An insurer uses the Poisson distribution with mean of 4 as the model for the number of claims per month on a particular policy. Each claim results in a payment of 1 by the insurer. Find the probability that the total payment by the insurer in a given month is less than one standard deviation above the mean monthly payment.

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

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

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

7,24

step-by-step explanation: