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 16,616 children. a. Assuming that this tumor occurs as usual, find the mean number of cases in groups of 16,616 children. b. Using the unrounded mean from part (a), find the probability that the number of tumor cases in a group of 16,616 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 (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 (Round to three decimal places as needed.) c. The probability of more than one case is (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? O A. Yes, because the probability of more than one case is very large. O B. No, because the probability of more than one case is very large. O C. Yes, because the probability of more than one case is very small. OD. No, because the probability of more than one case is very small.

Answer 1

a mean= np=166168 0.0000 LL = 0.182776 s 0:183 b) o od 1 = P(x=0)+ P(x=1) . By using ( Poission. distribt) e 0.182 776x (0 1827160 Ft 0 e - 0 182776 x (o 182716! = 0.83295 t o.15224 0 985 c) P(x-1)= P(x=0) + P(X21) - (0.985.1.9.). 0.015 = d) option a