On the average, 7 cars arrive at the drive-up window of a bank every hour. What’s the probability that exactly 5 cars will arrive in the next hour?

Guest May 30, 2015

#2**+15 **

If we assume that the number of cars arriving follows a Poisson distribution then the probability of exactly 5 cars appearing in the next hour is:

p = λ^{5}e^{-λ}/5! where λ is the mean arrival rate (= 7 here)

$${\mathtt{p}} = {\frac{{{\mathtt{7}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left(-{\mathtt{7}}\right)}}{{\mathtt{5}}{!}}} \Rightarrow {\mathtt{p}} = {\mathtt{0.127\: \!716\: \!668\: \!292\: \!289\: \!6}}$$

(For exactly k cars arriving the probability is λ^{k}e^{-λ}/k!)

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Alan May 31, 2015

#2**+15 **

Best Answer

If we assume that the number of cars arriving follows a Poisson distribution then the probability of exactly 5 cars appearing in the next hour is:

p = λ^{5}e^{-λ}/5! where λ is the mean arrival rate (= 7 here)

$${\mathtt{p}} = {\frac{{{\mathtt{7}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left(-{\mathtt{7}}\right)}}{{\mathtt{5}}{!}}} \Rightarrow {\mathtt{p}} = {\mathtt{0.127\: \!716\: \!668\: \!292\: \!289\: \!6}}$$

(For exactly k cars arriving the probability is λ^{k}e^{-λ}/k!)

.

Alan May 31, 2015