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A soft drink vending machine can be regulated so that it dispenses an average of \(\mu\) oz of soft drink per cup. a. If the ounces dispensed per cup are normally distributed with a standard deviation of 0.2 oz, find the setting for \(\mu\) that will allow a 6 -oz glass to hold (without overflowing) the amount dispensed \(99 \%\) of the time. b. Use a computer or calculator to simulate drawing a sample of 40 cups of soft drink from the machine (set using your answer to part a).

Short Answer

Expert verified
The mean setting for the machine that will allow a 6-oz glass to hold the drink without overflowing 99% of the time should be approximately 5.534 oz.

Step by step solution

01

Understand the Z-Score Concept

The Z-score follows a standard normal distribution with a mean of 0 and a standard deviation of 1. It represents the number of standard deviations a particular data is from its mean. We can formulate it as \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. In our case, a 6 -oz glass should hold the drink 99% of the time. So we refer to the Z-table to find the Z-score that corresponds to the 99th percentile, which is approximately 2.33.
02

Calculate the Mean Value (\(\mu\))

Let's input our known data into the Z-score equation. The data point (\(X\)) is 6 oz (glass capacity), the Z-score is 2.33 (99 percentile), and the standard deviation (\(\sigma\)) is 0.2 oz which gives \(2.33 = \frac{6 - \mu}{0.2}\) Now, all we have to do is solve this equation for \(\mu\), which gives \(\mu = 6 - 2.33 * 0.2 = 5.534\).
03

Simulate Drawing a Sample of 40 Cups

For part b, we would need to use a calculator or a computer simulation to draw a sample of 40 cups of soft drink from the machine. This part is done outside of this context and is dependent on the software or device used. However, in such a simulation, the mean would be 5.534 and the standard deviation 0.2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-Score
A Z-score is a statistical measurement that describes a value's position relative to the mean of a group of values. It is also an essential concept in the realm of normal distribution. The Z-score tells us how many standard deviations away a specific measurement is from the mean. This can be useful for determining differences between data sets and identifying outliers.

For example, if a data point has a Z-score of 2.33, it means the data point is 2.33 standard deviations above the mean. In this scenario, given the normal distribution of our data—ounces dispensed from the vending machine—a Z-score related to the 99th percentile would show us that only 1% of the data would fall above this point.

By using the Z-score formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
Where,
  • \(X\) is the data value (such as the 6 oz cup capacity),
  • \(\mu\) is the mean,
  • \(\sigma\) is the standard deviation.
We can solve for unknown quantities such as the mean \(\mu\), when values for \(X\), \(Z\), and \(\sigma\) are known.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In simple terms, it tells us how spread out the numbers are in a given data set. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

In the context of the vending machine exercise, the standard deviation is given as 0.2 oz, meaning this is the typical amount of deviation from the average amount a cup may be dispensed.

Understanding standard deviation helps us know how consistent the machine is at dispensing soft drinks. A smaller standard deviation would mean that most glasses receive an amount of drink close to the average, making it more predictable. This predictability is crucial when trying to ensure a 6 oz glass can be filled (without overflow) 99% of the time.
Percentiles
Percentiles offer a way to understand and interpret data by showing the relative standing of a value within a data set. They partition the data into 100 equal parts, essentially telling us what portion of the data falls below a given value.

For example, the 99th percentile means that 99% of the data points lie below this value, and only 1% lie above. In the exercise, ensuring that a 6 oz glass holds the soft drink 99% of the time indicates that the 99th percentile of ounces dispensed must not exceed 6 oz. This requires calculating the appropriate mean using other known values such as the Z-score and standard deviation.

This is why in the Z-score calculation we refer to the Z-table to translate the 99th percentile into its respective Z-score, which allows us to redefine the mean to balance out this high percentage of confidence for the cup volume being enough to prevent overflow in most cases.

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Most popular questions from this chapter

According to Federal Highway Administration 2007 statistics, the percent of licensed female drivers has just surpassed the percent of licensed male drivers. Of the drivers in the United States, \(50.2 \%\) are females. If a random sample of 50 drivers is to be selected for a survey, a. what is the probability that no more than half (25) of the drivers are female? b. what is the probability that at least three-fourths (38) of the drivers are female?

In order to see what happens when the normal approximation is improperly used, consider the binomial distribution with \(n=15\) and \(p=0.05 .\) since \(n p=0.75\) the rule of thumb \((n p>5 \text { and } n q>5)\) is not satisfied. Using the binomial tables, find the probability of one or fewer successes and compare this with the normal approximation.

The middle \(60 \%\) of a normally distributed population lies between what two standard scores?

The extraction force required to remove a cork from a bottle of wine has a normal distribution with a mean of 310 Newtons and a standard deviation of 36 Newtons. a. The specs for this variable, given in Applied Example \(6.13,\) were "\(300\mathrm{N}+100 \mathrm{N} /-150 \mathrm{N}\)" Express these specs as an interval. b. What percentage of the corks is expected to fall within the specs? c. What percentage of the tested corks will have an extraction force of more than 250 Newtons? d. What percentage of the tested corks will have an extraction force within 50 Newtons of \(310 ?\)

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