Leyland Motors can service an average 5 cars per hour, and the owner of Leyland Motors wants to know the probability of various customer arrival rates. Given the average arrival rate of 4 customers per hour, the owner uses the Poisson distribution to calculate the probabilities of various customer arrivals per hour. What is the probability that at least 2 customers will arrive within 1 hour

P(X ≥ 2) = 0.90842

Step-by-step explanation:

This is a Poisson distribution problem

And Poisson distribution formula is given by

P(X = x) = (e^-λ)(λˣ)/x!

where λ = mean = average arrival rate = 4 customers per hour

x = variable whose probability is required = at least 2. P(X ≥ 2)

P(X ≥ 2) = 1 - P(X < 2)

P(X < 2) = P(X=0) + P(X=1)

P(X=0) = (e⁻⁴)(4⁰)/0!

P(X=0) = 0.01832

P(X=1) = (e⁻⁴)(4¹)/1!

P(X=1) = 0.07326

P(X < 2) = P(X=0) + P(X=1) = 0.01832 + 0.07326 = 0.09158

P(X ≥ 2) = 1 - P(X < 2) = 1 - 0.09158 = 0.90842

Hope this Helps!!!

the answer is 28

step-by-step explanation: