meanpoisson distributionself-study
Suppose that the number of accidents occurring at a particular intersection each week is a Poisson random variable. It is estimated that the probability of having at least one accident in a week, at the intersection, is 0.98.
Calculate the average number of accidents occurring each week at the intersection.
Thanks in advance for the help!
What is the probability that at least two weeks will elapse between accident?
This uses the fact that time between Poisson events follows the exponential distribution (probability distribution that describes the time between events in a Poisson process - wikipedia./exponential distribution). When finding the probability of at least two weeks the lower bound is 2, and the distribution has a mean of 0.25. Hope this helps.
The following analysis illustrates one approach to obtaining a solution. At least it might help show how to work with the Poisson distribution.
To answer this question constructively and clearly, let's make a few simplifying assumptions to avoid getting bogged down in details that haven't been described. For instance, you might choose to
assume that a "breakdown" is an event with such a short duration that a machine is back in operation immediately after a breakdown; and
therefore the same machine could break down multiple times during a week (although this might be a rare event).
As apparently intended by the question, we will make some additional stronger assumptions. Some such (modeling) assumptions are needed to make any progress at all with the answer. Their chief purpose would be to give us a point of departure for eliciting additional information from the plant engineers so we could develop improved models and better answers:
All machines independently have the same chances of breaking down and
those chances do not vary over time.
These assumptions imply the number of breakdowns observed among any number $N$ of machines during any period of $x$ weeks has a Poisson$(\lambda N x)$ distribution, where $\lambda$ is a parameter common to all machines at all times. The question tells us about the breakdown rate for $x = 1$ week:
$$\lambda N 1\text{ weeks} = 2.$$
Therefore
$$\lambda = 2 / (N \text{ machine-weeks}).$$
In a random sample of $26$ such machines, the number of breakdowns in a week will have a Poisson distribution with parameter
$$\mu = \lambda\times (26\text{ machines})\times (1\text{ week}) = 26\lambda = 52/N.$$
From the formula for Poisson probabilities, the chance of no breakdowns among these $26$ machines is
$$e^{-\mu} 0! = e^{-\mu} = e^{-26\lambda} = e^{-52/N}.$$
Since $N\ge 26$, this value cannot exceed $e^{-52/26}=e^{-2}\approx 0.135$, but as $N$ grows large it could become arbitrarily small.
This is not a final answer. It only shows the implication of four assumptions that were made upon interpreting the question in terms of the chance of no breakdowns in a week. (Other interpretations of the question are possible, due to the contorted syntax used to pose it.) In particular, the dependence upon the unknown total number of machines is clear and explicit. This is about as far as one can go, given the limited information supplied in the question.
A simulation (covering almost 200 years of operation) illustrates the ideas. Its output consists of two histograms: the weekly breakdown counts for all $N$ machines and the counts for the sample of the machines. Here is an example for $N=60$:
On each histogram are drawn two vertical lines: a gray one indicating the location of the actual rate (as given by the preceding solution) and a dashed red one indicating the average rate during the simulation. In each case those lines are visibly coincident, showing that the simulation and the preceding analysis are in agreement.
Studying the R code that produced this simulation may help clarify the ideas.
n <- 60 # Number of machines
sample.size <- 26 # Must be less than or equal to n
weekly.mean <- 2 # Events per week, on average
n.iter <- 1e4 # Size of this simulation in weeks
set.seed(17) # Reproduce these results exactly
#
# Simulate all machines.
#
lambda <- weekly.mean/n # Weekly breakdown rate per machine
x <- matrix(rpois(n.iter*n, lambda), nrow=n) # Breakdowns by machine by week
weekly.breakdowns <- colSums(x) # Total breakdowns each week
sample.breakdowns <- colSums(x[1:sample.size, ]) # Total breakdowns in the sample
#
# Plot the results.
#
par(mfrow=c(1,2))
eps <- 0.99
hist(weekly.breakdowns, breaks=(-1):max(weekly.breakdowns)+eps,
freq=FALSE, cex.main=0.9)
abline(v=lambda * n, lwd=2, col="Gray")
abline(v=mean(weekly.breakdowns), col="Red", lwd=3, lty=3)
mu <- weekly.mean * sample.size / n
hist(sample.breakdowns, breaks=(-1):max(sample.breakdowns)+eps,
freq=FALSE, cex.main=0.9)
abline(v=mu * n, lwd=2, col="Gray")
abline(v=mean(sample.breakdowns), col="Red", lwd=3, lty=3)
Best Answer
Hint: if you knew the average, you could determine the probability of having exactly 0, exactly 1, exactly 2, and so forth.
So, what you need to do is rearrange the equation that you use so that if you know what you know (probability of 1+ accidents =.98) you can determine the average.