Question

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 moisture loss for a package of chicken breasts is normally distributed with mean value \(4.0 \%\) and standard deviation \(1.0 \% .\) (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.) Let \(x\) denote the moisture loss for a randomly selected package. 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 \(7.0 \%\) ? d. Find a number \(z^{*}\) such that \(90 \%\) of all packages have moisture losses below \(z^{*} \%\). e. What is the probability that moisture loss differs from the mean value by at least \(1 \%\) ?

Short answer

a. The probability that \(x\) is between \(3.0 \%\) and \(5.0 \%\) is 0.6826. b. The probability that \(x\) is at most \(4.0 \%\) is 0.5. c. The probability that \(x\) is at least \(7.0 \%\) is 0.0013. d. \(z^{*}\) such that \(90 \%\) of all packages have moisture losses below \(z^{*} \%\) is \(5.28\%. e. The probability that moisture loss differs from the mean value by at least \(1 \%\) is 0.3174.

Step-by-step solution

Probability of x between 3.0% and 5.0%

First, convert the percentages to z scores using the formula \(z = (x - \mu) / \sigma\) where \(x\) is the given value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. The z scores are \(z_1 = (3 - 4)/1 = -1\) and \(z_2 = (5 - 4)/1 = 1\). Look up these values in the z-table. The cumulative probability of a z score of -1 is 0.1587 and the cumulative probability of a z score of 1 is 0.8413. Therefore, the probability that \(x\) is between 3.0% and 5.0% is \(P(3\% \le x \le 5\%) = P(-1 \le z \le 1) = 0.8413 - 0.1587 = 0.6826.

Probability of x at most 4.0%

Standardize the value 4.0% to a z score using the same formula. Here, \(z = (4 - 4)/1 = 0\). The cumulative probability associated with a z score of 0 is 0.5. Therefore, the probability \(x\) is at most 4.0% is \(P(x \le 4\%) = P(z \le 0) = 0.5\).

Probability of x at least 7.0%

Standardize the value 7.0% to a z score, giving \(z = (7 - 4)/1 = 3\). The cumulative probability associated with a z score of 3 is 0.9987. But because the value must be at least 7.0%, we're interested in the area to the right of this z score, so the probability is \(P(x \ge 7\%) = 1 - P(z \le 3) = 1 - 0.9987 = 0.0013\).

Find z* such that 90% of packages have moisture losses below z*

This is a reverse lookup. In this case, we have the cumulative probability and are finding the corresponding z score. From the z-table, the z score corresponding to a cumulative probability of 0.90 is approximately 1.28. Convert this to a percentage using the formula \(x = \mu + z*\sigma\), which gives \(x = 4 + 1.28*1 = 5.28\). Therefore, \(z* = 5.28\%\).

Probability that moisture loss differs from the mean value by at least 1%

This problem is about finding the probability that the moisture loss is either less than 3.0% (mean - 1%) or more than 5.0% (mean + 1%). Thus, the required probability is \(P(x \le 3\% \text{ or } x \ge 5\%) = 1 - P(3\% \le x \le 5\%) = 1 - 0.6826 = 0.3174\).

Key concepts

Probability Calculation

Probability calculation is fundamental in statistics and it tells us the likelihood of a certain event occurring based on known parameters. In a normal distribution, probabilities of outcomes are determined by the distribution's mean and standard deviation. The area under the distribution curve represents the total probability, which is always equal to 1.

To find the probability of an event within a specific range, like in our exercise where we're looking to find the probability of moisture loss between 3.0% and 5.0%, we calculate the area under the curve between those values. This is done using the standard normal distribution, commonly referred to as a Z-table, after converting our values to z-scores.

In practical terms, the calculation in our step-by-step solution shows that the probability of the moisture loss being within that range is about 68.26%, indicating it's quite a likely event given the parameters of the normal distribution.

Z-Score

Understanding the z-score is crucial when working with the normal distribution in statistics. It is a measure that describes the position of a raw score in terms of its distance from the mean, measured in standard deviations.

The formula for the z-score is given by: \( z = \frac{{x - \mu}}{{\sigma}} \) where \( x \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. A z-score of 0 indicates that the value is exactly at the mean, while positive or negative scores indicate that the value lies above or below the mean, respectively.

In our chicken breasts example, calculating a z-score allows us to compare moisture loss percentages against the standard normal distribution to find probabilities.

Cumulative Probability

The cumulative probability in a normal distribution represents the probability that a random variable X will take a value less than or equal to a certain x. It's reflected in the area under the distribution curve to the left of x.

Calculation of cumulative probabilities is often done with the aid of standard normal distribution tables (z-tables), or with the help of statistical software or calculators. For instance, when we looked for the probability that moisture loss will be at most 4.0%, or the probability that it is at least 7.0%, we used the concept of cumulative probability.

In the exercise, finding the cumulative probability for a z-score of 3 tells us nearly all packages (99.87%) have less moisture loss than 7%, which is helpful in quality control and setting standards for packaging.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. 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.

Within the context of the normal distribution, the standard deviation plays a key role as it shapes the spread of the distribution curve. For the packaged meat example, a standard deviation of 1.0% tells us that most of the observed moisture loss values are within 1% above or below the mean value of 4.0%. It's also important in calculating z-scores and interpreting these scores; for example, a z-score of 1 would mean the moisture loss is one standard deviation above the mean.