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
had exactly 2 hits? 12 Disease Cluster Neuroblastoma, a rare form of cancer, occurs in 11...
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