Post-Operative ComplicationsDate: 04/25/2002 at 00:48:54 From: Jay Smith Subject: Statistics Q. The rate of post-operative complication in a complex vascular reconstruction procedure, ranging in severity from wound separation or infection to death, is 12%. (i) For 40 such procedures, what are the chances that the number of patients who suffer post-operative complications will be more than ten? Less than expected? (ii) In a series of 65 such procedures, determine the most probable number of patients who suffer post-operative complications. What is the probability? Date: 04/25/2002 at 10:55:57 From: Doctor Mitteldorf Subject: Re: Statistics The key to solving problems of this kind is the binomial distribution formula: C(n,r)*(p^r)*(1-p)^(n-r) This represents the probability that you will have r occurrences out of a sample of n, when the probability of each occurrence separately is p. The factor p^r represents probabilties for those r occurrences. The factor (1-p)^(n-r) is the corresponding probability factor for the remaining (n-r) instances where the complication does not occur. C(n,r) is the number of ways this can happen, equal to the number of ways to select r objects out of n total: n! -------------- (n-r)! r! For example, in your question, p = 0.12 and 1-p = 0.88. In part (i) you are talking about n = 40. The probability for more than 10 complications will be the sum of the binomial formula for r = 11 through 40. (You may find that if you calculate the first few terms after r = 11 that they are decreasing rapidly with r, and you can neglect most of these to a good approximation.) When they ask about "less than expected," they are referring to the expected number of occurrences out of 40, which is just 12% of 40 = 4.8. Thus "less than expected" is the sum for r = 0 through r = 4. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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