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 | 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

RRBBRRYR

.10

RRRGBRRB

.11

BYGGYGBR

.14

GRGYBRGG

.10

YYYRYGYY

.20

RYGRRBBY

.15

YYYYBGRR

.20

Total

1.00

 

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

.10

.10

.10

RRRGBRRB

.11

 

 

BYGGYGBR

.14

.14

.14

GRGYBRGG

.10

.10

.10

YYYRYGYY

.20

.20

 

RYGRRBBY

.15

.15

 

YYYYBGRR

.20

.20

 

Total

1.00

0.89

0.34

 

Prior: Pr{ Yellow Shows }

Pr{ Yellow Shows } = Pr{ One of RRBBRRYR, BYGGYGBR, GRGYBRGG, YYYRYGYY, RYGRRBBY or

YYYYBGRR Shows } = Pr{ RRBBRRYR } + Pr{ BYGGYGBR } + Pr{ GRGYBRGG } + Pr{ YYYRYGYY} +  Pr{ RYGRRBBY } + Pr{ YYYYBGRR } =  .10 + .14 + .10 + .20 + .15 + .20 = 0.89

 

Joint: Pr{ “BR” and Yellow Show }

Pr{ “BR” and Yellow Shows } = Pr{ One of RRBBRRYR, BYGGYGBR or GRGYBRGG Shows } =

Pr{ RRBBRRYR } + Pr{ BYGGYGBR } + Pr{ GRGYBRGG } =  .10 + .14 + .10 = 0.34

Pr{ “BR” Shows | Yellow Shows } = Joint/Prior = Pr{ “BR” and Yellow Show }/ Pr{ Yellow Shows } = .34/.89

 

Pr{ “RBB” Shows  | Yellow Shows }

 

Sequence*

Probability

Prior(Yellow)

Joint(“RBB” and Yellow)

RRBBRRYR

.10

.10

.10

RRRGBRRB

.11

 

 

BYGGYGBR

.14

.14

 

GRGYBRGG

.10

.10

 

YYYRYGYY

.20

.20

 

RYGRRBBY

.15

.15

.15

YYYYBGRR

.20

.20

 

Total

1.00

0.89

0.25

 

Prior: Pr{ Yellow Shows }

Pr{ Yellow Shows } = Pr{ One of RRBBRRYR, BYGGYGBR, GRGYBRGG, YYYRYGYY, RYGRRBBY or

YYYYBGRR Shows } = Pr{ RRBBRRYR } + Pr{ BYGGYGBR } + Pr{ GRGYBRGG } + Pr{ YYYRYGYY} +  Pr{ RYGRRBBY } + Pr{ YYYYBGRR } =  .10 + .14 + .10 + .20 + .15 + .20 = 0.89

 

Joint: Pr{ “RBB” and Yellow Show }

Pr{ “RBB” and Yellow Shows } = Pr{ RRBBRRYR } + Pr{ RYGRRBBY } =  .10 + .15 = 0.25

 

Pr{ “RBB” Shows | Yellow Show } = Joint/Prior = Pr{ “RBB” and Yellow Show }/ Pr{ Yellow Shows } = .25/.89

 

Pr{ Red Shows | Blue Shows }

 

 

Sequence*

Probability

Prior(Blue)

Joint(Red and Blue)

RRBBRRYR

.10

.10

.10

RRRGBRRB

.11

.11

.11

BYGGYGBR

.14

.14

.14

GRGYBRGG

.10

.10

.10

YYYRYGYY

.20

 

 

RYGRRBBY

.15

.15

.15

YYYYBGRR

.20

.20

.20

Total

1.00

0.80

0.80

 

Prior: Pr{ Blue Shows }

Pr{ Blue Shows } = Pr{ One of RRBBRRYR, RRRGBRRB, BYGGYGBR, GRGYBRGG, RYGRRBBY or

YYYYBGRR Shows } = Pr{ RRBBRRYR } + Pr{ RRRGBRRB } + Pr{ BYGGYGBR } + Pr{ GRGYBRGG } + Pr{ RYGRRBBY } + Pr{ YYYYBGRR } = .10 + .11 + .14 + .10 + .15 + .20 = 0.80

 

Joint: Pr{ Red and Blue Show }

Pr{ Red and Blue Show } = Pr{ One of RRBBRRYR, RRRGBRRB, BYGGYGBR, GRGYBRGG, RYGRRBBY or YYYYBGRR Shows } = Pr{ RRBBRRYR } + Pr{ RRRGBRRB } + Pr{ BYGGYGBR } + Pr{ GRGYBRGG } + Pr{ RYGRRBBY } + Pr{ YYYYBGRR } = .10 + .11 + .14 + .10 + .15 + .20 = 0.80

 

Pr{ Red Shows | Blue Shows } = Joint/Prior = Pr{ Red and Blue Show }/ Pr{ Blue Shows } = .80/.80

 

 

 

 

Case Two | Probability and Random Variables | Pair of Dice

 We have a pair of four-sided dice. We assume that the dice operate separately and independently of each other. Here are their probability models: 

1st Die

Probability

 

2nd Die

Probability

1

1/10

 

3

1/4

2

2/10

 

4

1/4

7

3/10

 

5

1/4

8

4/10

 

6

1/4

Total

10/10

 

Total

4/4

 

Our experiment consists of tossing the pair of dice, and noting the resulting pair of face values as (1st Die, 2nd Die).

 

List the possible face-pairs, and compute a probability for each pair.

 

 

1(1/10)

2(2/10)

7(3/10)

8(4/10)

3(1/4)

(1,3)

(2,3)

(7,3)

(8,3)

4(1/4)

(1,4)

(2,4)

(7,4)

(8,4)

5(1/4)

(1,5)

(2,5)

(7,5)

(8,5)

6(1/4)

(1,6)

(2,6)

(7,6)

(8,6)

 

Pr{(1,3)} = Pr{1 from 1st}*Pr{3 from 2nd} = (1/10)*(1/4)  = 1/40

Pr{(1,4)} = Pr{1 from 1st}*Pr{4 from 2nd} = (1/10)*(1/4)  = 1/40

Pr{(1,5)} = Pr{1 from 1st}*Pr{5 from 2nd} = (1/10)*(1/4)  = 1/40

Pr{(1,6)} = Pr{1 from 1st}*Pr{6 from 2nd} = (1/10)*(1/4)  = 1/40

 

Pr{(2,3)} = Pr{2 from 1st}*Pr{3 from 2nd} = (2/10)*(1/4)  = 2/40

Pr{(2,4)} = Pr{2 from 1st}*Pr{4 from 2nd} = (2/10)*(1/4)  = 2/40

Pr{(2,5)} = Pr{2 from 1st}*Pr{5 from 2nd} = (2/10)*(1/4)  = 2/40

Pr{(2,6)} = Pr{2 from 1st}*Pr{6 from 2nd} = (2/10)*(1/4)  = 2/40

 

Pr{(7,3)} = Pr{7 from 1st}*Pr{3 from 2nd} = (3/10)*(1/4)  = 3/40

Pr{(7,4)} = Pr{7 from 1st}*Pr{4 from 2nd} = (3/10)*(1/4)  = 3/40

Pr{(7,5)} = Pr{7 from 1st}*Pr{5 from 2nd} = (3/10)*(1/4)  = 3/40

Pr{(7,6)} = Pr{7 from 1st}*Pr{6 from 2nd} = (3/10)*(1/4)  = 3/40

 

Pr{(8,3)} = Pr{8 from 1st}*Pr{3 from 2nd} = (4/10)*(1/4)  = 4/40

Pr{(8,4)} = Pr{8 from 1st}*Pr{4 from 2nd} = (4/10)*(1/4)  = 4/40

Pr{(8,5)} = Pr{8 from 1st}*Pr{5 from 2nd} = (4/10)*(1/4)  = 4/40

Pr{(8,6)} = Pr{8 from 1st}*Pr{6 from 2nd} = (4/10)*(1/4)  = 4/40

 

 

Define HIGH as the higher of the two face values in the pair, and LOW as the lower of the two face values in the pair and MID as MID = (HIGH + LOW)/2. Compute the values and probabilities for MID. Show all work in full detail for full credit.

 

 

1(1/10)

2(2/10)

7(3/10)

8(4/10)

3(1/4)

(1,3) | 2

(2,3) | 2.5

(7,3) | 5

(8,3) | 5.5

4(1/4)

(1,4) | 2.5

(2,4) | 3

(7,4) | 5.5

(8,4) | 6

5(1/4)

(1,5) | 3

(2,5) | 3.5

(7,5) | 6

(8,5) | 6.5

6(1/4)

(1,6) | 3.5

(2,6) | 4

(7,6) | 6.5

(8,6) | 7

 

MID {(1,3)} = (3 + 1)/2 = 2 @ 1/40

MID {(1,4)} = (4 + 1)/2 = 2.5 @ 1/40

MID {(1,5)} = (5 + 1)/2 = 3 @ 1/40

MID {(1,6)} = (6 + 1)/2 = 3.5 @ 1/40

 

MID {(2,3)} = (3 + 2)/2 = 2.5 @ 2/40

MID {(2,4)} = (4 + 2)/2 = 3 @ 2/40

MID {(2,5)} = (5 + 2)/2 = 3.5 @ 2/40

MID {(2,6)} = (6 + 2)/2 = 4 @ 2/40

 

MID {(7,3)} = (7 + 3)/2 = 5 @  3/40

MID {(7,4)} = (7 + 4)/2 = 5.5 @  3/40

MID {(7,5)} = (7 + 5)/2 = 6 @  3/40

MID {(7,6)} = (7 + 6)/2 = 6.5 @ 3/40

 

MID {(8,3)} = (8 + 3)/2 = 5.5 @ 4/40

MID {(8,4)} = (8 + 4)/2 = 6 @ 4/40

MID {(8,5)} = (8 + 5)/2 = 6.5 @ 4/40

MID {(8,6)} = (8 + 6)/2 = 7 @ 4/40

 

Regrouping pairs by MID…

 

MID {(1,3)} = (3 + 1)/2 = 2 @ 1/40

 

MID {(1,4)} = (4 + 1)/2 = 2.5 @ 1/40

MID {(2,3)} = (3 + 2)/2 = 2.5 @ 2/40

 

MID {(1,5)} = (5 + 1)/2 = 3 @ 1/40

MID {(2,4)} = (4 + 2)/2 = 3 @ 2/40

 

MID {(1,6)} = (6 + 1)/2 = 3.5 @ 1/40

MID {(2,5)} = (5 + 2)/2 = 3.5 @ 2/40

 

MID {(2,6)} = (6 + 2)/2 = 4 @ 2/40

 

MID {(7,3)} = (7 + 3)/2 = 5 @  3/40

 

MID {(7,4)} = (7 + 4)/2 = 5.5 @  3/40

MID {(8,3)} = (8 + 3)/2 = 5.5 @ 4/40

 

MID {(7,5)} = (7 + 5)/2 = 6 @  3/40

MID {(8,4)} = (8 + 4)/2 = 6 @ 4/40

 

MID {(7,6)} = (7 + 6)/2 = 6.5 @ 3/40

MID {(8,5)} = (8 + 5)/2 = 6.5 @ 4/40

 

MID {(8,6)} = (8 + 6)/2 = 7 @ 4/40

 

Probability Model for MID

 

Pr{ MID = 2 } = Pr{ (1,3) } = 1/40

Pr{ MID = 2.5 } = Pr{ (1,4) or (2,3) Shows } = Pr{ (1,4) } + Pr{ (2,3) } = (1/40) + (2/40) = 3/40

Pr{ MID = 3 } = Pr{ (1,5) or (2,4) Shows } = Pr{ (1,5) } + Pr{ (2,4) } = (1/40) + (2/40) = 3/40

Pr{ MID = 3.5 } = Pr{ (1,6) or (2,4) Shows } = Pr{ (1,6) } + Pr{ (2,5) } = (1/40) + (2/40) = 3/40

Pr{ MID = 4 } = Pr{ (2,6) } = 2/40

Pr{ MID = 5 } = Pr{ (7,3) } = 3/40

Pr{ MID = 5.5 } = Pr{ (7,4) or (8,3) Shows } = Pr{ (7,4) } + Pr{ (8,3) } = (3/40) + (4/40) = 7/40

Pr{ MID = 6 } = Pr{ (7,5) or (8,4) Shows } = Pr{ (7,5) } + Pr{ (8,4) } = (3/40) + (4/40) = 7/40

Pr{ MID = 6.5 } = Pr{ (7,6) or (8,5) Shows } = Pr{ (7,6) } + Pr{ (8,5) } = (3/40) + (4/40) = 3/40

Pr{ MID = 7 } = Pr{ (8,6) } = 4/40

 

Case Three | Descriptive Statistics | 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

Compute and interpret the following statistics: sample size, p00, p25, p50, p75, p100,

(p50 – p00), (p75 – p25), (p100 – p50). Show your work. Completely discuss and interpret your results, as indicated in class and case study summaries.

 

n     p0      p25      p50      p75     p100     R20      R31     R42

50    8.25    10.2    11.775    12.59    18.1    3.525    2.39    6.325

 

There are 50 people who finished the 2009 Fictitious City Iron Man in our sample.

The fastest finisher in our sample finished the 2009 Fictitious City Iron Man in 8.25 hours.

Approximately 25% of the 2009 Fictitious City Iron Man finishers in our sample finished in 10.2 hours or less.

Approximately 50% of the 2009 Fictitious City Iron Man finishers in our sample finished in 11.775 hours or less.

Approximately 75% of the 2009 Fictitious City Iron Man finishers in our sample finished in 12.59 hours or less.

The slowest finisher in our sample finished the 2009 Fictitious City Iron Man in 18.10 hours.

 

range20 = p50 – p00 = 11.775 – 8.25 = 3.525

Approximately 25% of the 2009 Fictitious City Iron Man finishers in our sample finished in between 8.25 and 11.775 hours. The largest possible difference in finish time between any pair of finishers in our lower half sample is 3.525 hours.

 

Range31 = p75 – p25 = 12.59 – 10.2 = 2.39

Approximately 25% of the 2009 Fictitious City Iron Man finishers in our sample finished in between 10.2 and 12.59 hours. The largest possible difference in finish time between any pair of finishers in our middle half sample is 2.39 hours.

 

Range42 = p100 – p50 = 18.1 – 11.775 = 6.325

Approximately 25% of the 2009 Fictitious City Iron Man finishers in our sample finished in between 11.775 and 18.1 hours. The largest possible difference in finish time between any pair of finishers in our upper half sample is 6.325 hours.

 

Case Four | Confidence Interval, Population Proportion | 2009 Fictitious City Iron Man

 

Using the context and data of Case Three, consider the proportion of finishers of the 2009 Fictitious City Iron Man who finish in strictly less than 10 hours. Compute and interpret a 95% confidence interval for this population proportion. Show your work. Completely discuss and interpret your results, as indicated in class and case study summaries.

 

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

 

sample size n = 50

event=“2009 Fictitious City Iron Man finish time is strictly less than 10 hours”

event count = e = 10

p = e /n = 10/50 = 0.20

1 – p = 1 – (.20) = 0.80

sdp = sqrt(p*(1 – p)/n) = sqrt( .2*.8 / 50 ) ≈ 0.056569

 

from 2.00   0.02275    0.95450, Z ≈ 2.00

 

 lower95 = p – ( 2*sdp ) ≈ 0.20 – ( 2* 0.056569) ≈ 0.086863

upper95 = p + ( 2*sdp ) ≈ 0.20 + ( 2* 0.056569) ≈ 0.313137

Our population is the population of finishers of the 2009 Fictitious City Iron Man, and our population proportion is the population proportion of finishers with finish times strictly below 10 hours.

 Each member of the Family of Samples is a single random sample of 50 finishers of the 2009 Fictitious City Iron Man.

From each member sample, compute: the sample number e of finishers with finish times strictly below 10 hours, the sample proportion p = e / 50 and the standard error for proportion sdp = sqrt( p*(1 – p)/50 ). Then from the means table row 2.00   0.02275    0.95450, use Z ≈ 2.00. Then compute the interval

 

[lower95 = p – ( Z*sdp ), upper95 = p + ( Z*sdp )].

 

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 proportion. If our interval resides in this 95% supermajority of member intervals, then between 8.7% and 31.3% of finishers of the 2009 Fictitious City Iron Man have finish times strictly below 10 hours.

 

 

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