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Week 10 ( 11/27-12/1) questions Bio sections

3.12 The probability of the intersection is 0.02, the product of the probabilities is 0.01. The events are therefore not independent
3.15 P[A | B ] is the P[A intersected B]/P[A] = 0.02/0.10=0.20
3.16 Let B=A2 and A=A1 and Ac the complement of A. The conditional probability is P[B | Ac] = P[ B intersect Ac]/P[ A c] = P[B]-P[B intersect A]/(1-P[A]) = 0.10-0.02/(1-0.10) = 0.089
3.17 Let A be the event that a 77 year old man is affected and B be the event that a 76 year old man is affected and C the event that a 82 year old man is affected. By assumption, all these events are independent. The probability that three people taken from each group are affected is P[A intersect B intersect C] = 8.8/100000
3.24 Let A be the event that a woman in the age group 75-79 is affected and B be the event that a man in that age group is affected. We have P[B | A] = 0.0015/0.023 = 0.031. The unconditional probability P[B] = 0.049 is slightly larger indicating a correlation.
3.25 Let A be the event that a woman in the age group 75-79 is affected and B be the event that a man in that age group is affected. We have P[A | B] = 0.0015/0.049 = 0.031. The unconditional probability P[A] = 0.023 is slightly smaller. A common environmental factor associated with Alzheimer's desease would have the effect that the conditional probability is heigher then the unconditional probability.
3.28 The expected number of cases with Alzheimer's disease is 1000 0.061 = 61
3.76 Let A be the event that the difference of the automated Diastolic blood pressure has increased by bigger or equal than 10mmHG and B the event that the difference of the manual Diastolic blood pressure is bigger or equal than 10mmHG. Let N be the number of measurements. From Table 3.9, we see that the probability of A intersected with B is 6/N and the probability of B is P[B]=13/N and the probability of A is P[A]=21/N. The sensitivity is P[A | B] = P[ A intersect B ]/P[B] = (6/N)/(13/N) = 6/13 = 0.462.
3.77 The specificity of the automated Diastolic blood pressure measurements is P[Ac | Bc] = (51/N)/(66/N) = 51/66 = 0.773.
3.78 With the notation from above the predictive values positive (PV+) and predictive values negative (PV-) are
PV+ = P[ B | A] = P[B intersect A]/P[A]=6/21 = 0.286
PV- = P[ Bc | Ac] = 51/58 = 0.879
3.83 The probabilities in the prevalence columns are P[Yi]. The conditional probabilities P[Xi|Yj] are given in the matrix of Table 3.11. Look up the entry in the 4'th row and 5'th column: P[X5|Y4] = 0.10.
3.89 Use Bayes rule to determine P[Y2 | X1 intersect X7] = 0.187
4.10 While commen sense tells the right answer 3=0.2*15 immediatly, it is useful to formulate the problem precicely using the language of probability theory.
Let D be the group of all grade school students and A be the population of grade school students who develop influenca. Assume D contains N students. The probability of A, P[A] is known to be 0.2. Let B be the members of the class under consideration and C is the subset of students in B which develop influenca.
We have C = A intersect B. It is an assumption that B is independent of A. Therefore, we have P[C] = P[A intersect B] = P[A] P[B] = 0.2 * P[B] = 0.2 * 15/N. The expected number of students is N P[C ] = 0.2*15 = 3.
4.14 X is the Binominal distributed random variable giving the number of inner-city newbotrns with HIV positive test results. The probability p to be HIV positive is 30/3741. The probability 1-p to be HIV negative is 3711/3741. The Binominal formula (which assumes that each birth event is independent from the others) gives P[X=5] = 3741!/(5! 3736!) p5 (1-p)3711 = 0.1573.
4.15 P[X > 4] = 1- P[X < 5]. This probability can be determined easily with Mathematica (see a guide how to use Mathematica at Harvard from any computer. The computer does not need to have Mathematica installed but have access to the Web).
<< Statistics`DiscreteDistributions`
N[1- CDF[BinomialDistribution[500, 30/3741], 4]] 
which gives 0.373.
4.20 As in 4.14. The Binominal formula gives P[X3=5] = 0.0004

Last update, 12/18/2000