Math 240 Name: the early 1990s, a leukemia cluster was identified in the Massach
ID: 2933250 • Letter: M
Question
Math 240 Name: the early 1990s, a leukemia cluster was identified in the Massachusetts town of Woburn. An andisual numbers of cases of leukemia appeared in this small town. Was it evidence of a problem in the town, or was it chance? That question led to a famous 80-day trial in which the families of cight leukemia victims sued the local chemical companies, and became the basis of the book and movie A Civil Action. It we assume that the events occur independently, we can use a Binomial model to find the probability that a cluster of cases of size x could occur in Woburn due to chance. 1) Use the national average of leukemia incidence to get a value for p, and the population of Woburn as the value for m. In the United States in the early 1990s, there were about 30,800 new cases of leukemia each year and about 280,000,000 people. The population of Woburn was about 35,000. What is the value of p? What is the value of n? 2) With these values of p and n, about how many new cases of leukemia would you expect in a town of the size of Woburn? What is the standard deviation for the number of cases? 3) In a town of the size of Woburn, find the exact probability of exactly 0 cases: exactly 1 case: exactly 2 cases: exactly 3 cases: exactly 4 cases: exactly 5 cases: exactly 6 cases: exactly 7 cases: 4) Find the probability of 7 cases or less: 5) How unlikely would 8 or more cases be?Explanation / Answer
Q1.
p = 30800/35000 = 0.88
Q2.
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 35000 * 0.88
= 30800
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 35000 * 0.88 * 0.12
= 3696
III.
standard deviation = sqrt( variance ) = sqrt(3696)
=60.7947
Q3.
P( X = 0 ) = ( 8 0 ) * ( 0.88^0) * ( 1 - 0.88 )^8
= 0.000000043
ii.
P( X = 1 ) = ( 8 1 ) * ( 0.88^1) * ( 1 - 0.88 )^7
= 0.000002523
iii.
P( X = 2 ) = ( 8 2 ) * ( 0.88^2) * ( 1 - 0.88 )^6
= 0.000064746
iv.
P( X = 4 ) = ( 8 4 ) * ( 0.88^4) * ( 1 - 0.88 )^4
= 0.008704698