The auto parts department of an automotive dealership sends out a mean of 4.34.3 special orders daily. What is the probability that, for any day, the number of special orders se…

Question

asked Apr 25, 2021 110k views

1 Answer

Asyncwait · Answer 1 · 2021-04-30T19:36:18+0000

Answer:

P(X = 5) = 0.166

Explanation:

Given

Mean = 4.3

x = 5

Required

Determine the probability that the order is 5

This question can be answered using Poisson distribution.

$P(X = x) = (\alpha ^x e^(-\alpha))/(x!)$

Where

$\alpha$ is used to represent the mean

and

$\alpha = 4.3$

x = 5

e = 2.71828 ---- Euler's constant

So, we have:

$P(X = x) = (\alpha ^x e^(-\alpha))/(x!)$

P(X = 5) = (4.3^5 * 2.71828^(-4.3))/(5!)

P(X = 5) = (4.3^5 * 2.71828^(-4.3))/(5* 4 * 3 * 2 *1)

P(X = 5) = (4.3^5 * 2.71828^(-4.3))/(120)

P(X = 5) = (1470.08443 * 0.01356859825)/(120)

P(X = 5) = (19.9469850243)/(120)

P(X = 5) = 0.1662248752

P(X = 5) = 0.166 Approximated