Distributions 1. In September of 2000, heart transplantation at St. George\'s Ho
ID: 3360472 • Letter: D
Question
Distributions
1. In September of 2000, heart transplantation at St. George's Hospital in London was suspended because of concern that more patients were dying than previously. Newspapers reported that the 80% mortality rate in the last 10 cases at the hospital was of particular concern because it was over five times the national average. Let the random variable X represent the number of deaths in a random sample of 10 cases. Suppose that the probability of death at this hospital is equal to the national rate of 15%. (a) Identify the probability distribution of X (both its name and its parameter values). (b) Ifp = .15 at St. George's Hospital, determine the probability that in a random sample of io heart transplant cases that at least 8would result death (c) When analyzing data on all 371 patients who received a heart transplant at this hospital between 1986 and 2000, researchers found that 79 had died. Determine the probability of finding 79 or more deaths in a random sample of 371 patients when p.15 (d) Explain how these calculations provide strong evidence that the true underlying death rate at this hospital is higher than the national rate. (e) Suppose I continue to observe heart transplant operations at this hospital until the next death. What is the probability that there will be 0 successful operations before this first death? Five successful operations? (f What is the probability that there will be 10 successful operations before the 2nd death?Explanation / Answer
Question 1
(a) Here The sample size = 10 and p = 0.15 (initially assumed equals to national average) . GIven distribution is a binomial one.
(b) Here of p = 0.15 ; wrting the problem mathematically.
Pr (X <= 8; 10; 0.15) = 8.665 x 10-6
which is very less.
(c) Here number of patients = 371
so population proportion = 79/371 = 0.2129
Binomial Method :
so if p = 0.15
Pr(X >= 79; 371; 0.15) = 0.00074 [ from binomial table and calcultor]
Normal approximation to binomial
Expected number of deaths when p = 0.15 = 371 * 0.15 = 55.65
Standard deviation of number of deaths when p = 0.15 = sqrt(0.15 * 0.85 * 371) = 6.8777
so if p = 0.15
Pr(X >= 79; 55.65; 6.8777) = NORMAL (X > 78.5 ; 55.65; 6.8777) [We minus 0.5 for continutiy correction]
Z = (78.5 - 55.65)/ 6.8777 = 3.32
Pr(Z > 3.32) = 1 - 0.99955 = 0.00045
[Note instead of central limit theorem. We shall use normal approximation to binomial]
(d) As we can see that both probabilities are so low that it is very low to happen. So, we can say that the true underlying death rate at this hospital is higher than the national rate.
(e) Here now we have to find that next operation will be an unsucceful operations. The distribution is geometric distribution.
Here probability of failure of an operation and death is 0.2129
Probability of success in operation = 1 - 0.2129 = 0.7871
Pr(0 succeful operations) = Pr(death in the next operation) = 0.2129 * (1 - 0.7871)0 = 0.2129
Pr(five susseful operations) = Pr(5 consecutive succeful operations) * Pr(Death in the sixth one) = (1 - 0.2129)5 * 0.2129 = 0.0643
(f) The distribution is negative binomial distribution
Pr( 10 succeful operations before 2nd death) = Pr(1 death in 11 total operations) * Pr(11th operation is death) = 11C1 (0.2129)(1 - 0.2129)10 ( 0.2129)
= 0.04137