Winter, 2008 Tuesday, Feb. 5Stat 321 – Day 12Other Discrete Random Variables (3.3, 3.4)Example 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) If p = .15 at St. George’s Hospital, determine the probability that in a random sample of 10 heart transplant cases that at least 8 would result in deathWinter, 2008 Tuesday, Feb. 5 n=10.10 .200 0.34868 0.107371 0.73610 0.375812 0.92981 0.677803 0.98720 0.879134 0.99837 0.967215 0.99985 0.993636 0.99999 0.999147 1.00000 0.999928 1.00000 1.000009 1.00000 1.00000Winter, 2008 Tuesday, Feb. 5(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.Winter, 2008 Tuesday, Feb. 5A random variable X follows a negative binomial (p. 118) distribution if- Each trial has two possible outcomes (typically referred to as “success” and “failure”).- The probability of “success” remains constant on each trial (call it p).- The trials are independent.- The random variable of interest (call it X) is the number of failures before the rth success The probability distribution of a negative binomial random variable with parameter p is given by    xrpprxrxXP  111 for ,,1,0 x.It can be shown that expected value E(X)=r(1-p)/p and the variance V(X)=r(1-p)/p2When r = 1, this is also referred to as a geometric random variable.(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?(g) Consider selecting 10 operations from the 371 in this time period which saw 79 deaths. Whatis the probability that we will see 8 or more deaths in our sample of 10? Hint: Define the random variable. Is this a “bernoulli” process?(h) What if we had used the binomial probability distribution to calculate this probability?Winter, 2008 Tuesday, Feb. 5A random variable X follows a hypergeometric distribution (p. 117) if- Each trial has two possible outcomes (typically referred to as “success” and “failure”).- The random variable of interest (call it X) is the number of successes in a fixed number of“trials” n from a finite population of size N.Note: The trials are not independent since you are sampling without replacement.The probability distribution of a hypergeometric random variable with parameters M (number of successes in the population), n (sample size) and N (population size) is given by  nNxnMNxMxXP for ),min(,0max( MnxMNn .It can be shown that expected value E(X)=nM/N (see p. 118 for Variance).When the population size is much larger than the size of the sample (N > 20n), the binomial distribution can be used to approximate the hypergeometric distribution (p. 118).(i) Suppose the probability of a heart transplantation death in the general population is .0001. What is the probability that of at least one such death in a random sample of 5000 people? Data that can be considered as counts of rare events are often well-modeled by the Poisson probability distribution (p. 121). This distribution has probability mass function !xexXPx for x= 0, 1, 2, …, where ->0 is the parameter of the Poisson distribution. It can be shown that the mean and variance of a Poisson distribution are both equal to -: E(X)=---and V(X)=--. Note: You can approximate the binomial distribution with a Poisson with -=np for large n and small p (p < .01, n>100, np< 20). Note, n=5000 is not an option in Table A.1, but -=5 is in A.2.Examples: - Numbers of fumbles made by 110 Division IA college football teams in the games played on a particular weekend. - Number of horsekick fatalities in Prussian cavalry corps (1898) per unit per yearWinter, 2008 Tuesday, Feb. 5Example 2: Define the random variable, “let X represent…,” and identify the appropriate distribution family. (a) A claims officer at a Social Security office will have time to examine six claims during a particular day. There are ten claims on her desk, of which four concern disability and six concernold-age benefits. If the six claims to be examined that day are randomly selected from the ten, what is the probability that all disability claims will have been examined by the end of the day?(b) Sixty percent of a population of consumers is reputed to prefer Brand A toothpaste. If a group of consumers is interviewed, what is the probability that the sixth person interviewed is thefirst consumer encountered who prefers Brand A?(c) Suppose that the suicide rate in a certain state is 1 suicide per 100,000 inhabitants per month. What is the probability that in a city of 400,000 inhabitants within this state, there will be six or more suicides in a given month. Hint: You have two choices of distributions here. Did you compute the exact probability or an approximation?(d) Following up on the conditions of (c), what is the probability that there will be at least two months during a 12-month period that will have six or more suicides? (e) What assumption(s) are you making in