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Chapter 6: Problem 98

A soft-drink vending machine is supposed to pour 8 ounces of the drink into a paper cup. However, the actual amount poured into a cup varies. The amount poured into a cup follows a normal distribution with a mean that can be set to any desired amount by adjusting the machine. The standard deviation of the amount poured is always \(.07\) ounce regardless of the mean amount. If the owner of the machine wants to be \(99 \%\) sure that the amount in each cup is 8 ounces or more, to what level should she set the mean?

Short Answer

Expert verified
The owner should set the mean at approximately 8.16 ounces to be 99% sure that the machine pours at least 8 ounces.

Step by step solution

01

Find the z-value

Drawing upon the knowledge of one-tailed test in normal distribution, we can focus on the 99% probability in the positive direction. This translates to finding a z-value corresponding to a cumulative probability of 0.01 (as it's the remaining percentage from 100%). Using a standard normal distribution table or a z-score calculator, we find the z-value to be approximately -2.33.
02

Apply the z-score formula

We then apply the z-score formula for normal distribution: \(Z = \( (X - \mu) / \蟽 )\), where Z is the z-score, X is the value from the data set, \mu is the mean and \蟽 is the standard deviation. We substitute the known values: -2.33 = (8 - \mu) / 0.07. Solve this equation for the unknown mean.
03

Solve for the mean

Rearranging and solving this equation for \mu, we get \mu = 8 - (-2.33*0.07) = 8.16 ounces. Therefore, the mean should be set at 8.16 ounces to assure that 99% of the time, the machine pours a drink of 8oz or higher.

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Key Concepts

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

Standard Deviation
Standard deviation is a measure used to quantify the amount of variation or spread in a set of data values. In the context of the soft-drink vending machine, the standard deviation of 0.07 ounces shows how much the amount dispensed will vary from the mean amount.

A smaller standard deviation means that the values tend to be close to the mean, indicating more consistency.
  • If our standard deviation was larger, it would indicate more variability in the amount of drink poured.
  • With a standard deviation of 0.07 ounces, we can expect most pours to be within 0.07 ounces of the mean setting.
Understanding the standard deviation helps the owner fine-tune the machine to ensure that the amount poured stays close to the desired amount.
Z-Score
The z-score, also known as the standard score, is a way of describing a data point's relationship to the mean of a set of data. It shows how many standard deviations a given value (or score) is from the mean.

In our example, the vending machine's value of 8 ounces must be assessed using the standard deviation to find the z-score. This z-score helps identify what mean setting will ensure that the required 8 ounces are poured 99% of the time.
  • The formula to calculate the z-score is: \(Z = \frac{X - \mu}{\sigma}\)
  • Here, \(X\) is the data point (8 ounces), \(\mu\) is the mean (to be determined), and \(\sigma\) is the standard deviation (0.07 ounces).
  • Using these parameters lets us calculate the z-score necessary to solve for the desired mean in the machine adjustment.
Understanding the z-score helps leverage the power of standard deviation and normal distribution to make precise adjustments.
Probability
Probability is the measure of the likelihood that a particular event will occur. In terms of the vending machine, the probability of pouring exactly 8 ounces or more is what we're interested in.

Probability is central to deciding how the machine's mean should be set to meet the 99% requirement.
  • A one-tailed probability check at the 99% confidence level helps ensure the machine鈥檚 performance.
  • In this scenario, the key is to ensure that the lower tail of the normal distribution represents only 1% of occurrences that are less than 8 ounces.
  • This ensures that 99% of the time, the machine will pour at least 8 ounces or more.
Calculating the probability helps strategically set the mean for optimal performance.
One-Tailed Test
A one-tailed test in statistics is used when we want to determine if something is greater or less than a specific value. In the case of the vending machine, we are only interested in whether the amount poured is 8 ounces or more.

Using a one-tailed test helps focus our analysis on the minimum requirement.
  • We evaluate the probability of pouring 8 ounces or more.
  • This involves setting the mean in a way that the machine will fall into the acceptable range 99% of the time or more.
  • By calculating for one side ("the tail") of the distribution, it simplifies the decision.
  • This is why the calculation involved finding the z-score that corresponds to a 1% level of error (99% confidence).
Utilizing a one-tailed test streamlines our efforts in establishing a precise mean that meets operational demands.

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Most popular questions from this chapter

The management at Ohio National Bank does not want its customers to wait in line for service for too long. The manager of a branch of this bank estimated that the customers currently have to wait an average of 8 minutes for service. Assume that the waiting times for all customers at this branch have a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes. a. Find the probability that a randomly selected customer will have to wait for less than 3 minutes. b. What percentage of the customers have to wait for 10 to 13 minutes? c. What percentage of the customers have to wait for 6 to 12 minutes? d. Is it possible that a customer may have to wait longer than 16 minutes for service? Explain.

For a binomial probability distribution, \(n=25\) and \(p=.40\). a. Find the probability \(P(8 \leq x \leq 13)\) by using the table of binomial probabilities (Table I of Appendix C). b. Find the probability \(P(8 \leq x \leq 13)\) by using the normal distribution as an approximation to the binomial distribution. What is the difference between this approximation and the exact probability calculated in part a?

A charter bus company is advertising a singles outing on a bus that holds 60 passengers. The company has found that, on average, \(10 \%\) of ticket holders do not show up for such trips; hence, the company routinely overbooks such trips. Assume that passengers act independently of one another. a. If the company sells 65 tickets, what is the probability that the bus can hold all the ticket holders who actually show up? In other words, find the probability that 60 or fewer passengers show up. b. What is the largest number of tickets the company can sell and still be at least \(95 \%\) sure that the bus can hold all the ticket holders who actually show up?

Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is \(2.4\) minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a \(50 \%\) discount on the charges. The company wants to limit this discount to at most \(5 \%\) of the customers. What should the maximum guaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution.

Find the following areas under a normal distribution curve with \(\mu=20\) and \(\sigma=4\). a. Area between \(x=20\) and \(x=27\) b. Area from \(x=23\) to \(x=26\) c. Area between \(x=9.5\) and \(x=17\)
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