =
Pr{RRBBRRYR}
+ Pr{BYGGYGBR} + Pr{RYGRRBBY} + Pr{GRGYBRGG} =
0.21+0.14+0.35+0.10
= 0.80
Pr{
Blue Shows | Red Shows}= Pr{ Blue and Red Show}/ Pr{Red Shows} = 0.80/1
Work all four cases.