Let's break down the problem step by step.

Given Information:

• The diameter of large lids is Normally distributed.
• Mean ($\mu$) = 3.98 inches
• Standard deviation ($\sigma$) = 0.02 inches
• The acceptable range for the lid diameter is between 3.95 inches and 4.05 inches.

Part (a): Percentage of lids too small to fit

We need to find the percentage of lids with a diameter less than 3.95 inches.

1. Calculate the Z-score for 3.95 inches: $Z = \frac{X - \mu}{\sigma} = \frac{3.95 - 3.98}{0.02} = \frac{-0.03}{0.02} = -1.5$

2. Find the corresponding percentage (probability) for $Z = -1.5$: Using the standard normal distribution table, the probability that $Z$ is less than -1.5 is approximately 0.0668.

   This means that 6.68% of the large lids are too small to fit.

Part (b): Percentage of lids too big to fit

We need to find the percentage of lids with a diameter greater than 4.05 inches.

1. Calculate the Z-score for 4.05 inches: $Z = \frac{X - \mu}{\sigma} = \frac{4.05 - 3.98}{0.02} = \frac{0.07}{0.02} = 3.5$

2. Find the corresponding percentage (probability) for $Z = 3.5$: Using the standard normal distribution table, the probability that $Z$ is greater than 3.5 is very small, around 0.0002.

   This means that only about 0.02% of the large lids are too big to fit.

Part (c): Comparison and Analysis

• Small lids: 6.68% are too small.
• Large lids: 0.02% are too large.

Does it make sense for the lid manufacturer to try to make one of these values larger than the other?

It does not make sense for the lid manufacturer to deliberately make either of these values larger. Both undersized and oversized lids create problems for customers: undersized lids can lead to spills if they don’t fit securely, while oversized lids may not attach properly at all. The manufacturer should aim to minimize both probabilities as much as possible to ensure customer satisfaction and avoid product failures.

Do you want more details or have any questions?

Follow-up Questions:

1. How would the percentages change if the standard deviation were reduced to 0.01 inches?
2. What percentage of lids would be within the range of 3.96 to 4.04 inches?
3. If the mean were adjusted to 4.00 inches, what would the percentages of lids being too small or too large be?
4. How would the problem change if the lids followed a non-normal distribution?
5. How might the results influence the restaurant’s choice of lid suppliers?

Tip: In quality control, ensuring that the mean of the production process is centered within the tolerance limits is crucial to minimizing defective products.