I was given the following homework question:
If a company's computer has around 28.14 errors per week. What is the probability of less than 3 errors at one day ?
Can someone give me a little start up aid to solve the question? (no demand here, for solving my homework!).
What I got so far:
I shall estimate $p(X<3)$, therefore I could simply add the probabilities for $p(X=0)+p(X=1)+p(X=2)$.
But this is where I am struggling: How to get a useful information for any of those probabilities out of 28.14 per week?
$28.14/7 = 4.02$ So on average there is 4.02 errors per day. But how to know how likely it is that there's less or more than average on a day ?
Once again I am not asking for a solution but for a little help getting started.
Using the poisson distribution, you have a $Po(4)$ per day (approximating $4.02 \approx 4$)
So the requested probability is
$\mathbb{P}[X<3]=e^{-4}[1+4+\frac{4^2}{2!}]\approx 24\%$
it is understood that it is not forbidden to use exactly 4.02 a a mean errors per day....the calculation process is the same