Answer:
The probability that the insurer in a given month is less than one standard deviation above the mean monthly paymeent is 0.7851
Step-by-step explanation:
The mean monthly payment is 4, and the standard devitation, by virtue of being a Poisson distribution is √4 = 2. Thus, we want to know the probability of X being less than 6, where X is the total montlhy payment.
Note that P(X<6) = P(X=0) + P(X=1) + P(X=2)+ P(X=3)+P(X=4)+P(X=5)
Also
[tex]P(X=k) = \frac{e^{-4}4^k }{k!}[/tex]
Therefore,
P(X = 0) = e^{-4} * 1
P(X=1) = e^{-4} * 4
P(X=2) = e^{-4} * 4²/2 = e^{-4}*8
P(X=3) = e^{-4} * 4³/3! = e^{-4}*32/3
P(X=4) = e^{-4} * 4⁴/4! = e^{-4} * 32/3
P(X=5) = e^{-4} * 4⁵/5! = e^{-4} * 128/15
As a result
[tex]P(X<6) = e^{-4} *(1 + 4 + 8+32/3+32/3+128/15) = e^{-4} * 643/15 = 0.7851[/tex]
We conclude that the probability that the insurer in a given month is less than one standard deviation above the mean monthly paymeent is 0.7851