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Normal and uniform probability distribution.

  1. May 29, 2012 #1
    Am I right? Thank you!!


    Problem:
    A cereal brand of cereal claims that the mean number of raisins in each box is 80 with a standard deviation of 6. If the raisins are normally distributed, what are the chances that an arbitrary box has

    1) fewer than 70 raisins and
    2) more than 90 raisins.

    What should be your answers if the raisins are uniformly distributed.

    Solution:

    {70-80}/{6}=-1.67; we disregard the negative sign in order to find the value in the table. The value is 0.9520. 1-0.9520=0.048 or 4.80% chance that an arbitrary box has fewer than 70 raisins.

    {90-80}/{6}=1.67, which is also 0.9525.
    1-0.9520=0.048 or 4.80% chance that an arbitrary box has more than 90 raisins.


    If the raisins are uniformly distributed, the probability that an arbitrary box has fewer than 70 raisins is 1/80 * 70=0.875 or 87.5%.
    Probability that there are more than 90 raisins is 0..
     
  2. jcsd
  3. May 29, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    You need to find the uniform distribution with mean 80 and variance 62= 36. Do you know the formulas for mean and variance of a uniform distribution on the interval [a,b]? Once you have determined a and b the rest is straightforward.

    RGV
     
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