a) "\\lambda=2."
"P(X>3)=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)=\\\\\n=1-e^{-2}(\\frac{2^0}{0!}+\\frac{2^1}{1!}+\\frac{2^2}{2!}+\\frac{2^3}{3!})=0.1429."
b) "\\lambda=2*2=4."
"P(X>5)=1\u2212P(X=0)\u2212P(X=1)\u2212P(X=2)\u2212P(X=3)+\\\\+P(X=4)+P(x=5)=\\\\\n=1\u2212e^{\u22124}(\\frac{4^0}{0!}+\\frac{4^1}{1!}+\\frac{4^2}{2!}+\\frac{4^3}{3!}+\n\\frac{4^4}{4!}+\\frac{4^5}{5!})=0.2149."
c) Binomial distribution with "n=15, p=0.25."
"P(X\\ge5)=1-P(X<5)=1-C_{15}^00.25^00.75^5-C_{15}^10.25^10.75^4-\\\\\n-C_{15}^20.25^20.75^3-C_{15}^30.25^30.75^2-C_{15}^40.25^40.75^1=0.3135."
d) There are no customers in the first half of hour "\\lambda=\\frac{2}{2}=1" .
"P(X=0)=e^{-1}\\frac{1^0}{0!}=0.3679."
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