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Statistics: 90th percentile demand

  1. Mar 16, 2013 #1
    I'm a little confused on how you would do something like this:

    I have a cumulative probability distribution H(x):

    0 0.31042
    1 0.67825
    2 0.89149
    3 0.97196
    4 0.9942
    5 0.999
    6 0.99985
    7 0.99997
    8 1

    and I need to find the 90th percentile of demand..
    I understand that this means its the minimum x so that H(x) >= 90% but I need an example to make sure I'm doing this right.
    Would this mean x >= 3 in this case??
  2. jcsd
  3. Mar 16, 2013 #2
    You've got the right idea, but 3 is actually the 97(.196)th percentile whereas 2 is the 89(.149)th.

    You need to interpolate between these two points: usually this is a linear interpolation. Moving from 2 to 3 you get from 89.149 to 97.196; how much is this? How far would you have to move from 2 to get to 90?
  4. Mar 16, 2013 #3


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    Science Advisor

    Hey bookworm121.

    For this distribution, you are correct in say that x = 3 for the 90th percentile.

    It is just a minimum value where x = Min_i P(X <= i) >= p where is your percentile and i is the value corresponding to the value in the distribution and X is the random variable.

    Min_i is a function that returns the smallest value of i such that the expression holds.
  5. Mar 16, 2013 #4


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    Science Advisor

    I would not recommend this since you are losing information through the definition of the random variable.

    Linearity assumes an equal kind of weighting between bins and this may not be representative of the population (or sample).
  6. Mar 17, 2013 #5
    By your definition that is correct.
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