Dear Indra, I think you can probably calculate this from first principles using the normal probability distribution. I suspect that the average value of N will remain the same.

Thank you for your answer, Peter. I think that this problem is similar to computing the amount of time T required to build N items that take Tg +/- stdev_Tg to be built. In this case, T = N*Tg, so T is just a scaled version of Tg. It has the same distribution as Tg. In the case of setting T to a constant value, it is the inverse distribution, which only exists in certain conditions: http://en.wikipedia.org/wiki/Inverse_distribution#Relation_to_original_distribution

The answer depends on the specific distribution of the time intervals - mean and standard deviation are not enough. If the event rate is constant (e.g. does not increase or decrease as time passes since the last event), the time intervals will come from an exponential distribution, and the standard deviation will be equal to the mean (of time intervals between events). In that case, the number of events in some time interval T will come from a Poisson distribution, and the variance of events observed per time interval will equal the mean (with enough observations).

If the event rate changes as a function of time since the last event (think failure rate of car breaks) then you need to look at a different distribution - I think the Weibull distribution is probably the best choice. If the event rate is constant between events but changes randomly each time an even occurs, look at the Lomax distribution. In either case, the distribution of the events per time interval T may not have a closed form, although the integral which would give you the distribution you want can still be solved numerically or via MCMC.

Regardless of the time interval distribution, the sample standard deviation will approach the  standard deviation of the distribution as more samples of cases. For samples from a normal distribution the ratio of sample variance to distribution  is (N-1)/N. If you look up the derivation of that result (which is in all intro statistics textbooks and all over the internet) it should be straightforward to apply the method to other distributions.  And you can always check yourself with Monte Carlo.

Thank you for your detailed answer, John. I understand that if it can be assumed that the time to build an item is independent from the time it took any previous items to be built, the time intervals follow an exponential distribution (memoryless property), so the number of items built on a given time interval would follow a Poisson distribution.

You don't need the distribution to figure out the standard deviation.  If the times taken to produce the items are IID (independent and identically distributed) then the standard deviation of the sum of the first n times will be sqrt(n) times the individual standard deviation.  Or if you are working with averages, then the standard deviation of the average of the first n times will be 1/sqrt(n) times the individual standard deviation.  This is a standard result in statistics on IID random variables.  It follows directly from the moment properties of random variables.

If you would like to see this in the context of your model it is pretty simple.  It is common in many processes of this kind to assume that the time taken to produce an item has a gamma distribution (which subsumes the exponential), so that the time taken to produce multiple items also has a gamma distribution, with different parameters.

So, for example, you might do the following:

----------

Ti = Time taken to produce item i

E(Ti) = Mu

S(Ti) = Sigma

----------

Alpha = Gamma Shape parameter = E(Ti)^2 / V(Ti) = Mu^2 / Sigma^2

Theta = Gamma Scale parameter = V(Ti) / E(Ti) = Sigma^2 / Mu

Ti ~ Gamma ( Shape = Alpha , Scale = Theta )

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TTn = T1 + T2 + ... + Tn = Time taken to produce items 1-n

TTN ~ Gamma ( Shape = n * Alpha , Scale = Theta )

----------

E(TTn) = n * Alpha * Theta = n *Mu

V(TTn) = n * Alpha * Theta^2 = n * Sigma^2

S(TTn) = sqrt(n) * Sigma

----------

This gives the distribution for the individual time Ti and the distribution for the sum TTn.  As you can see, the standard deviation of the latter is sqrt(n) times the standard deviation of the former.

Hope that helps.

http://en.wikipedia.org/wiki/Gamma_distribution

http://en.wikipedia.org/wiki/Exponential_distribution

You should take a look at chapter 2 of van Kampen's book, "Random events", which studies in depth the kind of problem you are interested in. The basic stochastic process  in this respect is "shot noise" and the associated Poisson distribution, see 2.2 therein.

Gracias Antonio, he echado de nuevo un vistazo al proceso de Poisson donde me indicas, aquí: http://www.fulviofrisone.com/attachments/article/413/VanKampen.pdf#page=46

De nada :-)

If you are willing to assume that the inter-arrival times of consecutive items form a sequence of iid random variables, then your question becomes a standard basic question in Renewal Theory. The expected number of items, m(t), produced/arriving up to a fixed time t depends on the distribution F of the individual inter-arrival times and can be calculated as follows:

m(t) = SUM (from 1 to infinity) of F_n (t)

where F_n is the cumulative distribution function of the sum of n inter-arrival times (which is F convoluted with itself n times). In general, it is hard to find explicit expressions fo F_n, hence difficult to compute m(t) explicitely. It is however a consequence of the Law of Large Numbers that as t tends to infinity, the ratio t/N(t) converges (almost surely) to the expectation/average, A, of the inter-arrival time, so that for large t, N(t) is with high probability close t/A.

Similar, more complicated, expression (in terms F-n) can be given for the variance of N(t). 

I recommend to you to look up Wikipedia and other sources on elementary Renewal theory.