# A vending machine dispenses coffee into an eight-ounce cup. The amount

A vending machine dispenses coffee into an eight-ounce cup. The amounts of coffee dispensed into the cup are normally distributed, with a standard deviation of 0.03 ounce. You can allow the cup to overflow 1% of the time. What amount should you set as the mean amount of coffee to be dispensed?

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Cheryl King
Given:
$$\displaystyle{x}=\text{Ounces of cup}={8}$$
$$\displaystyle\sigma=\text{Standard deviation}={0.03}$$
The top 1% has the property that 1% of the data values are larger than the cutoff value and thus and thus $$\displaystyle{100}\%-{1}\%={99}\%$$ of the data values are smaller than the cutoff value.
We will use the standard normal table in the appendix to determine the z-score corresponding to an area of 99% or 0.99 or the closest area.
In the standard normal table in the appendix, we note that the value 0.99 lies closest to 0.9901.
The value .9901 is mentioned in the row starting with 2.3 and in the column starting with .03, which corresponds to a z-score of 2.33.
$$\displaystyle{z}={2.33}$$
However, we also know that the z-score is the value x decreased by the mean, divided by the standard deviation.
$$\displaystyle{z}={\frac{{{x}-\mu}}{{\sigma}}}$$
$$\displaystyle\mu={x}-{z}\sigma$$ Solve to $$\displaystyle\mu$$
$$\displaystyle={8}-{\left({2.33}\right)}{\left({0.03}\right)}$$ Substitute
$$\displaystyle={7.9301}$$
Thus the mean amount of coffee to be dispensed is 7.9301 ounces, which would then imply that the cup of 8 ounces will overflow about 1% of the time.
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Juan Spiller
Step 1
The amounts of coffee dispensed into the cup follow normal distribution with standard deviation 0.03 ounces. Also, the vending machine dispenses coffee into an eight-ounce cup and the cup to overflow is 1% of the time.
The random variable x represents the amounts of coffee dispensed.
The cup to overflow is 1% of the time represents the area of 0.01 to the right of x.
Thus, $$\displaystyle{P}{\left({X}{ < }{x}\right)}={0.99}$$
$$\displaystyle{P}{\left({X}−\frac{\mu}{\sigma}{ < }{x}−\frac{\mu}{\sigma}\right)}={0.99}$$ (by standardizing)
$$\displaystyle{P}{\left({Z}{ < }{x}−\frac{\mu}{\sigma}\right)}={0.99}$$
$$\displaystyle\Phi{\left({x}−\frac{\mu}{\sigma}\right)}={0.99}$$
Find the value 0.99 from the body of the standard normal table. The value is approximately 2.33.
Step 2
$$\displaystyle\Phi{\left({2.33}\right)}={0.99}$$.
Thus,
$$\displaystyle\Phi{\left({x}−\frac{\mu}{\sigma}\right)}=\Phi{\left({2.33}\right)}$$
$$\displaystyle\frac{{{x}-\mu}}{\sigma}={2.33}$$
$$\displaystyle{x}=\mu+{2.33}{\left(\sigma\right)}$$
Substitute $$\displaystyle\mu={8}\ {\quad\text{and}\quad}\ \sigma={0.03}$$.
$$\displaystyle{x}={8}+{2.33}{\left({0.03}\right)}$$
$$\displaystyle={8.0699}$$
The value can be also obtained using the Excel formula, =NORM.INV(0.99,8,0.03) as 8.0698.
Thus, the amount that should be set as the mean amount of coffee to be dispensed is 8.07 ounces.