Key

The 1st Hourly

Math 1107

Summer Term 2007

 

Protocol

 

You will use only the following resources: Your individual calculator; individual tool-sheet (single 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.

 

In each case, show complete detail and work for full credit. Follow case study solutions and sample hourly keys in presenting your solutions. Work all six 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. All of your work goes on one side each of the blank sheets provided. Space out your work. Do not share information with any other students during this hourly. Sign and Acknowledge:

 

I agree to follow this protocol.

 

______________________________________________________________________________________

Name (PRINTED)                                       Signature                                         Date

 

Case One

Pair of Dice

Random Variable

25 Points Maximum

We have a pair of dice – a fair d3 {faces 1, 3, 5} and a loaded d4 {faces 2, 4, 6, 7} – note the probability model for the d4 below. 

 Face

Probability

2

0.10

4

0.20

6

0.30

7

0.40

Total

1.00

We assume that the dice operate separately and independently of each other. Suppose that our experiment consists of tossing the dice, and noting the resulting face-pair.

1. List the possible pairs of face values, and compute a probability for each pair of face values.

15 Points Maximum

 (d4,d3)

2

4

6

7

1

(2,1)

(4,1)

(6,1)

(7,1)

3

(2,3)

(4,3)

(6,3)

(7,3)

5

(2,5)

(4,5)

(6,5)

(7,5)

 

The pairs are: (2,1), (4,1), (6,1), (7,1), (2,3), (4,3), (6,3), (7,3), (2,5), (4,5), (6,5) and (7,5).

 

Pr{(2,1)} = Pr{2 from d4}*Pr{1 from d3} = (1/10)*(1/3) = 1/30                                                              

Pr{(2,3)} = Pr{2 from d4}*Pr{3 from d3} = (1/10)*(1/3) = 1/30                                                            

Pr{(2,5)} = Pr{2 from d4}*Pr{5 from d3} = (1/10)*(1/3) = 1/30 [subtotal = 3/30]

Pr{(4,1)} = Pr{4 from d4}*Pr{1 from d3} = (2/10)*(1/3) = 2/30                                                             

Pr{(4,3)} = Pr{4 from d4}*Pr{3 from d3} = (2/10)*(1/3) = 2/30                                                             

Pr{(4,5)} = Pr{4 from d4}*Pr{5 from d3} = (2/10)*(1/3) = 2/30 [subtotal = 6/30]

Pr{(6,1)} = Pr{6 from d4}*Pr{1 from d3} = (3/10)*(1/3) = 3/30                                                               

Pr{(6,3)} = Pr{6 from d4}*Pr{3 from d3} = (3/10)*(1/3) = 3/30                                                             

Pr{(6,5)} = Pr{6 from d4}*Pr{5 from d3} = (3/10)*(1/3) = 3/30 [subtotal = 9/30]

Pr{(7,1)} = Pr{7 from d4}*Pr{1 from d3} = (1/10)*(1/3) = 4/30                                                              

Pr{(7,3)} = Pr{7 from d4}*Pr{3 from d3} = (1/10)*(1/3) = 4/30                                                             

Pr{(7,5)} = Pr{7 from d4}*Pr{5 from d3} = (1/10)*(1/3) = 4/30 [subtotal = 12/30]

Check: (3/30)+(6/30)+(9/30)+(12/30) = (3+6+9+12)/30 = 30/30 = 1

 

2. When both faces in the pair are odd, define THING as the product of the face values in the pair. When exactly one face in the pair is odd, define THING as the sum of the face values in the pair. Compute and list the possible values for the variable THING, and compute a probability for each value of THING.

10 Points Maximum

 

(d4,d3)

2

4

6

7

1

(2,1) One Odd Face: THING = (2+1) = 3

(4,1) One Odd Face: THING = (4+1) = 5

(6,1) One Odd Face: THING = (6+1) = 7

(7,1) Both Faces Odd: THING = 7*1=7

3

(2,3) One Odd Face: THING = (2+3) = 5

(4,3) One Odd Face: THING = (4+3) = 7

(6,3) One Odd Face: THING = (6+3) = 9

(7,3) Both Faces Odd: THING = 7*3=21

5

(2,5) One Odd Face: THING = (2+5) = 7

(4,5) One Odd Face: THING = (4+5) = 9

(6,5) One Odd Face: THING = (6+5) = 11

(7,5) Both Faces Odd: THING = 7*5=35

The pairs are: (2,1), (4,1), (6,1), (7,1), (2,3), (4,3), (6,3), (7,3), (2,5), (4,5), (6,5) and (7,5).

THING{(2,1)} = 2 + 1 = 3

THING{(2,3)} = 2 + 3 = 5

THING{(2,5)} = 2 + 5 = 7

THING{(4,1)} = 4 + 1 = 5

THING{(4,3)} = 4 + 3 = 7

THING{(4,5)} = 4 + 5 = 9

THING{(6,1)} = 6 + 1 = 7

THING{(6,3)} = 6 + 3 = 9

THING{(6,5)} = 6 + 5 = 11

THING{(7,1)} = 7*1 = 7

THING{(7,3)} = 7*3 = 21

THING{(7,5)} = 7*5 = 35

 

Pr{THING=3} = Pr{(2,1)} = 1/30

Pr{THING=5} = Pr{One of (2,3), (4,1) Shows} = Pr{(2,3)} + Pr{(4,1)} = (1/30) + (2/30) = 3/30

Pr{THING=7} = Pr{One of (2,5), (4,3), (6,1), (7,1) Shows} = Pr{(2,5)} + Pr{(4,3)} + Pr{(6,1)} + Pr{(7,1)} = (1/30) + (2/30) + (3/30) + (4/30) = 10/30

Pr{THING=9} = Pr{One of (4,5), (6,3) Shows} = Pr{(4,5)} + Pr{(6,3)} = (2/30) + (3/30) = 5/30

Pr{THING=11} = Pr{(6,5)} = 3/30

Pr{THING=21} = Pr{(7,3)} = 4/30

Pr{THING=35} = Pr{(7,5)} = 4/30

 

Check: (1/30) + (3/30) + (10/30) + (5/30) + (3/30) + (4/30) + (4/30) = (1+3+10+5+3+4+4)/30 = 30/30 = 1

 

Show all work and full detail for full credit.

 

Case Two

Color Slot Machine

Conditional Probabilities

25 Points Maximum

Here is our slot machine – on each trial, it produces a six color sequence, using the table below:

Sequence*

Probability

BGBYYG

0.10

GBBRRG

0.15

YRBGGR

0.11

YRGGBB

0.35

YBYYRR

0.23

BBYYRY

0.05

BBBBBY

0.01

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 5th6th )

Compute the following conditional probabilities.

 

1. Pr{“BB” Shows | Yellow Shows}

9 Points Maximum

 

Sequence*

Probability

BGBYYG

0.10

YRBGGR

0.11

YRGGBB

0.35

YBYYRR

0.23

BBYYRY

0.05

BBBBBY

0.01

Total

0.85

 

Pr{Yellow Shows} = Pr{One of BGBYYG, YRBGGR, YRGGBB, YBYYRR, BBYYRY, BBBBBY Shows} =

Pr{BGBYYG} + Pr{YRBGGR + Pr{YRGGBB + Pr{YBYYRR + Pr{BBYYRY + Pr{BBBBBY} =

0.10 + 0.11 + 0.35 + 0.23 + 0.05 + 0.01 = 0.21 + 0.58 + 0.06 = 0.79 + 0.06 = .85

 

 

Sequence*

Probability

YRGGBB

0.35

BBYYRY

0.05

BBBBBY

0.01

Total

0.41

 

Pr{“BB” and Yellow Show} = Pr{One of YRGGBB, BBYYRY, BBBBBY Shows} =

Pr{YRGGBB} + Pr{BBYYRY} + Pr{BBBBBY} = 0.35 + 0.05 + 0.01 = 0.41

 

Pr{“BB” Shows | Yellow Shows} = Pr{“BB” and Yellow Show}/ Pr{Yellow Shows} = 0.41/0.85

 

2. Pr{Yellow Shows  | “BR” Shows }

8 Points Maximum

 

 

Sequence*

Probability

GBBRRG

0.15

Total

0.15

 

Pr{“BR” Shows} = Pr{GBBRRG} = 0.15

Sequence*

Probability

Total

0

 

Pr{Yellow and “BR” Show} = Pr{No Events Qualify} = 0

 

Pr{Yellow Shows  | “BR” Shows } = Pr{Yellow and “BR” Show}/ Pr{ “BR” Shows} = 0/0.15 = 0

 

3. Pr{Green Shows | Red Shows}

8 Points Maximum

 

Sequence*

Probability

GBBRRG

0.15

YRBGGR

0.11

YRGGBB

0.35

YBYYRR

0.23

BBYYRY

0.05

Total

0.89

 

Pr{Red Shows} = Pr{One of GBBRRG, YRBGGR, YRGGBB, YBYYRR, BBYYRY Shows} =

Pr{GBBRRG} + Pr{YRBGGR} + Pr{YRGGBB} + Pr{YBYYRR} + Pr{BBYYRY} =

0.15 + 0.11 + 0.35 + 0.23 + 0.05 = 0.26 + 0.58 + 0.05 = 0.84 + 0.05 = 0.89

 

 

Sequence*

Probability

GBBRRG

0.15

YRBGGR

0.11

YRGGBB

0.35

Total

0.61

 

 

Pr{Green and Red Show} = Pr{One of GBBRRG, YRBGGR, YRGGBB Shows} =

Pr{GBBRRG} + Pr{YRBGGR} + Pr{YRGGBB} = 0.15 + 0.11 + 0.35 = 0.26 + 0.35 = 0.61

 

Pr{Green Shows | Red Shows} = Pr{Green and Red Show} / Pr{Red Shows} = 0.61/0.89

 

Show all work and full detail for full credit.

 

Case Three

Long Run Argument and Perfect Samples

Traumatic Brain Injury (TBI) and Glasgow Coma Scale (GCS)

25 Points Maximum

 

Traumatic brain injury (TBI) is an insult to the brain from an external mechanical force, possibly leading to permanent or temporary impairments of cognitive, physical, and psychosocial functions with an associated diminished or altered state of consciousness. The Glasgow Coma Scale (GCS) is the most widely used system for scoring the level of consciousness of a patient who has had a traumatic brain injury. GCS is based on the patient's best eye-opening, verbal, and motor responses. Each response is scored and then the sum of the three scores is computed. The total score varies from 3 to 15. The GCS categories are Mild (for GCS scores between 13 and 15), Moderate (for GCS scores between 9 and 12) and Severe (for GCS scores between 3 and 8). It is not unusual for people to die with TBI before they can be treated or evaluated. We can augment the GCS categories by adding a PAD (Pre-admission Death, TBI Noted) category. Suppose that the probabilities tabled below apply to TBI cases:

 

TBI Severity

Probability

Mild

0.10

Moderate

0.15

Severe

0.60

PAD

0.15

Total

1.00

 

1. Interpret each probability using the Long Run Argument.

10 Points Maximum

 

In long runs of sampling with replacement, approximately 10% of sampled TBI cases are mild.

In long runs of sampling with replacement, approximately 15% of sampled TBI cases are moderate.

In long runs of sampling with replacement, approximately 60% of sampled TBI cases are severe.

In long runs of sampling with replacement, approximately 15% of sampled TBI cases are pre-admission deaths(PAD).

 

2. Compute and discuss Perfect Samples for n=1200.

15 Points Maximum

 

9 Points Maximum

EMild = 1200*PMild = 1200*0.10 = 120

EModerate = 1200*PModerate = 1200*0.15 = 180

ESevere = 1200*PSevere = 1200*0.60 = 720

EPAD = 1200*PPAD = 1200*0.15 = 180

 

Check: EMild + EModerate + ESevere + EPAD = 120 + 180 + 720 +180 =300 + 900 = 1200

 

6 Points Maximum

Random samples of TBI cases of size 1200 yield approximately 120 mild cases.

Random samples of TBI cases of size 1200 yield approximately 180 moderate cases.

Random samples of TBI cases of size 1200 yield approximately 720 severe cases.

Random samples of TBI cases of size 1200 yield approximately 180 PAD cases.

 

Show all work and full detail for full credit. Provide complete discussion for full credit.

Case Four

Color Slot Machine

Probability Rules

25 Points Maximum

Here is our slot machine – on each trial, it produces a six color sequence, using the table below:

Sequence*

Probability

BGBYYG

0.10

GBBRRG

0.15

YRBGGR

0.11

YRGGBB

0.35

YBYYRR

0.23

BBYYRY

0.05

BBBBBY

0.01

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 5th6th )

Compute the following conditional probabilities.

 

1. Pr{“BY” Shows }

9 Points Maximum

 

Sequence*

Probability

BGBYYG

0.10

YBYYRR

0.23

BBYYRY

0.05

BBBBBY

0.01

Total

0.39

 

Pr{“BY” Shows } = Pr{One of BGBYYG, YBYYRR, BBYYRY, BBBBBY Shows} =

Pr{BGBYYG} + Pr{YBYYRR} + Pr{BBYYRY} + Pr{BBBBBY} = 0.10 + 0.23 + 0.05 + 0.01 = 0.33 + 0.06 = 0.39

 

2. Pr{ Green Shows 4th }

8 Points Maximum

 

Sequence*

Probability

YRBGGR

0.11

YRGGBB

0.35

Total

0.46

 

Pr{Green Shows 4th} = Pr{One of YRBGGR, YRGGBB Shows} =

Pr{YRBGGR} + Pr{YRGGBB} = 0.11 + 0.35 = 0.46

 

3. Pr{ Yellow Shows } – Use the Complementary Rule

8 Points Maximum

Failure to Use CR: 6 points off

 

Sequence*

Probability

GBBRRG

0.15

Total

0.15

 

Other Event = “Yellow does not Show”

Pr{Yellow does not Show} = Pr{GBBRRG} = 0.15

Pr{Yellow Shows} = 1 – Pr{Yellow does not Show} = 1 – 0.15 = 0.85

 

Check:

Sequence*

Probability

BGBYYG

0.10

YRBGGR

0.11

YRGGBB

0.35

YBYYRR

0.23

BBYYRY

0.05

BBBBBY

0.01

Total

0.85

 

Pr{Yellow Shows} = Pr{One of BGBYYG, YRBGGR, YRGGBB, YBYYRR, BBYYRY, BBBBBY Shows} =

Pr{BGBYYG} + Pr{YRBGGR} + Pr{YRGGBB} + Pr{YBYYRR} + Pr{BBYYRY} + Pr{BBBBBY} =

0.10 + 0.11 + 0.35 + 0.23 + 0.05 + 0.01 = 0.21 + 0.58 + 0.06 = 0.79 + 0.06 = 0.85

 

Show all work and full detail for full credit.

 

Case Five

Design Fault Spot

25 Points Maximum