Answer to Question #215873 in Statistics and Probability for Sampson Nsiah

Question #215873

Q1

A mine hauls, on average, three dump truck of waste per hour to a waste dump. For a given hour, find the probability that it will haul the following number of truck loads:

a)     At most 3 trucks.

b)     At least 3 trucks.

c)      Five or more 


1
Expert's answer
2021-07-13T05:06:52-0400

Let "X=" the number of trucks per hour: "X\\sim Po(\\lambda)."

Given "\\lambda=3."


a)

"P(X\\leq 3)=P(X=0)+P(X=1)+P(X=2)"

"+P(X=3)=\\dfrac{e^{-\\lambda}\\cdot\\lambda^0}{0!}+\\dfrac{e^{-\\lambda}\\cdot\\lambda^1}{1!}"

"+\\dfrac{e^{-\\lambda}\\cdot\\lambda^2}{2!}+\\dfrac{e^{-\\lambda}\\cdot\\lambda^3}{3!}"

"=\\dfrac{e^{-3}(6+18+27+27)}{6}=13e^{-3}\\approx0.647232"


b)

"P(X\\geq 3)=1-P(X=0)-P(X=1)-P(X=2)"

"=1-\\dfrac{e^{-\\lambda}\\cdot\\lambda^0}{0!}-\\dfrac{e^{-\\lambda}\\cdot\\lambda^1}{1!}-\\dfrac{e^{-\\lambda}\\cdot\\lambda^2}{2!}"

"=1-\\dfrac{e^{-3}(2+6+9)}{2}=1-8.5e^{-3}\\approx0.576810"




c)

"P(X\\geq 5)=1-P(X=0)-P(X=1)-P(X=2)"

"-P(X=3)-P(X=4)=1-\\dfrac{e^{-\\lambda}\\cdot\\lambda^0}{0!}"

"-\\dfrac{e^{-\\lambda}\\cdot\\lambda^1}{1!}-\\dfrac{e^{-\\lambda}\\cdot\\lambda^2}{2!}-\\dfrac{e^{-\\lambda}\\cdot\\lambda^3}{3!}-\\dfrac{e^{-\\lambda}\\cdot\\lambda^4}{4!}"

"=1-\\dfrac{e^{-3}(24+72+108+108+81)}{24}"


"=1-\\dfrac{131}{8}e^{-3}\\approx0.184737"


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