Question

The number of calls received by an office on Monday morning between 8:00 AM and 9:00 AM has a mean of 6. Calculate the probability of getting at least 1 call between eight and nine in the morning. Round your answer to four decimal places.

Answer 1

Answer:

0.9975 = 99.75% probability of getting at least 1 call between eight and nine in the morning.

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 interval

The number of calls received by an office on Monday morning between 8:00 AM and 9:00 AM has a mean of 6.

This means that $\mu = 6$

Calculate the probability of getting at least 1 call between eight and nine in the morning.

Either you get no calls, or you get at least one call. The sum of the probabilities of these events is decimal 1. Mathematically, we have that:

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$. So

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = \frac{e^{-6}*(6)^{0}}{(0)!} = 0.0025$

Then

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0025 = 0.9975[tex]

0.9975 = 99.75% probability of getting at least 1 call between eight and nine in the morning.