April 19th, 2010 at 11:06:33 AM
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I sell sculptures. On average, out of every 7 Sculpture sales, one will be a turtle the rest will be other types of sculpture, how many turtles do i need to have in stock if I want a 90% chance of not running out in the next 100 sculpture sales?

Is there an equation or an approximation?

I can do the reverse using the binomial theorem, i can find the probability of running out if i know how often they sell and how many I have but that's a clumsy method.

Is there an equation or an approximation?

I can do the reverse using the binomial theorem, i can find the probability of running out if i know how often they sell and how many I have but that's a clumsy method.

April 19th, 2010 at 12:23:08 PM
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So here's a non-math guy adding a question -- isn't it important to know the variability in your turtle sales? I mean, it could be that your typical experience is to go years without a turtle sale and then encounter a buyer who wants a large batch of them. It seems to me that could still give you 1/7 of your sales but have an impact on the 90% probability of no out-of-stock in the next 100 sales. If you assume that the buyer's selection of a sculpture is random, then I think your answer can be calculated, but I would probably do it wrong.

April 19th, 2010 at 12:43:20 PM
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You are right, But my variability isn't that high.

April 19th, 2010 at 1:06:35 PM
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Quote:RbStimersI sell sculptures. On average, out of every 7 Sculpture sales, one will be a turtle the rest will be other types of sculpture, how many turtles do i need to have in stock if I want a 90% chance of not running out in the next 100 sculpture sales?

Is there an equation or an approximation?

I can do the reverse using the binomial theorem, i can find the probability of running out if i know how often they sell and how many I have but that's a clumsy method.

This is a good confidence interval kind of problem. In 100 sales the expected turtles sold will be 14.29. The standard deviation is sqrt(100*(1/7)*(6/7)) = 3.50.

Let t be the number of turtles made, and x the number sold.

pr(x<=t)=0.9

pr(x-14.29<=t-14.29)=0.9

pr((x-14.29)/3.5)<=(t-14.29)/3.5))=0.9

The left side of the inequality follows a standard normal distribution (mean of 0, standard deviation of 1). This next step takes an introductory statistics course, or some faith, to accept.

(t-14.29)/3.5 = normsinv(0.9) This is the Excel function.

(t-14.29)/3.5 = 1.282

t-14.29 = 4.4870

t = 18.77

Nobody is likely to buy 0.77 of a turtle statue, so I would round up to 19. According to the binomial distribution, the probability of selling 18 or less is 88.35%, and 19 or less is 92.74%.

It's not whether you win or lose; it's whether or not you had a good bet.

April 19th, 2010 at 1:28:26 PM
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While, mathematically that's a valid point, in the real world it's not.Quote:DocI mean, it could be that your typical experience is to go years without a turtle sale and then encounter a buyer who wants a large batch of them.

Unless I'm missing something, this is an item that is typically purchased one at time.

That being the case, a single buyer that wants to purchase a large quantity will by predestined to do one or more of the following:

- Go directly to a wholesaler,

- Expect them to not be in stock,

- Haggle over price and/or delivery time.

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April 19th, 2010 at 7:32:08 PM
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THANK YOU SO MUCH!

the numbers are just like my experience... Now I can quit being a combination of over stocked and under stocked all at the same time. I have way too many of my frequent sellers and way to little of my infrequent sellers, this is awesome! THANK YOU!

the numbers are just like my experience... Now I can quit being a combination of over stocked and under stocked all at the same time. I have way too many of my frequent sellers and way to little of my infrequent sellers, this is awesome! THANK YOU!