Question

A service center receives an average of 0.6 customer complaints per hour. Management's goal is to receive fewer than three complaints each hour. Assume the number of complaints follows the Poisson distribution. Determine the probability that exactly four complaints will be received during the next eight hours.

Answer 1

Answer:

0.18203 = 18.203% probability that exactly four complaints will be received during the next eight hours.

Step-by-step explanation:

We have the mean during a time-period, which means that the Poisson distribution is used to solve this question.

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 interval.

A service center receives an average of 0.6 customer complaints per hour.

This means that $\mu = 0.6h$, in which h is the number of hours.

Determine the probability that exactly four complaints will be received during the next eight hours.

8 hours means that $h = 8, \mu = 0.6(8) = 4.8$.

The probability is P(X = 4).

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

$P(X = 4) = \frac{e^{-4.8}*4.8^{4}}{(4)!} = 0.18203$

0.18203 = 18.203% probability that exactly four complaints will be received during the next eight hours.