4.1 A "probability-mass function" for X is Rosner's terminology for a probability distribution.

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Mercy Bryan
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1 Probability - Chapter 4 NOTE: Pc=complement of P 4.1 A "probability-mass function" for X is Rosner's terminology for a probability distribution. there can be 0, 1, or 2 hypertensive adults... A = mother s DBP >= 95 P(A) =.1 B = father s DBP >= 95 P(B) =.2 neither mom or dad... P(0) = Pc(A) Pc(B) =.9 x.8 =.72 either mom or dad... P(1) = (P(A) Pc(B)) + (Pc(A) P(B)) = (.1.8) + (.9.2) = =.26 both mom and dad... P(2) = P(A) P(B) =.1.2 =.02 X probability The expected value of X is the probability-weighted sum of the possible outcomes 0, 1, and 2...from definition E(X) = 0 (.72) + 1 (.26) + 2 (.02) = There expected value of the variance of X is the probability weighted sum of the squared differences of each possible outcome from the mean (E(X)). There is a short cut version of the formula...equation Var(X) = 0 2 (.72) (.26) (.02) Mean 2 Var(X) = Var(X) =.25

2 4.4 The cumulative-distribution function is the cumulative sum of the probability-mass function calculated in problem cumulative X probability probability P(0<= X < 1) P(1<= X < 2) P(X => 2) 4.5 When order of selection does matter, use the PERMUTATION formula... npk = n!/(n-k) n = total number k = subset = When order of selection does not matter, use the COMBINATION formula... nck = n! / k! (n-k!) = 50!/ 5!(45)! = / = 254,251,200 / 120 = 2,118, These are all COMBINATIONS... 10C 0 = 1 10C 1 = 10 10C 2 = 10 9 / 2 1 = 45 10C 3 = / = C 4 = / = C 5 = / = C 6 = 10 C 4 = C 7 = 10 C 3 = C 8 = 10 C 2 = 45 10C 9 = 10 C 1 = 10 10C 10 = 10 C 0 = This is a FACTORIAL = 362,880

3 4.9 The 20% nationwide flu rate is considered as the probability of developing the flu. Using the binomial table with n=15, P= P(x=0) =.0352 P(x=1) =.1319 P(x=2) =.2309 P(x=3) =.2501 P(x=4) =.1876 P(x=5) =.1032 The cumulative probability up to and including P(x=5) = P(X<=5) = 1 - ( ) The probability of 6 or more cases is... P(X>=6) = =.0611 If one uses P<=.05 or less to identify a rare event,.0611 would not be considered unusual and 6 cases is not a rare event The expected number of students in the class who will develop influenza...equation is... E(X) = n x P = 15 x.20 = Probability of exactly 6 events... Using the Poisson probability table: k=6, mu=4.0 = Probability of at least 6 events... from Poisson table...mu=4.0 P(x=0) =.0183 P(x=1) =.0733 P(x=2) =.1465 P(x=3) =.1954 P(x=4) =.1954 P(x=5) =.1563 P(X>=6) = 1 P(X<=5) = 1 - ( ) P(X>=6) = =.2148

4 4.13 The expected value, variance, and mu all are What is the probability that among 10 true hypertensives at least 50% are being treated appropriately and are complying with the treatment? P(hypertensive 1 being treated appropriately 1 complying with treatment the probability) =.125 P(X=>5) using STATCRUNCH... binomial distribution(.125,10,=>5) = What is the probability that at least 7 out of 10 hypertensives know they have high blood pressure? from Binomial table...n=10, P=.5 P(x=7) =.1172 P(x=8) =.0439 P(x=9) =.0098 P(x=10) =.0010 P(X=>7) = =.1719

5 4.38 If the preceding 50% rates were reduced to 40% by mass education what effect would this change have on the overall mortality rate among true hypertensives? P(hypertensive 1 being treated appropriately 1 complying with treatment the probability) =.216 assume that the annual mortality rate for untreated hypertensives PRIOR to rate reduction is X if treatment lowers the rate by 20%, the rate for treated hypertensives (who presumably are complying) is.8x...so... current mortality = treated mortality + untreated mortality = (proportion treated rate) + (proportion not treated rate) =.125(.8X) +.875X =.975X assume the 50% rates are lowered...same logic... current mortality = treated mortality + untreated mortality = (proportion treated rate) + (proportion not treated rate) =.216(.8X) +.784X =.957X.957X /.975X = = 98.1% thus, if rates are lowered, the mortality rate would be 98.1% of the current value (or 1.9% of mortality would be prevented) NOTE: admittedly, this problem was a bit harder than the others

6 4.63 Significant excess...of a RARE event... X = number of cases of Down's syndrome what is the probability of observing 27 cases (X) given that the expected value (MU) is 19 P(X=>27 MU=19)... use the Poisson table with MU = 19...sum the probabilities of k=27, k=28... and you get approximately or using STATCRUNCH... poisson distribution (19,=>27) = since is less than 0.05, you can conclude that there is a significant excess 4.64 Exact number of cases... X = number of cases of cleft palate what is the probability of observing exactly 12 cases (X) given that the expected value (MU) is 7 P(X=12 MU=7)... use the Poisson table with MU = 7...the probability is Significant excess... X = number of cases of cleft palate what is the probability of observing 12 cases (X) given that the expected value (MU) is 7 P(X=>12 MU=7)... use the Poisson table with MU = 7...sum the probabilities of k=12, k=13... and you get approximately or use STATCRUNCH... poisson distribution (7, =>12) = since is quite close to 0.05, you might conclude that there is a significant excess

Probability and Probability Distributions. Dr. Mohammed Alahmed

Probability and Probability Distributions 1 Probability and Probability Distributions Usually we want to do more with data than just describing them! We might want to test certain specific inferences about