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Chapter 7: Problem 46

Accurate labeling of packaged meat is difficult because of weight decrease resulting from moisture loss (defined as a percentage of the package's original net weight). Suppose that the normal distribution with mean value \(4.0 \%\) and standard deviation \(1.0 \%\) is a reasonable model for the variable \(x=\) moisture loss for a package of chicken breasts. (This model is suggested in the paper "Drained Weight Labeling for Meat and Poultry: An Economic Analysis of a Regulatory Proposal," Journal of Consumer Affairs \([1980]: 307-325 .)\) a. What is the probability that \(x\) is between \(3.0 \%\) and \(5.0 \%\) ? b. What is the probability that \(x\) is at most \(4.0 \%\) ? c. What is the probability that \(x\) is at least \(7 \%\) ? d. Describe the largest \(10 \%\) of the moisture loss distribution.

Short Answer

Expert verified
a. The probability that \(x\) is between 3.0% and 5.0% is approximately 68.27%. b. The probability that \(x\) is at most 4.0% is approximately 50%. c. The probability that \(x\) is at least 7% is approximately 0.13%. d. The 10% of highest moisture losses are those greater than 5.28%.

Step by step solution

01

Identify Details

From the exercise, the following details are obtained: mean \(\mu = 4.0%\) and standard deviation \(\sigma = 1.0%\). These are related to a normal distribution.
02

Compute Probability Between 3.0% and 5.0%

To find the probability that \(x\) is between 3.0% and 5.0%, convert these values into z-scores using the formula: z = (x - \(\mu\))/\(\sigma\). For x = 3.0%, z = (3.0 - 4.0) / 1.0 = -1.0 and for x = 5.0%, z = (5.0 - 4.0) / 1.0 = 1.0. Using a z-table, find the probability between these two z-scores, which is approximately 0.6827 or 68.27%.
03

Compute Probability at Most 4.0%

To find the probability that \(x\) is at most 4.0%, z-score is 0 (since z = (4.0 - 4.0) / 1.0 = 0). Using the z-table, the probability that z is less than or equal to 0 is 0.5000 or 50%.
04

Compute Probability At Least 7%

To find the probability that \(x\) is at least 7%, Find z-score for x = 7.0, z = (7.0 - 4.0) / 1.0 = 3.0. The cumulative probability from z-table for 3.0 is 0.9987 or 99.87%. However, we need the probability that x is at least 7.0, which is 1 - 0.9987 = 0.0013 or 0.13%.
05

Identify largest 10% of Moisture Loss Distribution

To find the largest 10%, you look for the z-score associated with the 90 percentile (since the largest 10% equals 100% - 90%). The z-score from the z-table for 90 percentile is approximately 1.28. Convert this z-score to x using the formula x = z*sigma + mean. Therefore, x = 1.28 * 1.0 + 4.0 = 5.28. So, the largest 10% of moisture loss is above 5.28%.

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Key Concepts

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

Moisture Loss
Moisture loss in packaged meat, such as chicken breasts, can significantly affect its labeled weight. This concept is crucial in understanding how products like these may lose weight due to evaporation or other factors during storage. In the given exercise, moisture loss is expressed as a percentage of the original weight of the product.

This percentage helps identify the weight decrease over time and aids in calculating the correct label weights for consumers. When moisture loss follows a pattern, like a normal distribution in this case, it becomes easier to estimate expected losses. Therefore, understanding the nature of moisture loss allows businesses to analyze product conditions better and efficiently manage quality control.
Z-scores
Z-scores are a standard way of expressing a value's relation to the mean of a group of values. If the value is the same as the mean, the z-score is 0. If it's above the mean, the z-score is positive; if below, it's negative. This scaling of data helps in determining how unusual a particular observation is within a distribution.

In the context of moisture loss, calculating z-scores allows us to convert a specific moisture percentage into this standardized form, where \( z = \frac{(x - \mu)}{\sigma} \). Here, \(x\) is the moisture percentage, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
  • A z-score of -1, for example, means the observation is 1 standard deviation below the mean.
  • Z-score of 1 implies it is 1 standard deviation above the mean.

Understanding z-scores helps in assessing the position of specific data points in relation to the entire data set.
Probability Calculations
Probability calculations help assess the chance that a certain event will occur. In this exercise, we employ these calculations to find probabilities related to moisture loss percentages and their z-scores.

Here's a brief insight:
  • To find the probability that moisture loss is between 3.0% and 5.0%, we find the z-scores for these two values. The calculations show a 68.27% probability between these values.
  • Probability of moisture loss at most 4.0% calculates to exactly 50%, because 4.0% corresponds to the mean of the distribution.
  • Probability of at least 7% loss translates to a very low 0.13%, as such extreme losses are far from the mean.

Using a z-table or statistical software is crucial for accurately determining these probabilities.
Standard Deviation
Standard deviation is a measure of how spread out numbers are in a dataset. It quantifies the amount of variation or dispersion in a set of values.

In the context of moisture loss, the standard deviation helps us understand the variance around the average expectation of moisture loss, which is 4.0% in this case. With a standard deviation of 1.0%, we know without any complicated graphing that most moisture loss observations fall within a 1% range either side of the mean.

The concept is vital because it offers insight into the consistency or variability of moisture loss.
  • A small standard deviation indicates that the data points tend to be close to the mean.
  • A large standard deviation suggests a wider spread and more unpredictable moisture loss results.

Understanding standard deviation equips us with the ability to compare different distributions easily and grasp the level of risk or certainty present in predictions.

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Most popular questions from this chapter

Consider babies bom in the "normal" range of 37-43 weeks gestational age. Extensive data support the assumption that for such babies born in the United States, birth weight is normally distributed with mean \(3432 \mathrm{~g}\) and standard deviation \(482 \mathrm{~g}\) ("Are Babies Normal," The \(-302\) ). (The investigad data from a particular s intervals, the hisbut after further investi1 that this was due ht in grams and others measuring to the nearest ounce and then converting to grams. A modified choice of class intervals that allowed for this measurement difference gave a histogram that was well described by a normal distribution.) a. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(4000 \mathrm{~g}\) ? is between 3000 and \(4000 \mathrm{~g}\) ? b. What is the probability that the birth weight of a randomly selected baby of this type is either less than \(2000 \mathrm{~g}\) or greater than \(5000 \mathrm{~g}\) ? c. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(7 \mathrm{lb}\) ? (Hint: \(1 \mathrm{lb}=453.59 \mathrm{~g} .\) ) d. How would you characterize the most extreme \(0.1 \%\) of all birth weights?

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean \(45 \mathrm{~min}\) and standard deviation \(5 \mathrm{~min}\). a. If \(50 \mathrm{~min}\) is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if we wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all students to complete the exam?

Based on past history, a fire station reports that \(25 \%\) of the calls to the station are false alarms, \(60 \%\) are for small fires that can be handled by station personnel without outside assistance, and \(15 \%\) are for major fires that require outside help. a. Construct a relative frequency bar chart that represents the distribution of the variable \(x=\) type of call, where type of call has three categories: false alarm, small fire, and major fire. What is the underlying population for this variable? b. Based on the given information, we can write \(P(x=\) false alarm) \(=.25\). Use the other two relative frequencies shown in the bar chart from Part (a) to write two other probability statements.

Because \(P(z<0.44)=.67,67 \%\) of all \(z\) values are less than \(0.44\), and \(0.44\) is the 67 th percentile of the standard normal distribution. Determine the value of each of the following percentiles for the standard normal distribution (Hint: If the cumulative area that you must look for does not appear in the \(z\) table, use the closest entry): a. The 91st percentile (Hint: Look for area \(.9100 .\) ) b. The 77 th percentile c. The 50 th percentile d. The 9 th percentile e. What is the relationship between the 70 th \(z\) percentile and the \(30 \mathrm{th} z\) percentile?

Determine the following standard normal ( \(z\) ) curve areas: a. The area under the \(z\) curve to the left of \(1.75\) b. The area under the \(z\) curve to the left of \(-0.68\) c. The area under the \(z\) curve to the right of \(1.20\) d. The area under the \(z\) curve to the right of \(-2.82\) e. The area under the \(z\) curve between \(-2.22\) and \(0.53\) f. The area under the \(z\) curve between \(-1\) and 1 g. The area under the \(z\) curve between \(-4\) and 4
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