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The weights of canned hams processed at 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?

Short Answer

Expert verified
About 21.19% of hams weigh less than the label; reducing the standard deviation is better.

Step by step solution

01

Understanding the Problem

We are given a normal distribution of ham weights with a mean (\(\mu=9.20\) pounds) and a standard deviation (\(\sigma=0.25\) pounds). We need to find the proportion of hams that weigh less than the label weight (9.00 pounds). Additionally, we are to compare two proposals: increasing the mean to 9.25 pounds or reducing the standard deviation to 0.15 pounds, to determine which is more effective in reducing the proportion of underweight hams.
02

Calculating Proportion Below Label Weight

To find the proportion of hams weighing less than 9.00 pounds, we calculate the z-score: \(z = \frac{X - \mu}{\sigma}\). Substituting the given values: \(z = \frac{9.00 - 9.20}{0.25} = \frac{-0.20}{0.25} = -0.80\). Using a z-table, find the probability corresponding to \(z = -0.80\). This value represents the proportion of hams weighing less than 9.00 pounds.
03

Proportion Calculation Result

Using the z-table, the probability corresponding to \(z = -0.80\) is approximately 0.2119. Thus, about 21.19% of the hams weigh less than the amount claimed on the label.
04

Option 1: Increasing Mean Weight

If the mean weight is increased to 9.25 pounds, calculate the new z-score: \(z = \frac{9.00 - 9.25}{0.25} = \frac{-0.25}{0.25} = -1.00\). Check the probability in a z-table for \(z = -1.00\), which shows the new proportion of hams weighing less than 9.00 pounds.
05

Option 1 Result Calculation

For \(z = -1.00\), the z-table gives a probability of approximately 0.1587. This means approximately 15.87% of hams would weigh less than the label weight if the mean is increased to 9.25 pounds.
06

Option 2: Reducing Standard Deviation

If the standard deviation is reduced to 0.15 pounds but the mean remains at 9.20 pounds, recalculate the z-score: \(z = \frac{9.00 - 9.20}{0.15} = \frac{-0.20}{0.15} \approx -1.33\). Find the probability from the z-table for \(z = -1.33\).
07

Option 2 Result Calculation

The z-table shows a probability of approximately 0.0918 for \(z = -1.33\). This indicates that about 9.18% of hams would weigh less than the label weight if the standard deviation is reduced to 0.15 pounds.
08

Recommendation Based on Analysis

Comparing the proportions, reducing the standard deviation to 0.15 pounds results in a lower proportion (9.18%) of underweight hams compared to increasing the mean to 9.25 pounds (15.87%). Reducing the standard deviation is more effective in minimizing the proportion of underweight hams.

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Key Concepts

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

Z-score
A Z-score, a fundamental concept in statistics, helps to determine how far a specific value is from the mean in terms of standard deviations.
When analyzing normally distributed data, like the weights of hams at Henline Ham Company, the Z-score allows us to find the proportion of hams that weigh less than the label weight.

The formula to calculate the Z-score is as follows:
  • \( z = \frac{X - \mu}{\sigma} \)
Where:
  • \( X \) is the value of interest (in this case, 9.00 pounds for the label weight).
  • \( \mu \) is the mean of the distribution (9.20 pounds).
  • \( \sigma \) is the standard deviation (0.25 pounds).
For this scenario, the Z-score calculates to \(-0.80\), implying that a weight of 9.00 pounds is 0.80 standard deviations below the mean.
This Z-score can then be used with a Z-table to find the proportion of the distribution that falls below this point — about 21.19% in this case.
Standard Deviation
In statistics, the standard deviation measures the dispersion or variability within a set of data.
It tells us how much individual data points differ from the mean.

In the context of the Henline Ham Company, a standard deviation of 0.25 pounds means that the weights of the hams vary by this amount around the average weight of 9.20 pounds.
A smaller standard deviation indicates that the data points (ham weights) are closer to the mean, leading to a more consistent product.

When Glen Henline considers reducing the standard deviation to 0.15 pounds while maintaining the current mean, he would achieve a tighter clustering of ham weights around the mean.
  • This means fewer hams would weigh substantially less than the label weight.
  • Calculating with the reduced standard deviation, a Z-score of \(-1.33\) is obtained for a 9.00-pound weight.
This adjustment results in only 9.18% of hams weighing less than 9.00 pounds.
Mean Weight
The mean, often referred to as the average, is the sum of all values divided by their number.
It provides a central value for a set of data points.

At Henline Ham Company, the mean weight for processed hams is set at 9.20 pounds.
This number signifies the average weight expected across all produced hams, balancing between extremes.
For quality improvement, Glen Henline considers increasing the mean weight to 9.25 pounds.
This change would shift the entire distribution of weights by 0.05 pounds upward.
  • Recomputing the Z-score for a 9.00-pound weight gives \(-1.00\).
  • Approximately 15.87% of hams would then weigh less than the label weight.
While shifting the mean upward reduces underweight hams, it's less effective than reducing standard deviation at minimizing those falling below label weight.
However, it still provides an easier adjustment to implement for small scale improvements.

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