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## Week 9 questions Bio sections

 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. 3.87 Let X7 be the screening criterion for Y2. The sensitivity is P[Xy | Y2] = 0.7. 3.88 P[X7] = 0.43325, and P[X7c| Y2c ] = 0.605 3.89 P[Y2 | X1 intersect X7] = 0.187 4.1 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. 4.2 As in 4.14. The Binominal formula gives P[X3=5] = 0.0004

 Last update, 11/29/2001