Question 1192724
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This uses the Poisson Distribution.
More info can be found here:
<a href = "https://online.stat.psu.edu/stat414/book/export/html/678">https://online.stat.psu.edu/stat414/book/export/html/678</a>


{{{lambda}}} = greek letter lambda
lambda = 1.5 = average number of arrivals per hour


P(x) = probability of having exactly x arrivals (x = 0, x = 1, x = 2, ...)
P(x) = {{{(e^(-lambda)*lambda^x)/(x!)}}}
The exclamation mark indicates factorial. Example: 5! = 5*4*3*2*1 = 120.
The 'e' refers to the special constant e = 2.718...


Plug in lambda = 1.5 and x = 6
P(x) = {{{(e^(-lambda)*lambda^x)/(x!)}}}


P(6) = {{{(e^(-1.5)*(1.5)^6)/(6!)}}}


P(6) = 0.00352998886172 which is approximate.


Answer: <font color=red>Approximately 0.0035</font>
This is about a 0.35% chance.
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