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Mean and standard deviation of a normal distribution

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.

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.

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