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# 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 t

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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
+26619
+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 = λ5e/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 λke/k!)

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Alan  May 31, 2015
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#1
+92164
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I don't think that you have given enough information :/

Melody  May 31, 2015
#2
+26619
+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 = λ5e/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 λke/k!)

.

Alan  May 31, 2015

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