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The Jen and Perry Ice Cream company makes a gourmet ice cream. Although the law allows ice cream to contain up to \(50 \%\) air, this product is designed to contain only \(20 \%\) air. Because of variability inherent in the manufacturing process, management is satisfied if each pint contains between \(18 \%\) and \(22 \%\) air. Currently two of Jen and Perry's plants are making gourmet ice cream. At Plant \(\mathrm{A}\), the mean amount of air per pint is \(20 \%\) with a standard deviation of \(2 \%\). At Plant \(\mathrm{B}\), the mean amount of air per pint is \(19 \%\) with a standard deviation of \(1 \%\). Assuming the amount of air is normally distributed at both plants, which plant is producing the greater proportion of pints that contain between \(18 \%\) and \(22 \%\) air?

Short Answer

Expert verified
It would be Plant A that is producing a greater proportion of pints that contain between 18% and 22% air because the acceptable range at Plant A lies within 1 standard deviation from the mean, whereas at Plant B it extends up to 3 standard deviations.

Step by step solution

01

Calculate Z-Scores for Plant A

We start by calculating the Z-scores for the 18% and 22% air points in Plant A. The Z-score is calculated using the formula \( Z = (X - \mu) / \sigma \) where X is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. So the Z-score when X is 18% and 22% would be \( Z1_A = (18 - 20) / 2 = -1 \) and \( Z2_A = (22 - 20) / 2 = 1 \). This means that the amounts of air between 18% and 22% lie within 1 standard deviation from the mean in both directions.
02

Calculate Z-Scores for Plant B

Next, we calculate the Z-scores for the 18% and 22% air points in Plant B using the same formula. So the Z-score when X is 18% and 22% would be \( Z1_B = (18 - 19) / 1 = -1 \) and \( Z2_B = (22 - 19) / 1 = 3 \). This means that the amounts of air between 18% and 22% lie within 3 standard deviations from the mean in one direction and 1 standard deviation in the other.
03

Compare the proportions

We know from the properties of the normal curve that about 68% of values will lie within 1 standard deviation of the mean, about 95% will lie within 2 standard deviations, and about 99.7% will lie within 3 standard deviations. So, Plant A, where the acceptable range lies within 1 standard deviation of the mean, should be producing the greater proportion of pints within the acceptable range.

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Key Concepts

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

Understanding Z-scores
A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It tells us how many standard deviations a data point is from the mean.
  • If the Z-score is 0, it indicates that the data point is exactly at the mean value.
  • A positive Z-score signifies that the data point is above the mean.
  • A negative Z-score indicates that the data point is below the mean.
In our exercise, calculating the Z-scores helps us understand how far the air content percentages at each plant are from the mean value. By knowing the Z-scores for specific air percentages, such as 18% and 22%, we can determine the proportion of pints that lie within a specific range. A key step in this process is using the formula: \[ Z = \frac{(X - \mu)}{\sigma} \] where:
  • \(X\) is the value of interest (e.g., 18% or 22% air).
  • \(\mu\) is the mean air content of the plant.
  • \(\sigma\) is the standard deviation of the air content.
By applying this formula, we can better understand the variability in the air content and how it affects the overall quality of the ice cream produced at each plant.
The role of standard deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In simpler terms, it shows how much individual values in a data set differ from the mean.
  • If the standard deviation is small, it means the values are close to the mean.
  • A large standard deviation indicates that the values are spread out over a wider range.
In the context of the Jen and Perry Ice Cream Company, the standard deviation plays a crucial role in determining the consistency of the air content in pints produced at different plants.
For instance, at Plant A, the standard deviation is 2%. This means that most of the pints will have air content relatively close to the 20% mean. Meanwhile, at Plant B with a 1% standard deviation, air content remains even closer to the 19% mean.
Understanding standard deviation helps us predict the proportion of pints that seat within an acceptable air content range. It is a valuable tool in quality control and ensuring each pint meets the desired specifications.
Mean and its significance
The mean, often referred to as the average, is a statistical measure that represents the central point of a data set. It is calculated by summing up all the values and then dividing by the total number of values.
The mean is crucial in the analysis of a normally distributed data set since it serves as a point of reference. For the Jen and Perry Ice Cream Company, the mean air content at each plant gives the expected average air per pint.
  • For Plant A, the mean is 20%, indicating that on average, pints have 20% air.
  • For Plant B, the mean is 19%, suggesting that on average, pints contain 19% air.
When combined with standard deviation and Z-scores, the mean aids in evaluating the production process's success in meeting the desired air content range. It is essential to comprehend that while both plants might strive for a similar air content percentage, their means denote slightly different production targets around which variability is assessed. Understanding the mean and using it effectively enables manufacturers to ensure their product consistently meets quality standards.

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Most popular questions from this chapter

During the 2009 edition of the reality show Britain's Got Talent, runner-up and Internet singing sensation Susan Boyle obtained \(20.2 \%\) of the first- place votes. Suppose that this percentage would hold true for all potential voters (note: the population of interest would be all viewers of ITV, which carries the show). Find the probability that, in a random sample of 250 potential voters, the number who would vote for Susan Boyle is a. exactly 57 b. 35 to 41 c. at least 60

According to the 2007 American Time Use Survey by the Bureau of Labor Statistics, employed adults living in households with no children younger than 18 years engaged in leisure activities for \(4.4\) hours a day on average (Source: http://www.bls.gov/news.release/atus,nr0.htm). Assume that currently such times are (approximately) normally distributed with a mean of \(4.4\) hours per day and a standard deviation of \(1.08\) hours per day. Find the probability that the amount of time spent on leisure activities per day for a randomly chosen individual from the population of interest (employed adults living in households with no children younger than 18 years) is a. between \(3.0\) and \(5.0\) hours per day b. less than \(2.0\) hours per day

A company that has a large number of supermarket grocery stores claims that customers who pay by personal check spend an average of \(\$ 87\) on groceries at these stores with a standard deviation of \(\$ 22\). Assume that the expenses incurred on groceries by all such customers at these stores are normally distributed. a. Find the probability that a randomly selected customer who pays by check spends more than \(\$ 114\) on groceries. b. What percentage of customers paying by check spend between \(\$ 40\) and \(\$ 60\) on groceries? c. What percentage of customers paying by check spend between \(\$ 70\) and \(\$ 105\) ? d. Is it possible for a customer paying by check to spend more than \(\$ 185\) ? Explain.

Determine the following probabilities for the standard normal distribution. a. \(P(-1.83 \leq z \leq 2.57)\) b. \(P(0 \leq z \leq 2.02)\) c. \(P(-1.99 \leq z \leq 0)\) d. \(P(z \geq 1.48)\)

In a 2007 survey of consumer spending habits, U.S. residents aged 45 to 54 years spent an average of \(9.32 \%\) of their after-tax income on food (Source: ftp://ftp.bls.gov/pub/special.requests/ce/standard/2007/ age.txt). Suppose that the current percentage of after-tax income spent on food by all U.S. residents aged 45 to 54 years follows a normal distribution with a mean of \(9.32 \%\) and a standard deviation of \(1.38 \% .\) Find the proportion of such persons whose percentage of after-tax income spent on food is a. greater than \(11.1 \%\) b. between \(6.0 \%\) and \(7.2 \%\)

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