Crunchy Corn Chips are packaged in bags labeled "net weight 10 ounces." In fact, the weight of a bag of Crunchy Corn Chips follows a normal distribution with mean ì ounces and standard deviation ó ounces. Worried about customer complaints and potential lawsuits, the manufacturers decided that no more than 0.5% of bags of Crunchy Corn Chips should be underweight. There are two possible adjustments that can be made: change the mean setting on the filling machines while leaving the standard deviation constant, or leave the mean at 10.4 ounces and try to adjust the standard deviation of 0.26 ounces. Describe the necessary adjustments under each of these two plans. Which do you think is more feasible? Explain.

Stephanie

Here is a really neat applet that will take the place of your z-score tables

http://davidmlane.com/hyperstat/z_table.html

You can experiment with both graphs.
Since you want .5% of the bags to be underweight, you are looking for an
"area" of .495 (I took .5 - .005)

On the second graph, enter 0.495 in the area
and 10.4 and .26 into the mean and sd boxes.
click on "below"
Change the mean and sd entries and see what happens.

To determine the necessary adjustments under each plan, we need to find the corresponding z-scores for the given percentiles. Let's assume the mean weight of a bag of Crunchy Corn Chips is denoted by μ, and the standard deviation is denoted by σ.

1. Plan 1: Change the mean setting on the filling machines while leaving the standard deviation constant.
- We need to find the z-score corresponding to the 0.5% percentile (or the left-tail area of 0.005) in a standard normal distribution.
- Using a standard normal distribution table or calculator, we find that the z-score for a left-tail area of 0.005 is approximately -2.57.
- To achieve the desired result, we need the mean weight (μ) to be such that -2.57 standard deviations below the mean equals an underweight bag.
- Mathematically, μ - 2.57σ = x, where x represents the desired underweight weight.
- By solving this equation, we can find the necessary adjustment to the mean setting on the filling machine.

2. Plan 2: Leave the mean at 10.4 ounces and adjust the standard deviation to 0.26 ounces.
- Similarly, we need to find the z-score corresponding to the 0.5% percentile (or the left-tail area of 0.005) using a standard normal distribution table or calculator.
- Let's denote this z-score as z.
- To achieve the desired result, we need the standard deviation (σ) to be such that -z standard deviations below the mean equals an underweight bag.
- Mathematically, μ - zσ = x, where x represents the desired underweight weight.
- By substituting the given values (μ = 10.4, σ = 0.26) and solving this equation, we can find the necessary adjustment to the standard deviation.

Now, let's evaluate which plan is more feasible.

To determine feasibility, we need to consider practical limitations. Adjusting the mean setting on the filling machines may require recalibration and could affect the production process. On the other hand, adjusting the standard deviation involves modifying the consistency of product weights, which might require process optimization.

Comparing the two plans, it is generally more feasible to adjust the mean setting on the filling machines while leaving the standard deviation constant. This adjustment does not require modifications to the production process, as it only involves changing the filling machines' mean setting. However, the specific feasibility depends on the technical capabilities and constraints of the manufacturing process.

Therefore, plan 1 (changing the mean setting) is usually the more feasible option, unless there are specific considerations that make plan 2 (changing the standard deviation) a better choice for the manufacturer.

To determine the necessary adjustments, we need to calculate the target specifications for both mean and standard deviation.

1. Adjusting the mean:
To find the target mean, we need to consider the desired proportion of bags that should be underweight. In this case, the manufacturers want no more than 0.5% of the bags to be underweight. Since we are dealing with a normal distribution, we need to find the z-score corresponding to the desired proportion.

Using a standard normal distribution table or a calculator, we can find that the z-score for a 0.5% proportion is approximately -2.81.

Since the desired target mean is the value that corresponds to a z-score of -2.81, we can use the formula:
Target mean = Mean + (z-score * standard deviation)

In this case, the target mean = ì + (-2.81 * ó)

2. Adjusting the standard deviation:
To find the target standard deviation, we need to calculate the z-score corresponding to the desired proportion of bags underweight. Following the same reasoning as before, we find that the z-score is approximately -2.81.

Using the formula:
Target standard deviation = Standard deviation * (desired z-score / current z-score)

In this case, the target standard deviation = 0.26 * (2.81 / current z-score)

Which adjustment is more feasible?
To determine which adjustment is more feasible, we need to consider the practicality of making changes in the mean or the standard deviation.

Adjusting the mean setting on the filling machines while leaving the standard deviation constant can be more feasible. This adjustment involves making changes to the machine settings, which can be relatively easier to implement compared to adjusting the standard deviation, which might require process optimization or altering the production conditions. Additionally, modifying the mean setting has a direct impact on the outcome, making it a more straightforward approach to achieve the desired specifications.

However, it's important to note that the feasibility of each adjustment ultimately depends on the specific circumstances and constraints of the manufacturing process.