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Suppose that the weight of navel oranges is normally distributed with mean=8 ounces and standard deviation...

Suppose that the weight of navel oranges is normally distributed with mean=8 ounces and standard deviation = 1.5 ounces.

What is the weight of the navel orange larger than only 10% of navel oranges.

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Answer #1

for top 10 % will fall at 90th percentile:

for 90th percentile critical value of z= 1.28
therefore corresponding value=mean+z*std deviation= 9.92 ounces
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