Answer:
44.80% probability that in a given week he will sell 2 or more policies but less than 4 policies.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
A life insurance salesman sells on the average 3 life insurance policies per week.
This means that [tex]\mu = 3[/tex]
Calculate the probability that in a given week he will sell 2 or more policies but less 4 policies.
[tex]P(2 \leq X < 4) = P(X = 2) + P(X = 3)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240[/tex]
[tex]P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240[/tex]
[tex]P(2 \leq X < 4) = P(X = 2) + P(X = 3) = 0.2240 + 0.2240 = 0.4480[/tex]
44.80% probability that in a given week he will sell 2 or more policies but less than 4 policies.
