Key

The Comprehensive Final Examination | Math 1107 | Spring 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 toolsheet, 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 | Random Variable, Pair of Dice

We have a pair of dice– note the probability models for the dice below. 

1^st d4		2^nd d4
Face	Probability	Face	Probability
1	1/10	1	4/10
2	2/10	2	3/10
3	3/10	3	2/10
4	4/10	4	1/10
Total	10/10=1	Total	10/10=1

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-value-pair.

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

Define HIGHTIE as the highest of the face values in the pair, or common value if tied. Define LOWTIE as the lowest of the face values in the pair or common value if tied. Define DIFFSQ =( (HIGHTIE - LOWTIE)/2)².

Compute and list the possible values for the variable DIFFSQ

Pairs

(1^st , 2^nd )	1(1/10)	2(2/10)	3(3/10)	4(4/10)
1(4/10)	(1,1)	(2,1)	(3,1)	(4,1)
2(3/10)	(1,2)	(2,2)	(3,2)	(4,2)
3(2/10)	(1,3)	(2,3)	(3,3)	(4,3)
4(1/10)	(1,4)	(2,4)	(3,4)	(4,4)

(1,1), (2,1), (3,1), (4,1), (1,2), (2,2), (3,2), (4,2), (1,3), (2,3), (3,3), (4,3), (1,4), (2,4), (3,4),(4,4)

Probabilities per Pair

Pr{(1,1)} = Pr{1 from 1^st die}*Pr{1 from 2^nd die} = (1/10)*(4/10) = 4/100

Pr{(2,1)} = Pr{2 from 1^st die}*Pr{1 from 2^nd die} = (2/10)*(4/10) = 8/100

Pr{(3,1)} = Pr{3 from 1^st die}*Pr{1 from 2^nd die} = (3/10)*(4/10) = 12/100

Pr{(4,1)} = Pr{4 from 1^st die}*Pr{1 from 2^nd die} = (4/10)*(4/10) = 16/100

Pr{(1,2)} = Pr{1 from 1^st die}*Pr{2 from 2^nd die} = (1/10)*(3/10) = 3/100

Pr{(2,2)} = Pr{2 from 1^st die}*Pr{2 from 2^nd die} = (2/10)*(3/10) = 6/100

Pr{(3,2)} = Pr{3 from 1^st die}*Pr{2 from 2^nd die} = (3/10)*(3/10) = 9/100

Pr{(4,2)} = Pr{4 from 1^st die}*Pr{2 from 2^nd die} = (4/10)*(3/10) = 12/100

Pr{(1,3)} = Pr{1 from 1^st die}*Pr{3 from 2^nd die} = (1/10)*(2/10) = 2/100

Pr{(2,3)} = Pr{2 from 1^st die}*Pr{3 from 2^nd die} = (2/10)*(2/10) = 4/100

Pr{(3,3)} = Pr{3 from 1^st die}*Pr{3 from 2^nd die} = (3/10)*(2/10) = 6/100

Pr{(4,3)} = Pr{4 from 1^st die}*Pr{3 from 2^nd die} = (4/10)*(2/10) = 8/100

Pr{(1,4)} = Pr{1 from 1^st die}*Pr{4 from 2^nd die} = (1/10)*(1/10) = 1/100

Pr{(2,4)} = Pr{2 from 1^st die}*Pr{4 from 2^nd die} = (2/10)*(1/10) = 2/100

Pr{(3,4)} = Pr{3 from 1^st die}*Pr{4 from 2^nd die} = (3/10)*(1/10) = 3/100

Pr{(4,4)} = Pr{4 from 1^st die}*Pr{4 from 2^nd die} = (4/10)*(1/10) = 4/100

and compute a probability for each value of DIFFSQ.

DIFFSQ per Pair

DIFFSQ {(1,1)} = ((HT – LT)/2)² = ((1 –1)/2)² = ((0)/2)² = 0;

DIFFSQ {(2,2)} = ((HT – LT)/2)² = ((2 –2)/2)² = (0/2)² =0;

DIFFSQ {(3,3)} = ((HT – LT)/2)² = ((3 –3)/2)² = (0/2)² = 0;

DIFFSQ {(4,4)} = ((HT – LT)/2)² = ((4 –4)/2)² = (0/2)² = 0;

DIFFSQ {(2,1)} = ((HT – LT)/2)² = ((2 –1)/2)² = (1/2)² = 1/4;

DIFFSQ {(1,2)} = ((HT – LT)/2)² = ((2 –1)/2)² = (1/2)² =1/4;

DIFFSQ {(3,2)} = ((HT – LT)/2)² = ((3 –2)/2)² = (1/2)² = 1/4;

DIFFSQ {(2,3)} = ((HT – LT)/2)² = ((3 –2)/2)² = (1/2)² =1/4;

DIFFSQ {(4,3)} = ((HT – LT)/2)² = ((4 –3)/2)² = (1/2)² = 1/4;

DIFFSQ {(3,4)} = ((HT – LT)/2)² = ((4 – 3)/2)² = (1/2)² = 1/4;

DIFFSQ {(3,1)} = ((HT – LT)/2)² = ((3 –1)/2)² = (2/2)² = 1;

DIFFSQ {(4,2)} = ((HT – LT)/2)² = ((4 –2)/2)² = (2/2)² = 1;

DIFFSQ {(1,3)} = ((HT – LT)/2)² = ((3 –1)/2)² = (2/2)² =1;

DIFFSQ {(2,4)} = ((HT – LT)/2)² = ((4 –2)/2)² = (2/2)² =1;

DIFFSQ {(4,1)} = ((HT – LT)/2)² = ((4 –1)/2)² = (3/2)² = 9/4;

DIFFSQ {(1,4)} = ((HT – LT)/2)² = ((4 –1)/2)² = (3/2)² =9/4;

Probabilities for DIFFSQ

Pr{DIFFSQ = 0} = Pr{One of (1,1), (2,2), (3,3) or (4,4) Shows} =

Pr{(1,1)} +Pr{ (2,2) } +Pr{ (3,3) } +Pr{ (4,4)} =

(4/100)+(6/100)+(6/100)+(4/100) = 20/100

Pr{DIFFSQ = 1/4} = Pr{One of (2,1),(1,2),(3,2),(2,3),(4,3) or (3,4) Shows} =

Pr{(2,1)} + Pr{(1,2)} + Pr{(3,2)} + Pr{(2,3)}+ Pr{(4,3)}+Pr{(3,4)} = ( 8/100) +

(3/100) + (9/100) + (4/100) + (8/100) + (3/100) = 35/100

Pr{DIFFSQ = 1} = Pr{One of (3,1),(4,2),(2,4),(1,3) Shows} =

Pr{(3,1)} + Pr{(4,2)} + Pr{(2,4)} + Pr{(1,3)} = (12/100) +

(12/100) + (2/100) + (2/100) = 28/100

Pr{DIFFSQ = 9/4} = Pr{One of (4,1),(1,4) Shows} =

Pr{(1,4)} + Pr{(4,1)} = (1/100) + (16/100) = 17/100

Case Two | Color Slot Machine, Computation of Conditional Probabilities

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

Sequence*	Probability
BRYRRR	.10
RGBRRB	.10
BBYGBR	.15
GRBRGG	.10
BGYGYY	.25
RRYYGY	.10
YYGBYY	.20
Total	1.00

*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1^st to 6^th , from left to right: (1^st 2^nd 3^rd 4^th 5^th6^th)

Compute the following conditional probabilities:

Pr{Blue Shows | Yellow Shows}

Pr{Yellow} = Pr{One of BRYRRR, BBYGBR, BGYGYY, RRYYGY or YYGBYY Shows} =

Pr{BRYRRR} + Pr{BBYGBR} + Pr{BGYGYY} + Pr{RRYYGY} + Pr{YYGBYY } =

.1 + .15 + .25 + .10 + .20 = .80

Pr{Blue and Yellow} = Pr{One of BRYRRR, BBYGBR, BGYGYY or YYGBYY Shows} =

Pr{BRYRRR} + Pr{BBYGBR} + Pr{BGYGYY} + Pr{YYGBYY } =

.1 + .15 + .25 + .20 = .70

Pr{B|Y} = Pr{BandY}/Pr{Y} = .7/.8

Pr{Green Shows  | “BR” Shows}

Pr{“BR”} = Pr{One of BRYRRR, RGBRRB, BBYGBR or GRBRGG Shows} =

Pr{ BRYRRR} + Pr{RGBRRB} + Pr{ BBYGBR} + Pr{GRBRGG} = .1 + .1 + .15 + .1 = .45

Pr{Green and “BR”} = Pr{One of RGBRRB, BBYGBR or GRBRGG Shows} =

Pr{RGBRRB} + Pr{ BBYGBR} + Pr{GRBRGG} = .1 + .15 + .1 = .35

Pr{Green|”BR”} = Pr{Green and “BR”}/Pr{“BR”} = .35/.45

Pr{ Yellow Shows 1^st | Red Shows}

Pr{Red} = Pr{One of BRYRRR, RGBRRB, BBYGBR, GRBRGG, RRYYGY} =

Pr{BRYRRR}+ Pr{RGBRRB}+ Pr{BBYGBR}+ Pr{GRBRGG}+ Pr{RRYYGY} =

.1 + .1 + .15 + .10 + .1 = .55

Pr{Yellow 1^st and Red} = Pr{No Sequences } = 0

Pr{Yellow 1^st | Red} = Pr{Yellow 1^st and Red}/Pr{Red} = 0/.55

Case Three | Descriptive Statistics | Framingham Heart Study

The objective of the Framingham Heart Study (FHS) is to identify the common factors or characteristics that contribute to Cardiovascular disease (CVD) by following its development over a long period of time (since 1948)  in a large group of participants who had not yet developed overt symptoms of CVD or suffered a heart attack or stroke. Blood pressure is a measurement of the force applied to the walls of the arteries as the heart pumps blood through the body. Blood pressure readings are measured in millimeters of mercury (mm Hg) and usually given as two numbers: the systolic blood pressure (SBP) reading, representing the maximum pressure exerted when the heart contracts and the diastolic blood pressure (DBP) reading, representing the pressure in the arteries when the heart is at rest.

Consider the systolic to diastolic blood pressure ratio R = SBP/DBP. 

A sample of FHS adult subjects yields the following ratios:

1.88         1.71         1.43         1.62         1.62         1.59         1.53         1.42         1.55         1.75         1.95         1.58

2.36         1.69         1.25         1.65         1.61         1.45         1.72         1.50         2.06         1.36         1.78         1.56

1.62         1.80         1.78         1.52         1.83         1.55         1.54         1.80         2.13         1.67         1.86         1.60

1.48         1.43         1.45         1.24         1.75         2.05         1.50         1.79         1.82         1.59         1.74         1.86

1.64         1.54         2.00         1.91         1.22         1.94         1.54         1.67         2.00         1.63         1.82         1.49

Compute and interpret the following statistics: sample size, p00, p25, p50, p75, p100, (p100 – p50), (p75 – p25), (p50 – p00). 

1.22	Min
1.54	P25
1.64	Median
1.81	P75
2.36	Max
0.72	Range42
0.27	Range31
0.42	Range20

There are 55 Framingham Heart Study subjects in the sample.

The smallest systolic to diastolic ratio in the sample is 1.22.

Approximately 25% of the FHS subjects in the sample have systolic to diastolic ratios of 1.54 or less.

Approximately 50% of the FHS subjects in the sample have systolic to diastolic ratios of 1.64 or less.

Approximately 75% of the FHS subjects in the sample have systolic to diastolic ratios of 1.81 or less.

The largest systolic to diastolic ratio in the sample is 2.36.

Approximately 50% of the FHS subjects in the sample have systolic to diastolic ratios between 1.22 and 1.64. The largest possible difference in systolic to diastolic ratios between any pair of subjects in the lower5 half sample is 0.42.

Approximately 50% of the FHS subjects in the sample have systolic to diastolic ratios between 1.54 and 1.81. The largest possible difference in systolic to diastolic ratios between any pair of subjects in the middle half sample is 0.27.

Approximately 50% of the FHS subjects in the sample have systolic to diastolic ratios between 1.64 and 2.36. The largest possible difference in systolic to diastolic ratios between any pair of subjects in the lower5 half sample is 0.72.

Case Four | Confidence Interval Proportion | Gestational Age

Gestational age is the time spent between conception and birth, usually measured in weeks. In general, infants born after 36 or fewer weeks of gestation are defined as premature, and may face significant challenges in health and development. Infants born after 37-40 weeks of gestation are generally viewed as full term, and those born after 41 or more weeks of gestation are generally viewed as post term. Suppose that a random sample of 2005 US resident live born infants yields the following gestational ages (in weeks):

24, 25, 25, 26, 27, 29, 30, 33, 35, 36, 36, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 37, 38, 38, 38, 38, 38

38, 38, 38, 38, 38, 38, 38, 39, 39, 39, 39, 39, 40, 40, 40, 40, 40, 40, 40, 40 , 40, 40, 41, 41, 41, 42, 42, 42, 43, 43

Define the event “2005 US resident live born infant is full term.” Estimate the population proportion for this event with 96% confidence. That is, compute and discuss a 96% confidence interval for this population proportion. Provide concise and complete details and discussion as demonstrated in the case study summaries.

Event = “Year 2005 Resident Live Born Infant is Full Term”

Sample size = n = 56

Event Count = e = 36

p = e/n = 36/56 » 0.642857

sdp=sqrt((36/56)*(20/56)/56) » 0.06403

From 2.10 0.017864 0.96427, Z= 2.10

lower96 = p – z*sdp » 0.642857 – (2.1*0.06403) » 0.508394

upper96 = p + z*sdp » 0.642857 + (2.1*0.06403) » 0.77732

Our population consists of all year 2005 US resident live births. Our population proportion is the proportion of those births with 37-40 weeks of gestation.

Each member of the family of samples is a single random sample of 56 live births. The family of sample consists of all possible samples of this type.

From each member of the family of samples, compute the number e of full term (gestation at 37-40 weeks) sample births and then:

p = e/n and sdp=sqrt(p*(1 – p)/n)

From 2.10 0.017864 0.96427, Z= 2.10 and then finally the interval: lower96 = p – z*sdp,

upper96 = p + z*sdp

Computing this for each member of the family of samples yields a family of intervals, approximately 96% of which capture the true population proportion. If our interval resides in this 96% super majority, then between 50.8% and 77.7% of year 2005 US Resident live births are full term(have between 37 and 40 weeks of gestation).

Table 1. Means and Proportions

Z(k) PROBRT PROBCENT 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 0.088508 0.82298 1.40 0.080757 0.83849 1.45 0.073529 0.85294 1.50 0.066807 0.86639 1.55 0.060571 0.87886 1.60 0.054799 0.89040 1.65 0.049471 0.90106 1.70 0.044565 0.91087 1.75 0.040059 0.91988 1.80 0.035930 0.92814 1.85 0.032157 0.93569 1.90 0.028717 0.94257 1.95 0.025588 0.94882 2.00 0.022750 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