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This is driving me crazy...

"A Machine fills a 3 pound coffee can with a measured amount of coffee. The Weight has a normal distribution with a mean of 3.1 pounds. The standard deviation is 2.0 OUNCES. What is the probability that a can picked at random will contain less than 3 pounds.

Ok so first I converted the SD to pounds (2.0 Ounces/ 16 Ounces = 0.125 Pounds)

I calculated the Z score, $$\frac{2.9 - 3.1}{0.125} = -1.6$$

I looked up the standard area table -> 0.0548

Wouldn't this be 5.48% or am I messing something up? Or this isn't probability?

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1 Answer 1

Your problem is that you calculated your Z-score as $\frac{2.9-3.1}{0.125}$. If the can contains 2.9999 pounds of coffee, that's still less than 3 pounds. So the correct Z-score is given by $$\frac{3 - 3.1}{0.125} =-0.8$$ which gives a probability of about 21%.

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