## Elementary Statistics (12th Edition)

a) mean=$0.000011\cdot 12429=0.136719.$ b)$P=P(X=0)+P(X=1)=\frac{0.136719^0\cdot e^{-0.136719}}{0!}+\frac{0.136719^1\cdot e^{-0.136719}}{1!}=0.872+0.119=0.991.$ c)$P(X>1)=1-P(X\leq2)=1-0.991=0.009.$ d) The probability of having more than 1 is really small, therefore the probability of having 4 is even smaller, therefore the answer is no.