RbStimers
RbStimers
Joined: Apr 19, 2010
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April 19th, 2010 at 11:06:33 AM permalink
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.
Doc
Doc
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April 19th, 2010 at 12:23:08 PM permalink
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.
RbStimers
RbStimers
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April 19th, 2010 at 12:43:20 PM permalink
You are right, But my variability isn't that high.
Wizard
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Wizard
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April 19th, 2010 at 1:06:35 PM permalink
Quote: RbStimers

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.



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%.
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DJTeddyBear
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April 19th, 2010 at 1:28:26 PM permalink
Quote: Doc

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.

While, mathematically that's a valid point, in the real world it's not.

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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RbStimers
RbStimers
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April 19th, 2010 at 7:32:08 PM permalink
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!

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