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.

Student X

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.

Best Regards,

Mr Koh

Categories: Math Queries

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