Question

The auto parts department of an automotive dealership sends out a mean of 6.3 special orders daily. What is the probability that, for any day, the number of special orders sent out will be exactly 3

Answer 1

0.0765 = 7.65% probability that, for any day, the number of special orders sent out will be exactly 3

Explanation:

We have the mean, 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) = (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 6.3 special orders daily.

This means that
`\mu = 6.3`

What is the probability that, for any day, the number of special orders sent out will be exactly 3?

This is P(X = 3). So

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 3) = (e^(-6.3)*6.3^(3))/((3)!) = 0.0765`

0.0765 = 7.65% probability that, for any day, the number of special orders sent out will be exactly 3