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 15,086 children.

a. Assuming that this tumor occurs as usual, find the mean number of cases in groups of 15,086 children.

b. Using the unrounded mean from part (a ), find the probability that the number of tumor cases in a group of

15,086 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?

Answer 1

a) mean number of cases in groups of 15,086 children =np=15086*0.000011=0.165946

b)

probability that the number of tumor cases in a group of 15,086 children is 0 or 1 =P(X=0)+P(X=1)

=e^-0.165946*0.165649⁰/0!+e^-0.165946*0.165649¹/1! =0.987664

c)

probability of more than one case =1-P(at most one case)=1-0.987664=0.012336

d)P(X>=4)=1-P(X<=3)=1-(P(X=0)+P(X=1)+P(X=2)+P(X=3))=0.000028

as probability of 4 or more cases is significantly low ; therefore this does not appear to occur by random chance

there is a chance that probability is higher than 0.000011.