Question

The meat department at a local supermarket specifically prepares its " 1 -pound" packages of ground beef so that there will be a variety of weights, some slightly more and some slightly less than 1 pound. Suppose that the weights of these "1pound" packages are normally distributed with a mean of 1.00 pound and a standard deviation of .15 pound. a. What proportion of the packages will weigh more than 1 pound? b. What proportion of the packages will weigh between .95 and 1.05 pounds? c. What is the probability that a randomly selected package of ground beef will weigh less than .80 pound? d. Would it be unusual to find a package of ground beef that weighs 1.45 pounds? How would you explain such a large package?

Short answer

Answer: 50% of the packages weigh more than 1 pound.

Step-by-step solution

Calculate the z-score for 1 pound

The z-score is calculated as: z = (X - μ) / σ where X is the weight (1 pound), μ is the mean (1.00 pound), and σ is the standard deviation (0.15 pound). z = (1 - 1) / 0.15 = 0

Find the probability from the z-table

Since z = 0, we get P(z > 0) = 0.5. This means that 50% of the packages will weigh more than 1 pound. #b. Find the proportion of packages weighing between 0.95 and 1.05 pounds#

Calculate the z-scores for 0.95 and 1.05 pounds

For 0.95 pound: z = (0.95 - 1) / 0.15 = -0.33 (approximately) For 1.05 pound: z = (1.05 - 1) / 0.15 = 0.33 (approximately)

Find the probabilities from the z-table

P(-0.33 < z < 0.33) = P(z < 0.33) - P(z < -0.33) = 0.6293 - 0.3707 = 0.2586 This means that approximately 25.86% of the packages will weigh between 0.95 and 1.05 pounds. #c. Find the probability that a package weighs less than 0.80 pound#

Calculate the z-score for 0.80 pound

z = (0.80 - 1) / 0.15 = -1.33 (approximately)

Find the probability from the z-table

P(z < -1.33) = 0.0918. This means that there is a 9.18% chance that a randomly selected package weighs less than 0.80 pound. #d. Determine if a package weighing 1.45 pounds is unusual#

Calculate the z-score for 1.45 pounds

z = (1.45 - 1) / 0.15 = 3.00 (approximately)

Determine if the z-score is unusual

Since the z-score is 3.00, this is more than 2 standard deviations away from the mean. It is generally considered unusual if the z-score is larger than 2 or smaller than -2. Therefore, it would be unusual to find a package of ground beef that weighs 1.45 pounds. An explanation for such a large package could be a mistake during packaging, or an unusually large piece of meat accidentally getting included in the pack.

Key concepts

Understanding Z-Score Calculations

A z-score represents how many standard deviations a data point is from the mean. In the context of our ground beef example, z-scores help us understand where a package's weight falls relative to the average (mean) package weight.

To calculate the z-score, you subtract the mean from the data point (in this case, the weight of the beef package) and then divide by the standard deviation. It's expressed as:
\[ z = \frac{{X - \mu}}{{\sigma}} \]
where \( X \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. This formula converts the weights into a standardized form, and we can then use the z-score to compare different weights on the same scale.

For example, a z-score of 0 indicates that the package weight is exactly at the mean. A positive z-score means the weight is above the mean, while a negative z-score indicates it's below the mean. Using z-scores, we can then consult z-tables, which give us the probability of a value falling within a certain range, allowing us to answer questions about the proportions of packages with specific weights.

Determining Probability with Normal Distribution

Probability in the context of normal distribution is all about finding the likelihood of an event occurring within a specific range. The normal distribution is a bell-shaped curve where most of the data points lie around the mean, and the probability of occurrences decreases as you move away from it.

Using our previous z-score calculations, we can refer to a z-table to find out the probabilities corresponding to those scores. For instance, a z-score of 0 (the mean) has a probability of 0.5, meaning there's a 50% chance that a beef package will be at or above the mean weight. The higher (or lower) the z-score, the lower the probability, as more extreme weights are less common.

By determining the z-scores for specific weights, such as 0.95 and 1.05 pounds, and then subtracting the lower z-score's probability from the higher one's, we get the probability for a package falling within that range. This allows supermarkets to ensure their weight variability meets customer expectations while maintaining quality control.

Importance of Standard Deviation in Data Analysis

Standard deviation is a measure of the spread or dispersion of 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. In the ground beef example, the standard deviation is 0.15 pounds, signifying a small variation from the mean of 1 pound.

When we say a z-score of more than 2 (or less than -2) is unusual, we're acknowledging that most data in a normal distribution lie within 2 standard deviations of the mean. This encompasses about 95% of the data. Therefore, anything beyond this range is considered uncommon and prompts an examination of potential causes, such as inconsistencies in meat packaging processes or errors in measurement. Understanding standard deviation helps us make sense of the variability in data, ensuring quality control, and setting standards in various industries and studies.