Key | The Comprehensive Final Examination | Math 1107
| Fall Semester 2010 | CJ Alverson
Protocol
You will use only the
following resources: Your
individual calculator; Your individual tool-sheets (two (2) 8.5 by 11 inch
sheets); Your writing utensils; Blank Paper (provided by me); This copy of the
hourly and the tables provided by me. 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. When you’re done: Print your name on a blank sheet of
paper. Place your tool-sheets, test and work under this sheet, and turn it all
in to me. Do not share information with any other students during this
test.
Sign and Acknowledge: I
agree to follow this protocol. Initial: ______
______________________________________________________________________________
Name
(PRINTED)
Signature
Date
Case One | Probability and Random Variables | Color Slot Machine
Here is our slot machine
– on each trial, it produces an 8-color sequence, using the table below:
|
Sequence* |
Probability |
|
RRBBRRYR |
.11 |
|
RRGRBRRB |
.10 |
|
BYGGYGBR |
.14 |
|
GRGYBRGG |
.10 |
|
YYYRYGRY |
.25 |
|
RYGRRBBY |
.15 |
|
YYBYBGRR |
.15 |
|
Total |
1.00 |
Define ThingBC as the number of times that blue
shows on a sequence. Compute the values and probabilities for ThingBC.
|
Sequence* |
Probability |
ThingBC |
|
RRBBRRYR |
.11 |
2 |
|
RRGRBRRB |
.10 |
2 |
|
BYGGYGBR |
.14 |
2 |
|
GRGYBRGG |
.10 |
1 |
|
YYYRYGRY |
.25 |
0 |
|
RYGRRBBY |
.15 |
2 |
|
YYBYBGRR |
.15 |
2 |
|
Total |
1.00 |
|
Pr{ ThingBC = 0 } = Pr{
YYYRYGRY } = 0.25
Pr{ ThingBC = 1 } = Pr{
GRGYBRGG } = 0.10
Pr{ ThingBC = 2 } = Pr{ ThingBC
≠ 0 and ThingBC ≠ 1 } = 1 – 0.25 – 0.10 = 0.65
Or the long way …
Pr{ ThingBC = 2 } = Pr{ One of
RRBBRRYR, RRGRBRRB, BYGGYGBR, RYGRRBBY or
YYBYBGRR Shows } = Pr{ RRBBRRYR } +
Pr{ RRGRBRRB } + Pr{ BYGGYGBR } +
Pr{ RYGRRBBY } + Pr{ YYBYBGRR }
= .11 + .10 + .14 + .15 + .15 = .35 +
.30 = 0.65
Define RVBY as 1 if “BY” shows on a sequence,
and as 0 if “BY” does not show on a sequence. Compute the values and
probabilities for RVBY.
|
Sequence* |
Probability |
RVBY |
|
RRBBRRYR |
.11 |
0/No |
|
RRGRBRRB |
.10 |
0/No |
|
BYGGYGBR |
.14 |
1/Yes |
|
GRGYBRGG |
.10 |
0/No |
|
YYYRYGRY |
.25 |
0/No |
|
RYGRRBBY |
.15 |
1/Yes |
|
YYBYBGRR |
.15 |
1/Yes |
|
Total |
1.00 |
|
Pr{RVBY = 1(Yes) / “BY” Shows} =
Pr{One of BYGGYGBR, RYGRRBBY or YYBYBGRR Shows } =
Pr{ BYGGYGBR } + Pr{ RYGRRBBY } + Pr{ YYBYBGRR } = 0.14 + 0.15 + 0.15 = 0.44
Pr{ RVBY = 0(No) / “BY”
Does Not Show } = 1 – Pr{RVBY = 1(Yes) / “BY” Shows} = 1 – 0.44 = 0.56
Or the long way …
Pr{RVBY = 0(No) / “BY” Does Not
Show} =
Pr{One of RRBBRRYR, RRGRBRRB,
GRGYBRGG or YYYRYGRY Shows } =
Pr{ RRBBRRYR } + Pr{ RRGRBRRB } +
Pr{ GRGYBRGG } + Pr{ YYYRYGRY } =
0.11 + 0.10 + 0.10 + 0.25 = 0.31 +
0.25 = 0.56
Show all work in full detail for full credit.
Case Two | Confidence Interval,
Population Mean | 2009 Fictitious
City Iron Man
An Iron Man
event comprises a 2.4 mile swim, 112 mile
bike course and a full marathon (26.2 mile run). Suppose that we have a random sample of finishers of the
2009 Fictitious City Iron Man, whose finishing times (in hours) are listed
below:
8.25, 8.56, 8.75, 9.10, 9.12, 9.20, 9.22, 9.25,
9.45, 9.85, 10.15, 10.18, 10.20, 10.25, 10.30, 10.46, 10.56, 10.80, 11.05
11.23, 11.30, 11.52, 11.56, 11.60, 11.75, 11.80, 11.95, 12.05, 12.15, 12.18,
12.20, 12.25, 12.28, 12.30, 12.35, 12.45 12.50 12.59, 12.60, 12.70, 12.73,
12.85, 12.90, 13.10, 13.25, 14.15, 15.25, 16.75, 17.30, 18.10
Consider the population
mean finish time for the 2009 Fictitious City Iron Man. Compute and interpret a 95% confidence interval
for this population mean. Show your work. Completely discuss and interpret your results, as
indicated in class and case study summaries.
lower95 =
m – ( Z*sd / sqrt(n) ) ≈ 11.6878 – (2*2.10451/sqrt(50)) ≈ 11.0926
upper95 = m + ( Z*sd / sqrt(n) ) ≈ ≈ 11.6878 – (2*2.10451/sqrt(50)) ≈ 12.2830
Our population is the
population of people who finished the 2009 Fictitious City Iron Man, and our
population mean is the population mean finish time.
Each member of the Family
of Samples is a single random sample of 50 people who finished the 2009
Fictitious City Iron Man.
From
each member sample, compute: sample mean finish time m,
sample standard deviation sd, then from the means table row 2.00 0.02275
0.95450, use Z = 2. Then compute the interval
[lower95
= m – ( Z*sd / sqrt(n) ), upper95 = m +
( Z*sd / sqrt(n) )].
Doing this for each member of the family of
samples yields a family of intervals.
Approximately 95% of the intervals in the family
cover the unknown population mean. If our interval resides in this 95%
supermajority of member intervals, then the true population mean finish time for the 2009 Fictitious City Iron
Man is between 11.1 and 12.3 hours.
Case Three | Summary Intervals | 2009 Fictitious City Iron
Man
Using the context and data of
Case Two, let m denote the sample mean, and sd the sample
standard deviation. Compute and interpret the intervals m±2sd and m±3sd, using Tchebysheff’s Inequalities and the
Empirical Rule. Be specific and complete. Show your work, and discuss
completely for full credit.
lower2 = m – (2*sd)
≈ 11.6878
– (2*2.10451) ≈ 7.47878
upper2 = m + (2*sd)
≈ 11.6878
+ (2*2.10451) ≈ 15.8968
lower3 = m – (3*sd)
≈ 11.6878
– (3*2.10451) ≈ 5.37427
upper3 = m + (3*sd)
≈ 11.6878
+ (3*2.10451) ≈ 18.0013
There are 50 people who
finished the 2009 Fictitious City Iron Man in our sample. At least 75% of the
people in our sample finished the 2009 Fictitious City Iron Man in between 7.48
and 15.90 hours. At least 89% of the people in our sample finished the 2009
Fictitious City Iron Man in between 5.37 and 18.00 hours.
If the finish times for
the 2009 Fictitious City Iron Man cluster symmetrically around a central value,
becoming rare as distance from the center increases, then: Approximately 95% of
the people in our sample finished the 2009 Fictitious City Iron Man in between
7.48 and 15.90 hours and approximately 100% of the people in our sample
finished the 2009 Fictitious City Iron Man in between 5.37 and 18.00 hours.
Case Four | Conditional Probability | Color Slot Machine
Using the color slot
machine from Case One, compute the following
conditional probabilities: Show full work
and detail for full credit.
Pr{ “BR” Shows | Yellow
Shows }
|
Sequence* |
Probability |
Prior(Yellow) |
Joint(“BR” and Yellow) |
|
RRBBRRYR |
.11 |
.11 |
.11 |
|
RRGRBRRB |
.10 |
|
|
|
BYGGYGBR |
.14 |
.14 |
.14 |
|
GRGYBRGG |
.10 |
.10 |
.10 |
|
YYYRYGRY |
.25 |
.25 |
|
|
RYGRRBBY |
.15 |
.15 |
|
|
YYBYBGRR |
.15 |
.15 |
|
|
Total |
1.00 |
0.90 |
0.34 |
Prior: Pr{ Yellow Shows }
Pr{ Yellow Shows } = Pr{
One of RRBBRRYR, BYGGYGBR, GRGYBRGG, YYYRYGRY, RYGRRBBY
or YYBYBGRR Shows } = Pr{
RRBBRRYR } + Pr{ BYGGYGBR } + Pr{ GRGYBRGG } +
Pr{ YYYRYGRY } + Pr{
RYGRRBBY } + Pr{ YYBYBGRR } = .11 + .14 + .10 + .25 + .15 + .15 = 0.90
Joint: Pr{ “BR” and
Yellow Show }
Pr{ “BR” and Yellow Shows
} = Pr{ One of RRBBRRYR, BYGGYGBR, GRGYBRGG Shows } =
Pr{ RRBBRRYR } +
Pr{ BYGGYGBR } + Pr{ GRGYBRGG } = .11 + .14 + .10 = 0.34
Pr{ “BR” Shows | Yellow
Shows } = Joint/Prior = Pr{ “BR” and Yellow Show }/ Pr{ Yellow Shows } =
.34/.90
Pr{ “RBB” Shows |
Yellow Shows }
|
Sequence* |
Probability |
Prior(Yellow) |
Joint(“RBB” and Yellow) |
|
RRBBRRYR |
.11 |
.11 |
.11 |
|
RRGRBRRB |
.10 |
|
|
|
BYGGYGBR |
.14 |
.14 |
|
|
GRGYBRGG |
.10 |
.10 |
|
|
YYYRYGRY |
.25 |
.25 |
|
|
RYGRRBBY |
.15 |
.15 |
.15 |
|
YYBYBGRR |
.15 |
.15 |
|
|
Total |
1.00 |
0.90 |
0.26 |
Prior: Pr{ Yellow Shows }
Pr{ Yellow Shows } = Pr{
One of RRBBRRYR, BYGGYGBR, GRGYBRGG, YYYRYGRY, RYGRRBBY
or YYBYBGRR Shows } = Pr{
RRBBRRYR } + Pr{ BYGGYGBR } + Pr{ GRGYBRGG } +
Pr{ YYYRYGRY } + Pr{
RYGRRBBY } + Pr{ YYBYBGRR } = .11 + .14 + .10 + .25 + .15 + .15 = 0.90
Joint: Pr{ “RBB” and
Yellow Show }
Pr{ “RBB” and Yellow
Shows } = Pr{ RRBBRRYR or RYGRRBBY Shows } = Pr{ RRBBRRYR
} +
Pr{ RYGRRBBY } =
.11 + .15 = 0.26
Pr{ “RBB” Shows | Yellow
Shows } = Joint/Prior = Pr{ “RBB” and Yellow Show }/ Pr{ Yellow Shows } = .26/.90
Pr{ Red Shows | Blue
Shows }
|
Sequence* |
Probability |
Prior(Blue) |
Joint(Red and Blue) |
|
RRBBRRYR |
.11 |
.11 |
.11 |
|
RRGRBRRB |
.10 |
.10 |
.10 |
|
BYGGYGBR |
.14 |
.14 |
.14 |
|
GRGYBRGG |
.10 |
.10 |
.10 |
|
YYYRYGRY |
.25 |
|
|
|
RYGRRBBY |
.15 |
.15 |
.15 |
|
YYBYBGRR |
.15 |
.15 |
.15 |
|
Total |
1.00 |
0.75 |
0.75 |
Prior: Pr{ Blue Shows }
Pr{ Blue Shows } = Pr{
One of RRBBRRYR, RRGRBRRB, BYGGYGBR, GRGYBRGG, RYGRRBBY or
YYBYBGRR Shows } = Pr{ RBBRRYR
} + Pr{ RRGRBRRB } + Pr{ BYGGYGBR } +
Pr{ GRGYBRGG } + Pr{ RYGRRBBY
} + Pr{ YYBYBGRR } = .11 + .10 + .14 + .10 + .15 + .15 = 0.75
Joint: Pr{ Red and Blue Show
}
Pr{ Red and Blue Show} =
Pr{ One of RRBBRRYR, RRGRBRRB, BYGGYGBR, GRGYBRGG, RYGRRBBY or YYBYBGRR Shows }
= Pr{ RBBRRYR } + Pr{ RRGRBRRB } + Pr{ BYGGYGBR } +
Pr{ GRGYBRGG } + Pr{ RYGRRBBY
} + Pr{ YYBYBGRR } = .11 + .10 + .14 + .10 + .15 + .15 = 0.75
Pr{ Red Shows | Blue
Shows } = Joint/Prior = Pr{ Red Shows }/ Pr{ Blue Shows } = .75/.75
Table 1: Means and Proportions
|
Z(k) PROBRT ROBCENT 0.05 0.48006
0.03988 0.10 0.46017
0.07966 0.15 0.44038
0.11924 0.20 0.42074
0.15852 0.25 0.40129
0.19741 0.30 0.38209
0.23582 0.35 0.36317
0.27366 0.40 0.34458
0.31084 0.45 0.32636
0.34729 0.50 0.30854
0.38292 0.55 0.29116
0.41768 0.60 0.27425
0.45149 0.65 0.25785
0.48431 0.70 0.24196
0.51607 0.75 0.22663
0.54675 0.80 0.21186
0.57629 0.85 0.19766
0.60467 0.90 0.18406
0.63188 0.95 0.17106
0.65789 1.00 0.15866
0.68269 |
Z(k) PROBRT PROBCENT 1.05 0.14686
0.70628 1.10 0.13567
0.72867 1.15 0.12507
0.74986 1.20 0.11507 0.76986 1.25 0.10565
0.78870 1.30 0.09680
0.80640 1.35 0.08850
0.82298 1.40 0.08075
0.83849 1.45 0.07352
0.85294 1.50 0.06680
0.86639 1.55 0.06057
0.87886 1.60 0.05479
0.89040 1.65 0.04947
0.90106 1.70 0.04456
0.91087 1.75 0.04005
0.91988 1.80 0.03593
0.92814 1.85 0.03215
0.93569 1.90 0.02871
0.94257 1.95 0.02558
0.94882 2.00 0.02275
0.95450 |
Z(k) PROBRT PROBCENT 2.05 0.020182
0.95964 2.10 0.017864
0.96427 2.15 0.015778
0.96844 2.20 0.013903
0.97219 2.25 0.012224
0.97555 2.30 0.010724
0.97855 2.35 0.009387
0.98123 2.40 0.008198
0.98360 2.45 0.007143
0.98571 2.50 0.006210
0.98758 2.55 0.005386
0.98923 2.60 0.004661
0.99068 2.65 0.004025
0.99195 2.70 .0034670
0.99307 2.75 .0029798
0.99404 2.80 .0025551
0.99489 2.85 .0021860
0.99563 2.90 .0018658
0.99627 2.95 .0015889
0.99682 3.00 .0013499
0.99730 |