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A vending machine dispenses coffee into a twelve ounce cup. The amount of coffee dispensed into the cup is normally distributed...Asked by Shamin
A vending machine dispenses coffee into an eight-ounce cup. The amount of coffee dispensed into the cup is normally distributed with a standard deviation of .03 ounce. You can allow the cup to OVERFILL 1% of the time. What amount should you set as the mean amount of coffee to be dispensed?
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Answered by
Bernd
x=8, δ=0.03
x=μ+zδ
μ=x-zδ
You have to find the mean that makes the range of 0oz to 8oz equal to 0.99 or the 99th percentile (because everything above 8oz should occur only in 1% of the time).
You can look up the z-value for 0.99 in a Standard Normal Distribution table.
z=2.32
μ=8-2.32*0.03=7.93
You get a mean of 7.93oz.
x=μ+zδ
μ=x-zδ
You have to find the mean that makes the range of 0oz to 8oz equal to 0.99 or the 99th percentile (because everything above 8oz should occur only in 1% of the time).
You can look up the z-value for 0.99 in a Standard Normal Distribution table.
z=2.32
μ=8-2.32*0.03=7.93
You get a mean of 7.93oz.
Answered by
n
Thank for putting it plainly!
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