The auto parts department of an automotive dealership sends out a mean of 3.2 special orders daily. What is the probability that, for any day, the number of special orders sent out will be more than 3?

0.603

Step-by-step explanation:

Given that mean = 3.2

Probability that special order sent out will be no more than 3

Using poisson :

P(x) = (e^λ * λ^x) / x!

λ= 3.2

P(x≤3) : = p(x = 0) + p(x = 1) + p(x= 2) + p(x =3)

P(x = 0) = (e^-3.2 * 3.2^0) / 0! = 0.0407622

P(x = 1) = (e^-3.2 * 3.2^1) / 1! = 0.1304390

P(x = 2) = (e^-3.2 * 3.2^2) / 2 = 0.2087024

P(x = 3) = (e^-3.2 * 3.2^3) / 3! = 0.2226159

Hence ;

0.0407622 +0.1304390 + 0.2087024 + 0.2226159 = 0.6025195

= 0.603

answer: 4 meters

step-by-step explanation: to you understand 2/3 better you can do 1 divided by 3 = . then multiply that by the 2 which is .66666666666666667 then you would do 6 x .66666666666666667 which is 4.      

hope this .