Question

A rare form of malignant tumor occurs in 11 children in a milion, so its probability is 0,000011. Four cases of this tumor occurred in a a. Assuming that this tumor occurs as usual, find the mean number of cases in groups of ch had 19,824 children 19,824 children b. Using the unrounded mean from part (a), find the probability that the number of tumor cases in a group of 19,824 children is 0 or 1. d. Does the ckuster 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)

Answer 1

a)

mean number of cases =np=19824*0.000011=0.218

b)P(X<=1)=P(X=0)+P(X=1)=e^-0.218*0.218⁰/0!+e^-0.218*0.218¹/1!=0.9794

c)P(X>1)=1-P(X<=1)=1-0.9794=0.0206

d)

probability of gettng 4 or more cases =P(X>=4)=0.0001

as probability of this event is highly unlikely ; therefore this does not appear to be attributed to random chance,