Tricky math question

[thorgrim29]

Hi, so I'm working on a physical inventory procedure for my internship (business accounting major), since we have lots of different products, assets and such in inventory and it's a bit of a mess.

Right now the idea is to make a count of a percentage of the total inventory each month, then make a total count at year end if it seems necessary. Problem is, I'd like to take the percentage each month from the whole thing, including what was already counted, to get an hopefully statistically significant view of the thing each month. I know it makes sence from an accounting point of view, but I have no idea how the math works.

Basically, I'd like to know what percentage of the total to take each month to reasonably sure that a large part (say 85%) of the assets have been counted come year end.

Thanks.

[Science Officer]

I don't know very much about statistics, but this seems like it wouldn't be too difficult to simulate.
If you know any computer programming, or some Excel magic, you could run a bunch of tests until you find the sample rate you want.

If you don't, your time might be better spent on researching statistics.
Or, you know, waiting for someone else here to answer your question :smallbiggrin:


EDIT: in an attempt to look less dumb

[RS14]

Let us pretend that each month, items are sampled with probability $p$, since that is easier. I'm sure there's a better way to do this where a constant proportion is sampled monthly, but I've not studied probability in depth, so I don't know.

Let us desire that $m=.85$ of the inventory is sampled at year end with probability $q=.95$. Let there be $N$ items to be inventoried. Let $r$ be the probability that an item is sampled in the first 11 months.


The number of items sampled over the first 11 months is given by a binomial distribution X with parameters $r$ and $N$. Furthermore, $(X-(N*r))/(N*r*(1-r))$ is approximately (http://en.wikipedia.org/wiki/Binomial_distribution#Normal_approximation) the standard normal distribution, for reasonable values of N and r. Thus the sum of the probabilities of the events in which .85*N,.85*N+1...N items are inventoried is the integral of the normal pdf from $(.85-r)/(r*(1-r))$ to $\infty$ or $1-\phi((.85-r)/(r*(1-r)))$ where phi denotes the cdf of the normal distribution.

I don't really like this material, and I'm starting to lose interest in explaining this with full detail, so I'm just going to gloss through the rest, and trust that you can look up what you need in a probability textbook.

Now taking the inverse normal-cdf of $1-q$ allows us to determine an appropriate value for $r$ such that $1-\phi((.85-r)/(r*(1-r)) > q$. Then we wish to find $p$ such that $(1-p)^11=1-r$, as sampling each with probability $p$ every month will result in a net probability $r$ of being sampled at least once. Because of the aforementioned normal approximation, we also know that at least .85 of the inventory is sampled at least once with probability at least .95.

Adjust everything as necessary, of course.

[thorgrim29]

Wow... I do not understand english algebra at all.... But thanks for the help anyway, I'll try to figure it out tomorrow. I was actually thinking of simulating it with excel.