Question

The auto parts department of an automotive dealership sends out a mean of 3.8 special orders daily. What is the probability that, for any day, the number of special orders sent out will be exactly 4? Round your answer to four decimal places.

Answer 1

Answer:

0.1944 = 19.44% probability that, for any day, the number of special orders sent out will be exactly 4

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 auto parts department of an automotive dealership sends out a mean of 3.8 special orders daily.

This means that $\mu = 3.8$

What is the probability that, for any day, the number of special orders sent out will be exactly 4?

This is P(X = 4).

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

$P(X = 4) = \frac{e^{-3.8}*(3.8)^{4}}{(4)!} = 0.1944$

0.1944 = 19.44% probability that, for any day, the number of special orders sent out will be exactly 4