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    Mean and standard deviation of a normal distribution

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    In grading Oranges into "A","B" and "C", a large orchard uses weights to distinguish Oranges. Any Orange weighing more than 2 ounces is classified as grade "A", while an Orange weighing less than 0.75 ounces is classified as grade "C". If the day's pick shows 16.6% are grade "A" and 6.68% are grade "C", determine the Mean and Standard Deviation. Assume the weights are normally distributed.

    Hint: First calculate the appropriate z values for the probabilities given,then set up the equations to solve for the Mean and the Standard Deviation.

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    Solution Preview

    Let
    M = mean
    s = standard deviation
    Orange weighing less than 0.75 ounces is classified as grade "C
    6.68% are grade "C"
    z value corresponding to probability 6.68% is -1.500056
    Thus z1=(0.75-M )/s = -1.500056 --- Equation (1)
    Any Orange weighing more than 2 ounces is classified as grade "A"
    16.6% are grade "A"
    z value corresponding to ...

    Solution Summary

    Mean and standard deviation of a normal distribution are calculated using the data given.

    $2.19

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