Question

The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and mean 0.2 each day during the weekend. The number of phone calls that Ben receives on one day is independent of the number of phone calls that he receives on other days. If today is Monday, what is the probability that Ben receives a total of 2 phone calls in a week?

Answer 1

Answer:

There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

$e = 2.71828$ is the Euler number

$\mu$ is the mean in the given time interval.

The problem states that:

The number of phone calls that Actuary Ben receives each day has a Poisson distribution with mean 0.1 during each weekday and mean 0.2 each day during the weekend.

To find the mean during the time interval, we have to find the weighed mean of calls he receives per day.

There are 5 weekdays, with a mean of 0.1 calls per day.

The weekend is 2 days long, with a mean of 0.2 calls per day.

So:

$\mu = \frac{5(0.1) + 2(0.2)}{7} = 0.1286$

If today is Monday, what is the probability that Ben receives a total of 2 phone calls in a week?

This is $P(X = 2)$. So:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-0.1286}*0.1286^{2}}{(2)!} = 0.0073$

There is a 0.73% probability that Ben receives a total of 2 phone calls in a week.