Question

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 sent out will be exactly 55

Answer 1

Answer:

$P(X = 5) = 0.166$

Step-by-step explanation:

Given

$Mean = 4.3$

$x = 5$

Required

Determine the probability that the order is 5

This question can be answered using Poisson distribution.

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

Where

$\alpha$ is used to represent the mean

and

$\alpha = 4.3$

$x = 5$

$e = 2.71828$ ---- Euler's constant

So, we have:

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

$P(X = 5) = \frac{4.3^5 * 2.71828^{-4.3}}{5!}$

$P(X = 5) = \frac{4.3^5 * 2.71828^{-4.3}}{5* 4 * 3 * 2 *1}$

$P(X = 5) = \frac{4.3^5 * 2.71828^{-4.3}}{120}$

$P(X = 5) = \frac{1470.08443 * 0.01356859825}{120}$

$P(X = 5) = \frac{19.9469850243}{120}$

$P(X = 5) = 0.1662248752$

$P(X = 5) = 0.166$ Approximated