MATH SOLVE

4 months ago

Q:
# A service station has a pump that distributes diesel fuel to automobiles. The station owner estimates that only about 3.2 cars use the diesel pump every 2 hours. Assume the arrivals of diesel pump users are Poisson distributed.a. What is the probability that three cars will arrive to use the diesel pump during a 1-hour period?b. Suppose the owner needs to shut down the diesel pump for half an hour to make repairs. However, the owner hates to lose any business. What is the probability that no cars will arrive to use the diesel pump during a half-hour period?c. Suppose five cars arrive during a 1-hour period to use the diesel pump. What is the probability of five or more cars arriving during a 1-hour period to use the diesel pump?

Accepted Solution

A:

Answer: a) 13.78%b) 44.93%c) 2.39%Step-by-step explanation:First we need to use the correct formula. In this case, it would be this:Poisson distribution (where m = mean):
P(x) = e^(-m) m^x / x!Now for the values of m, we already know that every 2 hours, 3.2 cars use the pump, therefore, we can assume that in one hour, is half the cars:1/2* 3.2 = 1.6 cars every hour
1/4* 3.2 = 0.8 cars every half an hour.Now that we have the formula and the possible values of m, let's do the exercise:a) In this case, we use m = 1.6 and replace it in the formula, and we will use x = 3, because we have 3 cars so:P(3) = e^(-1.6) * 1.6^3 / 3!P(3) = 0,1378. If you transform this into percentage, it would be 13.78%b) Using x = 0 (because we want that no cars arrive during the time), and m = 0.8 (half an hour) we have:P(0) = e^(-0.8) * 0.8^0 / 0!P(0) = 0.4493 ---> 44.93%c) Finally for this part, we use m = 1.6 and the values of X = 0,1,2,3,4:P(0) = e^(-1.6) * 1.6^0 / 0! = 0.2018P(1) = e^(-1.6) * 1.6^1 / 1! = 0.3230P(2) = e^(-1.6) * 1.6^2 / 2! = 0.2584P(3) = 0.1378P(4) = e^(-1.6) * 1.6^4 / 4! = 0,0551Let's sum all of these values:0.2018 + 0.3230 + 0.2584 + 0.1378 + 0.0551 = 0.9761Now, if we want to know the probability that five or more cars arrive during this period of one hour:P = 1 - 0.9761 = 0.0239 ----> 2.39%