Poisson distribution CDF

Aufrufe: 53     Aktiv: 01.07.2021 um 20:54

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Hallo zusammen,

A company manager wants to build new unloading bays in order to avoid delivery lorries
having to wait. Currently, 5 bays are working. Suppose it takes a full day to unload a lorry and that the
number of lorries arriving each day is approximately a Poisson variable of parameter 4.
a) What is the probability of having no lorries waiting?
b) How many unloading bays would be needed to raise this probability to over 0:9?

a) ist mir klar Pr(X <= 5) = e^-4*0/0!....e^-4*5/5!
b) CDF: lamda^k * e^-lamda/k! >= 0.9 ist das so gemeint? Dann muss ich k herausfinden?

Vielen Dank euch!
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