Patricia Crane

Answered

2022-01-12

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?

Answer & Explanation

Cheryl King

Expert

2022-01-13Added 36 answers

So,

$x=\text{Ounces of cup}=8$

$\sigma =\text{Standard deviation}=0.03$

One-tenth of one percent (tenth) of the data values in the top 1% are larger than the cutoff threshold and thus $100\mathrm{\%}-1\mathrm{\%}=99\mathrm{\%}$ are less than the cutoff threshold for the data.

The z-score corresponding to an area of 99% or 0.99 or the nearest area will be found using the standard normal table in the appendix.

The value 0.99 in the appendix's standard normal table is the closest to 0.9901, as noted.

The value.9901, which has a z-score of 2.33, is given in the row beginning with 2.3 and the column beginning with.03.

$z=2.33$

The z-score is defined as the value of x divided by the mean and the standard deviation, though, as we also know.

$z=\frac{x-\mu}{\sigma}$

$\mu =x-z\sigma$ Solve to $\mu$

$=8-\left(2.33\right)\left(0.03\right)$ Substitute

$=7.9301$

As a result, the average volume of coffee to be delivered is 7.9301 ounces, meaning that the 8-ounce cup will typically overflow 1% of the time.

Juan Spiller

Expert

2022-01-14Added 38 answers

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,$P\left(X<x\right)=0.99$

$P(X-\frac{\mu}{\sigma}<x-\frac{\mu}{\sigma})=0.99$ (by standardizing)

$P(Z<x-\frac{\mu}{\sigma})=0.99$

$\mathrm{\Phi}(x-\frac{\mu}{\sigma})=0.99$

Find the value 0.99 from the body of the standard normal table. The value is approximately 2.33.

Step 2

$\mathrm{\Phi}\left(2.33\right)=0.99$ .

Thus,

$\mathrm{\Phi}(x-\frac{\mu}{\sigma})=\mathrm{\Phi}\left(2.33\right)$

$\frac{x-\mu}{\sigma}=2.33$

$x=\mu +2.33\left(\sigma \right)$

Substitute$\mu =8\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}\sigma =0.03$ .

$x=8+2.33\left(0.03\right)$

$=8.0699$

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.

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,

Find the value 0.99 from the body of the standard normal table. The value is approximately 2.33.

Step 2

Thus,

Substitute

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.

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