probability given poisson dist.

Question

Librarians in busy libraries can re-shelf quite a few books a day. Assume that for each librarian 4 books per day get placed on the wrong shelf on average. in the first 15 days of March, the librarian placed 48 books on the wrong shelf. what is the probability the librarian will have incorrectly shelved 100 books by the end of March. I know its a Poisson dist. but I don't know how to set this up and carry it out; I know it's $P(X=100\, \text{given} X>48)$ but I don't know how to carry it out

Answer 1

If errors are Poisson distributed, then they must be independent occurrences. Hence the Comment by @DavidK, simplifies the problem to asking for $P(X = 52),$ where $X \sim Pois(\lambda)$ and $\lambda$ is the error rate for 16 days.

The mistake rate is 4 per day, and thus the error rate is $\lambda = 4(16) = 64$ for the remaining period of 16 days. (In my experience, a major source of error in working problems with the Poisson distribution is failure to adjust the error rate exactly to match the problem at hand.)

The probability is computed as $P(X = 52) = e^{-\lambda}\lambda^52/52!.$

In R this is could be done by using dpois (the Poisson PDF) or by direct computation.

 dpois(52, 64)
 ## 0.01659065
 exp(-64)*(64)^52/factorial(52)
 ## 0.01659065

However, I'm wondering if the problem might have meant "... incorrectly shelved at least 100 books by the end of March." In that case, you'd want $P(X \ge 52) = 1 - P(X \le 51):$

 1 - ppois(51, 64)
 ## 0.9447987

The question arises whether you were expected to use the normal approximation to the Poisson. The 'best fitting' normal distribution to $Pois(64)$ is $Norm(\mu = 64, \sigma = 8)$. Using the continuity correction I get a normal approximation of $P(X \ge 51.5)$ to be about 94%.

Note: To get a reasonable normal approximation for $P(X = 52)$, you should begin with $P(51.5 < X < 52.5).$

Here is a plot of some of the probabilities in the distribution $Pois(64)$ along with the approximating normal density curve.

Addendum: One might wonder whether making 48 mistakes in the initial 15 days is consistent with an error rate of 4 per day. One 95% confidence interval for $\lambda$ (15-day rate) would be $X + 2 \pm 1.96\sqrt{X + 1}$ or $(36.28, 63.72)$. For the per day error rate, this is $(2.42, 4.25)$, which includes $\lambda =4$, as originally stated. So, in this respect, the librarian behaved according to the model for the first half of the month.