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! = 0.2018
P(1) = e^(-1.6) * 1.6^1 / 1! = 0.3230
P(2) = e^(-1.6) * 1.6^2 / 2! = 0.2584
P(3) = 0.1378
P(4) = e^(-1.6) * 1.6^4 / 4! = 0,0551
Let's sum all of these values:
0.2018 + 0.3230 + 0.2584 + 0.1378 + 0.0551 = 0.9761
Now, 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%
