Problem 75 - Calculus and its applications

Question

Business: does thy cup overflow? Suppose the mean amount of cappuccino, \(\mu\), dispensed by a vending machine can be set. If a cup holds \(8.5 \mathrm{oz}\) and the amount dispensed is normally distributed with \(\sigma=0.3\) oz, what should \(\mu\) be set at to ensure that only 1 cup in 100 will overflow?

Step-by-Step Solution

Answer

Set \(\mu\) to 7.801 oz.

Step 1: Understanding the Problem

We need to determine the mean amount of cappuccino, \(\mu\), to ensure just 1 out of 100 cups overflows. Given a cup holds 8.5 oz and the standard deviation \(\sigma\) is 0.3 oz, we will use the properties of the normal distribution to find \(\mu\).

Step 2: Using the Z-Score

Since only 1 out of 100 cups should overflow, this means we are looking at the 99th percentile of a normal distribution. Using a Z-table, find the Z-score for the 99th percentile, which is approximately 2.33.

Step 3: Setting Up the Equation

The formula for the Z-score is: \[ Z = \frac{X - \mu}{\sigma} \]Here, \(X = 8.5\) is the capacity of the cup, \(\mu\) is the mean we need to find, and \(\sigma = 0.3\). Substitute the values and solve for \(\mu\).

Step 4: Solving for \(\mu\)

Substitute the known values into the Z-score equation:\[ 2.33 = \frac{8.5 - \mu}{0.3} \]Multiply both sides by 0.3:\[ 0.699 = 8.5 - \mu \]Re-arrange to find \(\mu\):\[ \mu = 8.5 - 0.699 = 7.801 \]

Step 5: Conclusion

The mean amount of cappuccino, \(\mu\), should be set to approximately 7.801 oz to ensure that only 1 out of 100 cups overflows.

Key Concepts

Mean Calculation

Calculating the mean, often denoted as \(\mu\), is the process of finding the average value in a set of numbers. In the context of a normal distribution like our cappuccino example, the mean represents the center of the data distribution. Here, it's crucial because we need to set the mean to ensure a specific outcome - in this case, controlling the amount of cappuccino dispensed by the vending machine so that only 1 in 100 cups overflow.

To calculate the mean when dealing with such a situation:

• Identify the target criteria (here, only 1 out of 100 cups should overflow).
• Use the properties of the normal distribution, such as the standard deviation (\(\sigma\)). In our example, \(\sigma = 0.3\) oz.
• Utilize the Z-score to set the mean formula accordingly.

Once you know these parameters, you can rearrange the mathematical equation to solve for \(\mu\). Here, this process ensures our machine dispenses an optimized average amount, preventing unnecessary spills.

Z-Score

The Z-score is a statistical measure that tells us how many standard deviations an element is from the mean. It's a useful tool in assessing probabilities and is key to understanding data within a normal distribution.

In our task of setting the cappuccino vending machine's mean dispense rate, the Z-score helped us determine how much above or below the average our cap needs to be so that only the 1% of instances see overflow:

• The Z-score was calculated to be approximately 2.33, corresponding to the 99th percentile of a standard normal distribution.
• This means that the point where our cappuccino amount should meet or exceed the cup capacity of 8.5 oz is located 2.33 standard deviations from the desired mean.

Thus, by knowing the typical spread of our normal distribution (given by the standard deviation), we use the Z-score to backtrack and find the exact mean \(\mu\) that achieves our objective. This is especially critical in quality control and consumer satisfaction scenarios.

Percentiles

Percentiles rank data within a distribution and help in understanding where a particular score falls in comparison to others. For example, if a value lies in the 99th percentile, it means it's greater than 99% of the other data points.

In our cappuccino scenario, knowing that only 1 in 100 cups should overflow implies we are interested in the 99th percentile of the cappuccino amounts:

• Percentiles provide a way to set thresholds, such as the overflow limit, ensuring we control variability according to consumer expectations.
• By leveraging the relationship between percentiles and Z-scores, we applied this to a normal distribution to find our target mean \(\mu\).

This approach gives us a robust method to balance between overfill and average fill, optimizing for wastage or inconvenience.