Question

had exactly 2 hits? 12 Disease Cluster Neuroblastoma, a rare form of cancer, occurs in 11 children i which had 12.429 children a. Assuming that neuroblastoma occurs as usual, find the mean number of cases in probability is 0.000011. Four cases of neuroblastoma occurred in Oak Park, llinsi 12.429 children. ot b. Using the unrounded mean from part (a), find the probability that the number of toma cases in a group of 12,429 children is 0 or 1. c. What is the probability of more than one case of neuroblastoma? d. Does the cluster of four cases appear to be attributable to random chance? Why or

Answer 1

12)

a) expected number of intervals =np=12429*0.000011=0.1367

b)P(X is 0 or 1)=P(X<=1)(P(X=0)+P(X=1)=e^-0.1367*0.1367⁰/0!+e^-0.1367*0.1367¹/1! =0.9915

c)P(more then 1 )=P(X>1)=1-P(X<=1)=1-0.9915=0.0085

d)as probability of getting 4 or more cases is less then 0.05 level therefore this event is not usual Hence we can not attribute this to random chance,

Birth problem:

5)P(X=12)=e^-11.5863*11.5863¹²/12! =0.1135

6)P(X=9)=e^-11.5863*11.5863⁹/9! =0.0964

7)

P(X>=1)=1-P(X=0)=1-e^-11.5863*11.5863⁰/0! =0.999991

8)

P(X>=2)=1-P(X<=1)=1-P(X=0)-P(X=1)=1-e^-11.5863*11.5863⁰/0!-e^-11.5863*11.5863¹/1!=0.999883