Question

The auto parts department of an automotive dealership sends out a mean of 4.7 special orders daily. What is the probability that, for any day, the number of special orders sent out will be exactly 3? Round your answer to four decimal places.

Answer 1

Answer:

0.1574

Step-by-step explanation:

We are given that

Mean=4.7

We have to find the probability that for any day the number of special order sent out will  be exactly 3.

We know that by Poisson distribution

$P(X=k)=\frac{e^{-\lambda}(\lambda)^k}{k!}$

Where $\lambda=mean$

Using the formula

$P(X=3)=\frac{e^{-4.7}(4.7)^3}{3!}$

$P(X=3)=0.15738\approx 0.1574$

Hence, the probability that for any day,the number of special orders sent out will be exactly 3=0.1574