00:01
In this exercise, we have the number of customers arriving in a one -hour period, modeled by a poisson distribution with mean mu.
00:13
Now, it's believed that the mean is 15.
00:17
So for part a, we're asked to define a prior distribution for mean from the gamma family of distributions, such that the mean is 15, and the standard deviation is 5.
00:41
Now for a gamma distribution, the mean is equal to alpha times beta.
00:49
Remember the gamma distribution has this form.
01:17
And so returning to the question, then the standard deviation for a gamma distribution is equal to the square root of alpha times beta.
01:27
So we can solve for alpha and beta here if we take the mean divided by the standard deviation.
01:35
It gives the square root of alpha is equal to 3, which means that we can solve for alpha in beta here.
01:40
Means that alpha is equal to 9.
01:46
And then using either equation, we can solve for beta.
01:57
That'll be 15 over 9.
02:05
So that means our prior distribution, function of mu, is we just have to give it the gamma form, gamma distribution form, using our solved alpha and beta.
02:43
So that answer is part a.
02:46
Now for part b, we have the number of customers that have arrived in 10 randomly one hour, randomly selected one hour intervals.
02:56
And based on this new information, we're asked to find the posterior distribution for mu.
03:03
So remember the arrivals are modeled by a pluson distribution...