Combining Normal Distributions and Calculating Probabilities

#1
Hi,

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.
The duration it takes to charge a car follows a Normal distribution with a mean of 5 hours and a standard deviation of 1.5 hours. Two households share an electric outlet and plug the first car in at midnight. Assuming that they plug the second car in immediately once the first is fully charged, what is the probability that the second car is not fully charged at 8am?

Thanks!

As far as I understand I should consider a new Normal distribution with a mean=5+5=10 hours and a standard deviation s=((1.5)^2 + (1.5)^2)^1/2=2.12 hours. Then, should I calculate the probability between 5 and 8 hours?

I found that P(5hours<X<8hours)=16.42%
 
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#2
Use your new combined normal and test for a combined time of more than 8 hours, because you are asked for the probability of both not being charged.
(The combined mean is 10, so the second car must have a more than 50% chance of not being charged after 8 hours.).
 
#3
Hi katxt,

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.
 
#4
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).
Let's say that the probability of each car being charged within 8 hours is
P(A charged within 8)=P(B charged within 8)=97.72% (using normal distribution with mean=5, s=1.2).

By combing the 2 normal distributions we can find the probability of both cars to be fully charged within 8 hours.
combined mean=5+5=10 and s=((1.5)^2 + (1.5)^2)^1/2=2.12

So, we can calculate P(AandB charged within 8)=17.36% (using z-table)
Hence, the probability of the second car to be fully charged if the first car has already been fully charged is
P(B|A)=P(AandB charged within 8)/P(A)=17.36%/97.72%=17.77%

And hence, the probability of the second car not to be fully charged within 8 hours if the first car has already been fully charged is
1-P(B|A)=100%-17.77%=82.23%

Does it sound reasonable?

Thank you!
 
#5
OK. Ingenious but I think the problem is much less complicated. Imagine one car with two batteries and charging the second battery is automatically started the moment the first is finished. Then the second is finished N(10, 2.12) as you say. What is the probability that the second is not finished at t = 8?
z = (8-10)/2.12 = -0.943 P<8 = 17.3% from the table p>8 = 1-17.3% = 82.7%