90-786 Intermediate Statistics:  Homework 3 Sample Answers

(3.40 = 15 pnts; 3.57 = 15 pnts; 3.69 = 10 pnts; 3.82 = 25 pnts; Case = 35 pnts)

 

MBS:

 3.40    a.         Define the following events:

                        W:  {Player wins the game Go}

                        F:  {Player plays first (black stones)}

 

P(WÈF) = 319/577 = .553

 

b.                  P(WÈF|CA) = 34/34 = 1.0

P(WÈF|CB) = 69/79 = 0.873

P(WÈF|CC) = 66/118 = 0.559

P(WÈF|BA) = 40/54 = 0.741

P(WÈF|BB) = 52/95 = 0.547

P(WÈF|BC) = 27/79 = 0.342

P(WÈF|AA) = 15/28 = 0.536

P(WÈF|AB) = 11/51 = 0.216

P(WÈF|AC) = 5/39 = 0.128

 

c.                   There are three combinations where the player with the black stones (first) is ranked higher than the player with the white stones:  CA, CB, and BA.

P(WÈF|CA Ç CB Ç BA) =

(34 + 69 + 40)/(34 + 79 + 54) = 143/167 = 0.856

 

d.                  There are three combinations where the players are the same level:  CC, BB, and AA.

P(WÈF|CC Ç BB Ç AA) =

(66 + 52 + 15)/(118 + 95 + 28) = 133/241 = 0.552

 

3.57                 The key to drawing a random sample is that every element drawn from the population must have an equal probability of being selected for inclusion in the sample.  When the data is available in electronic form, the task is fairly simple: most statistical software has a built in function that allows random sampling from a dataset.

                        Unfortunately, when the form of the data necessitates putting together the random sample “by hand”, the task is not as straightforward (as you may have noticed!).  The important thing is insuring the sample is selected truly at random.  For this, a random number table such as that found in Appendix B is needed.

                        There are a variety of algorithms that could be employed to use the random number table (RN table) for sample selection.  Here is one example:  use the RN table to determine a column of the  NYSE table (including all columns on all pages as potential columns).  After finding the column, go to the next random number on the RN table to determine a row.  Count down in the selected column to the designated row, and use that entry in the sample.  Then go to the next random number and do the same thing, until you have obtained 20 samples.  Use only the last two or three digits of the random numbers, as necessary (but make sure all entries in the NYSE table have a chance of being selected), and simply skip any numbers in the RN table that are outside the range of columns or rows in the NYSE table.  When you are finished, you will have a truly random sample (and very tired eyes!).

 

 

3.69     a.         P(Southwest) = 1/10; P(Continental) = 1/10.

b.                  P(on time) = 0.641

P(late) = 1 - P(on time) = 0.359

            c.         The data should be treated as upper bounds.  This assumes that the airlines are doing no better than the data reflects, essentially refusing to give the airlines the benefit of the doubt.  Such an approach to the data views it with a “skeptical” eye, and thereby should assuage the critics’ concerns.

 

3.82     a.         Define the following events:

                        A1: {Component 1 works properly}

                        A2: {Component 2 works properly}

                        B1: {Component 3 works properly}

                        B2: {Component 2 works properly}

                        A: {Subsystem A works properly}

                        B: {Subsystem B works properly}

 

The probability that a component fails is 0.1, so the probability that a component works properly is 1 – (0.1) = 0.9.

 

Subsystem A works properly if both components 1 and 2 work properly:

                P(A) = P(A1 È A2)   = P(A1) P(A2) = 0.9(0.9) = 0.81

Similarly:

                 P(B) = P(B1 È B2)   = P(B1) P(B2) = 0.9(0.9) = 0.81

 

The system operates properly if either subsystem A or B operates properly.  Thus, the probability that the system operates properly is:

                                P(AÇB)   = P(A) + P(B) - P(A È B)

                                                = 0.81 + 0.81 – 0.81(0.81) = .9639

 

b.                  The probability exactly one subsystem fails is:

         P(A È BC) + P(ACÈ B)   = P(A)P(BC) + P(AC)P(B)

                                                = 0.81(1 – 0.81) + (1 – 0.81)(0.81)

                                                = 0.1539 + 0.1539 = 0.3078

 

c.                   The probability that the system fails is the probability that both subsystems fail, or:

    P(AC È BC) = P(AC)P(BC)   = (1 – 0.81)(1 – 0.81) = 0.0361

 

d.                  For the system to operate correctly 99% of the time means that it fails not more than 1% of the time.  The probability that one subsystem fails is 1 – 0.81 = 0.19.  Since each subsystem is independent, the probability that n of them fail is 0.19n.  Therefore, we want to find n such that:

0.19n £ 0.01

Either solving algebraically or through some quick trial and error, we find that n = 3.

 

CHS: Amniocentesis, Blood Tests, and Down’s Syndrome Case

 

1.                  Verify the calculations given in the case

 

Start by defining the following event probabilities.  (“D” stands for “Presence of Down’s Syndrome”, “ND” for “Absence of Down’s Syndrome”, “+” for “Positive test result” and “-“ for “Negative test result”.)


 



The probability that a woman tests positive given that Down’s syndrome is present involves the joint probability of (1) a positive test given that the woman is positive, and (2) the woman is positive:

 



The probability that a woman tests positive when Down’s syndrome is not present is calculated similarly:

The total probability of a positive test result, then, is the sum of the two:

 

P(positive test) = P(positive given Downs present Ç positive given Downs absent) =

P(+) = 0.0033 + 0.2491 = 0.2524


The proportion of positive results that are false positives is simply the probability of a positive result when Down’s syndrome is not present, divided by the total probability of a positive result:

 

False negatives are calculated analogously:


Sens

P(False +)

0

1

0.05

0.999257

0.1

0.998515

0.15

0.997774

0.2

0.997035

0.25

0.996296

0.3

0.995559

0.35

0.994822

0.4

0.994087

0.45

0.993353

0.5

0.99262

0.55

0.991888

0.6

0.991157

0.65

0.990427

0.7

0.989698

0.75

0.988971

0.8

0.988244

0.85

0.987518

0.9

0.986794

0.95

0.98607

1

0.985348