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The Poisson Distribution Watch

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    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.
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    (Original post by VTan)
    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.
    I'm a bit rusty on this but I thought when you approximate the distribution you would do Y ~ N (100, 0.99) (as the first parameter in the distribution is the mean and then the probability of success which in your case would be 99%)

    No?

    Edit: and lambda would be 1 week....number of trials is what you need to calculate!
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    (Original post by VTan)
    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.

    So yes 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 ≥ inverse Norm 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.
 
 
 
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