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colby2152 · #2 December 13th, 2007, 07:31 AM

Quote:

Originally Posted by eugene1687

Hi there,
Is anyone able to help me with this question on Poisson Distribution.

A shop sells a product at an average rate of 4 per week. Assume that the number sold in a week is a Poisson variable.
What number should be in stock at the beginning of a week to have a 95% assurance of being able to meet all demand during the week?
Someone please help me..
Thank you sooo much..

You want $P(X > n) = .05$

Poisson is discrete, so let's make it easy and go through this 1 by 1 starting off with a stock of 4.

$P(X > 4) = 1 - P(X \le 4)$
$P(X > 4) = 1 - P(0) - P(1) - P(2) - P(3) - P(4)$
$P(X > 4) = 1 - exp(-4)(1 + 4 + 8 + \frac{64}{6} + \frac{256}{24})$
$P(X > 4) = .3717$

We obviously have to go much higher, so let's try a beginning stock of 6.

$P(X > 6) = .3717 - P(5) - P(6)$
$P(X > 6) = .3717 - .26 = .1117$

Higher yet again, let's try stock of 7.

$P(X > 8) = .1117 - P(7)$
$P(X > 8) = .1117 - .0183(3.25)$
$P(X > 8) = .052225$

P(8) ends up being equal to .0297, which is enough to put us over that limit, so our stock cannot be GREATER than 8. Our initial weekly stock must be eight.