Problem 47 - Basic Statistics for Business and Economics

Question

The weights of canned hams processed at the Henline Ham Company follow the normal distribution, with a mean of 9.20 pounds and a standard deviation of 0.25 pounds. The label weight is given as 9.00 pounds. a. What proportion of the hams actually weigh less than the amount claimed on the label? b. The owner, Glen Henline, is considering two proposals to reduce the proportion of hams below label weight. He can increase the mean weight to 9.25 and leave the standard deviation the same, or he can leave the mean weight at 9.20 and reduce the standard deviation from 0.25 pounds to \(0.15 .\) Which change would you recommend?

Step-by-Step Solution

Answer

21.19% of the hams weigh less than 9.00 pounds; reducing standard deviation is better.

Step 1: Understanding the Normal Distribution

The problem involves a normal distribution with a given mean and standard deviation. In part (a), we determine the proportion of hams weighing less than 9.00 pounds.

Step 2: Calculating the Z-Score for Part (a)

The Z-score is calculated using the formula: \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the label weight (9.00 pounds), \( \mu \) is the mean weight (9.20 pounds), and \( \sigma \) is the standard deviation (0.25 pounds). Thus, \( Z = \frac{9.00 - 9.20}{0.25} = -0.80 \).

Step 3: Finding the Proportion Below 9.00 Pounds

The proportion of hams weighing less than 9.00 pounds can be found using the standard normal distribution table. A Z-score of -0.80 corresponds to a probability of approximately 0.2119, meaning 21.19% of hams weigh less than 9.00 pounds.

Step 4: Analyzing Proposal 1: Increasing Mean to 9.25 Pounds

If the mean is increased to 9.25 pounds and the standard deviation remains 0.25 pounds, calculate the new Z-score: \( Z = \frac{9.00 - 9.25}{0.25} = -1.00 \). The corresponding probability for \( Z = -1.00 \) is approximately 0.1587, implying 15.87% weigh less than 9.00 pounds.

Step 5: Analyzing Proposal 2: Reducing Standard Deviation to 0.15 Pounds

With mean weight 9.20 pounds and standard deviation reduced to 0.15, compute the Z-score: \( Z = \frac{9.00 - 9.20}{0.15} = -1.33 \). The probability for \( Z = -1.33 \) is about 0.0918, meaning 9.18% weigh less than 9.00 pounds.

Step 6: Recommendation Based on Analysis

Proposal 2, reducing the standard deviation to 0.15 pounds, results in a lower proportion (9.18%) of hams weighing less than 9.00 pounds compared to Proposal 1 (15.87%). Thus, reducing the standard deviation is more effective.

Key Concepts

Z-Score Calculation

Determining how many standard deviations away a value is from the mean is key in normal distribution analysis. A Z-score calculation allows us to place any data point within the context of its distribution. Here's how you can understand it better:
\[ Z = \frac{X - \mu}{\sigma} \]
In this formula:

• \(X\) is the value you are examining. In this exercise, it is the label weight of the ham, 9.00 pounds.
• \(\mu\) is the mean of the distribution, here given as 9.20 pounds.
• \(\sigma\) is the standard deviation, which measures the spread of the data, given as 0.25 pounds in the original scenario.

Applying these values gives a Z-score of -0.80. This indicates that 9.00 pounds is 0.80 standard deviations below the mean weight of the hams. Calculating the Z-score is an essential step to use the standard normal distribution table, which in turn is used to find the corresponding probability of a ham weighing less than the given weight.

Mean and Standard Deviation

The mean and standard deviation are foundational in understanding any distribution of data, especially in normal distribution.
The mean, \(\mu\), serves as the center point. It is the average of all data points. In our normal distribution, it's where the peak of the curve is. Here, the mean weight of the canned hams is 9.20 pounds.
The standard deviation, \(\sigma\), tells us how much the data points tend to vary from the mean. It determines the spread of the distribution. A smaller standard deviation means the data points are closer to the mean, while a larger one indicates they are more spread out. Initially, the standard deviation is 0.25 pounds. This spread influences the probability outcomes significantly, especially when determining how many hams weigh below the label weight (9.00 pounds).
In probability analysis, the mean and standard deviation can be adjusted to see how they affect outcomes, like in Step 4 and Step 5 of the original solution. Adjusting the mean to 9.25 or reducing the standard deviation to 0.15 highlights how these two key parameters interact with Z-scores and the resulting probabilities.

Probability Analysis

Probability analysis involves interpreting the Z-score by looking it up in the standard normal distribution table. This table provides the probability that a standard normal random variable will be less than or equal to the given Z-score.
In the exercise, the Z-score of -0.80 corresponds to a probability of approximately 0.2119. This value tells us that about 21.19% of the hams weigh less than 9.00 pounds, which is Glen Henline's concern.
Probability analysis becomes more insightful when considering Glen's proposals. By increasing the mean to 9.25 (Proposal 1), the Z-score changes, reducing the probability to 15.87%. However, reducing the standard deviation to 0.15 (Proposal 2), results in a new probability of 9.18%.
This kind of analysis helps in making strategic decisions. It allows one to choose the best method to achieve a desired outcome, like minimizing the number of hams below label weight. Understanding how different parameters affect the probability helps Glen select the more effective proposal.