I thought of something else:
The probability of each of the car to be fully charged using a dedicated electric outlet can be calculated by the normal distribution with mean=5 and s=1.5.
So, if P(A) and P(B) the probability of the first car and second car to be fully charged then P(A)=P(B)...
Thank you for replying.
However, I think the question is more complicated.
The problem asks what is the probability that the second car is not fully charged at 8am assuming that the first car has been charged before.
I am trying to figure out a solution to a problem with normal distributions and I need some help. The problem is:
In a town of 1000 households, each one has 1 electric car. Each night, every house plugs in the electric car for charging in order to be ready to use at 8am in the morning.