|Posted by White Group Mathematics on January 24, 2016 at 11:50 AM|
Sales of cooking oil bought in a shop during a week follow a Poisson distribution with mean 100. How many units should be kept in stock to be at least 99% certain that supply will be able to meet demand? Use normal approximation to Poisson distribution, if appropriate.
I have already approximated the distribution to Y ~ N (100, 100). Then, I have tried to use P(Y>y) = 0.99 . That didn't work so I'm trying to use P( Y = y) = 0.99. However, with this, I'm stuck with P [ (y-100.5)/10 < Z < (y - 99.5)/10] = 0.99
The correct answer is 123. Please help.
You kinda got your understanding of stuff mixed up. Firstly, recognize that it is the demand that varies, not the supply. It is the amount demanded that follows a Poisson distribution, which is subsequently approximated to that of a normal distribution.
The part about defining the original underlying distribution modelling demand as Y ~ N (100, 100) is correct.
If you let k be the number of units you wish to store as stock, then the required inequality satisfying the requirements of the question should be P(Y ≤ k) ≥ 0.99, which after adjusting for continuity correct would yield P(Y ≤ k+0.5) ≥ 0.99
As such, k+0.5 ≥ InvNorm of a normal distribution curve with mean and variance both =100 *
ie after standardization, solve (k+0.5-100)/ 10 ≥ InvNorm ( 0.99, 0, 1)
Hope this clarifies. Peace.
* The graphic calculator should help you compute the inequality in k here; unless you are using basic tables, then standardization is needed.
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