Question

A service center receives an average of 0.6 customer complaints per hour. Management's goal is to receive fewer than five complaints each eight-hour shift. Assume the number of complaints follows the Poisson distribution. Determine the probability that managment is unhappy with the number of complaints in the next eight hours.

Answer 1

Answer:

48.68% probability that managment is unhappy with the number of complaints in the next eight hours.

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.

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

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

Eight hours:

This means that $n = 8, \mu = 0.6*8 = 4.8$

Determine the probability that managment is unhappy with the number of complaints in the next eight hours.

They will be unhappy if they receive five or more complaints.

Either they receive less than five complaints, or they receive at least five. The sum of the probabilities of these events is 1. So

$P(X < 5) + P(X \geq 5) = 1$

We want $P(X \geq 5)$.

Then

$P(X \geq 1) = 1 - P(X < 5)$

In which

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

So

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

$P(X = 0) = \frac{e^{-4.6}*4.6^{0}}{(0)!} = 0.0101$

$P(X = 1) = \frac{e^{-4.6}*4.6^{1}}{(1)!} = 0.0462$

$P(X = 2) = \frac{e^{-4.6}*4.6^{2}}{(2)!} = 0.1063$

$P(X = 3) = \frac{e^{-4.6}*4.6^{3}}{(3)!} = 0.1631$

$P(X = 4) = \frac{e^{-4.6}*4.6^{4}}{(4)!} = 0.1875$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0101 + 0.0462 + 0.1063 + 0.1631 + 0.1875 = 0.5132$

Finally

$P(X \geq 1) = 1 - P(X < 5) = 1 - 0.5132 = 0.4868$

48.68% probability that managment is unhappy with the number of complaints in the next eight hours.