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At Jen and Perry Ice Cream Company, a machine fills 1 -pound cartons of Top Flavor ice cream. The machine can be set to dispense, on average, any amount of ice cream into these cartons. However, the machine does not put exactly the same amount of ice cream into each carton; it varies from carton to carton. It is known that the amount of ice cream put into each such carton has a normal distribution with a standard deviation of \(.18\) ounce. The quality control inspector wants to set the machine such that at least \(90 \%\) of the cartons have more than 16 ounces of ice cream. What should be the mean amount of ice cream put into these cartons by this machine?

Short Answer

Expert verified
The machine should be set to dispense, on average, approximately 16.2304 ounces of ice cream.

Step by step solution

01

Determine the z-score

First, find the z-score for 0.10 (10%) in a standard normal distribution table, as we want to find out the amount that will leave 90% to the right (or above). The z-score is approximately -1.28.
02

Calculate the mean

The z-score formula is \(Z = (x - μ)/σ\), where Z is the z-score, x is the value in the data set, μ is the mean and σ is the standard deviation. We can rearrange the formula to solve for μ: \(μ = x - σ*Z\). x is the desired weight (16 ounces), σ is 0.18, and Z is -1.28: Plug in the values to get: \(μ = 16 - (-1.28 * 0.18)\).
03

Solve for the mean

Calculating this yields: \(μ = 16 - (-0.2304) = 16.2304\). So, to ensure that at least 90% of cartons carry more than 16 ounces of ice cream, the machine should be set to dispense, on average, 16.2304 ounces of ice cream into the cartons.

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Key Concepts

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

Standard Deviation
The standard deviation is a measure that indicates how much individual data points deviate from the mean of a data set. In the context of the ice cream filling machine at Jen and Perry Ice Cream Company, this concept is crucial to understanding the variations in the amount of ice cream dispensed into each carton.

Standard deviation, represented as \( \sigma \), helps to quantify the spread or dispersion of a set of data points. In this particular problem, it is given as 0.18 ounces. This means that the difference in the amount of ice cream in each carton generally deviates by 0.18 ounces from the average (or mean).

A small standard deviation signifies that the data points are concentrated close to the mean, while a large standard deviation indicates more spread out data. In quality control processes such as this, understanding standard deviation allows you to predict how much variation from the average is typical and helps in setting appropriate machine parameters to reduce unwanted variability in the products.
Z-Score
A z-score is a statistical measure that describes a data point's relation to the mean of a group of points. By standardizing data points, the z-score reveals how many standard deviations a particular value is from the mean.

In this problem, to ensure at least 90% of cartons contain more than 16 ounces of ice cream, the appropriate z-score must be determined. This was found to be approximately -1.28, which corresponds to the lower tail of the normal distribution since we are interested in the bottom 10% threshold, leaving 90% to the right of this score.
  • Z-scores are particularly useful in quality control as they allow for comparison between different data sets.
  • The formula for calculating a z-score is \( Z = \frac{x - \mu}{\sigma} \), where \( x \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
These calculations enable the setting of the machinery to new means, ensuring optimal product filling and compliance with quality standards.
Quality Control
Quality control plays a pivotal role in ensuring that products meet predefined quality requirements. At Jen and Perry Ice Cream Company, the aim of quality control is to ensure that the majority of their 1-pound ice cream cartons exceed 16 ounces.

For this task, quality control utilizes statistical tools to set machine parameters such that product variability remains within acceptable limits.
  • Ensuring at least 90% of ice cream cartons weigh more than the target 16 ounces highlights their commitment to maintaining a certain product standard.
  • Use of normal distribution and z-scores helps in determining appropriate settings.
This allows the company to minimize waste, increase efficiency, and maintain customer satisfaction. In addition, setting precise mean values for machine dispensing is a significant component of their quality assurance processes.
Mean Calculation
Mean calculation is integral when setting machinery to disperse specific product quantities consistently. In the context of this problem, the mean \( \mu \) represents the average amount of ice cream the machine should dispense to ensure quality control standards are met.

The mean is calculated using the rearranged z-score formula: \( \mu = x - \sigma \times Z \). Here, \( x \) is the desired lower bound of 16 ounces, \( \sigma \) is the standard deviation of 0.18 ounces, and \( Z \) is the z-score of -1.28.
  • This calculation yields \( \mu = 16 - (-1.28 \times 0.18) = 16.2304 \) ounces.
  • The mean ensures most cartons will have more than the minimum desired weight.
Choosing this strategic mean ensures that variations in dispensing do not result in too many underfilled cartons, aligning production with quality and regulatory goals.

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Most popular questions from this chapter

Let \(x\) be a continuous random variable that is normally distributed with a mean of 25 and a standard deviation of 6 . Find the probability that \(x\) assumes a value a. between 29 and 36 b. between 22 and 35

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Determine the value of \(z\) so that the area under the standard normal curve a. in the right tail is \(.0500\) b. in the left tail is \(.0250\) c. in the left tail is \(.0100\) d. in the right tail is \(.0050\)

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.

For the standard normal distribution, what does \(z\) represent?

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