Question

A cappuccino vending machine is designed to dispense an average of ? oz per cup. if the ounces per fill are normally distributed, with a standard deviation of 0.4 oz, what value should ? be set at so 6 oz cups will overflow only 2% of the time? 6.82 oz 5.18 oz 5.60 oz 6.00 oz

Answer 1

Overflow 2% of the time means that the upper tail has an area of 2%, or the lower tail has an area of 100-2=98%.
From tables, P(X<Z)=0.98 => Z=2.0537
We want the maximum fill be 6 oz 98% of the time, or

Mean+Z*(standard deviation)=6 oz
mean = 6oz -2.0537*0.4=5.18 oz

Answer 2

5.18oz.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`X = 6, \sigma = 0.4`

Overflow only 2% of the time?

Value of
`\mu` when Z has a pvalue of 1-0.02 = 0.98. So X when Z = 2.055.

`Z = (6 - \mu)/(\sigma)`

`2.055 = (6 - \mu)/(0.4)`

`6 - \mu = 2.055*0.4`

`\mu = 5.18`

So the correct answer is:

5.18oz.