Question

A rare form of malignant tumor occurs in 11 children in a​ million, so its probability is 0.000011. Four cases of this tumor occurred in a certain​ town, which had 13 comma 280 children. a. Assuming that this tumor occurs as​ usual, find the mean number of cases in groups of 13 comma 280 children. b. Using the unrounded mean from part ​(a​), find the probability that the number of tumor cases in a group of 13 comma 280 children is 0 or 1. c. What is the probability of more than one​ case? d. Does the cluster of four cases appear to be attributable to random​ chance? Why or why​ not? a. The mean number of cases is nothing. ​(Type an integer or decimal rounded to three decimal places as​ needed.) b. The probability that the number of cases is exactly 0 or 1 is nothing. ​(Round to three decimal places as​ needed.) c. The probability of more than one case is nothing. ​(Round to three decimal places as​ needed.) d. Let a probability of 0.05 or less be​ "very small," and let a probability of 0.95 or more be​ "very large". Does the cluster of four cases appear to be attributable to random​ chance? Why or why​ not? A. ​Yes, because the probability of more than one case is very large. B. ​Yes, because the probability of more than one case is very small. C. ​No, because the probability of more than one case is very small. D. ​No, because the probability of more than one case is very large.

Answer 1

Answer:

a) The mean number of cases is 0.14608 cases.

b) The probability that the number of cases is exactly 0 or 1 is 0.990.

c) The probability of more than one case is 0.010

d) No, because the probability of more than one case is very small

Step-by-step explanation:

We can model this problem with a Poisson distribution, with parameter:

$\lambda=r*t=0.000011*13,280=0.14608$

a) The mean amount of cases is equal to the parameter λ=0.14608.

b) The probability of having 0 or 1 cases is:

$P(k=0)=\frac{\lambda^0 e^{-\lambda}}{0!}=\frac{1*0.864}{1} =0.864\\\\ P(k=1)=\frac{\lambda^1 e^{-\lambda}}{0!}=\frac{0.14608*0.864}{1} =0.126\\\\P(k\leq1)=0.864+0.126=0.990$

c) The probability of more than one case is:

$P(k>1)=1-P(k\leq 1)=1-0.990=0.010$

d) The cluster of 4 cases can not be due to pure chance, as it is a very high proportion of cases according to the average rate. Just having more than one case has a probability of 1%.