91影视

At some fast-food restaurants, customers who want a lid for their drinks get them from a large stack left near straws, napkins, and condiments. The lids are made with a small amount of flexibility so they can be stretched across the mouth of the cup and then snuggly secured. When lids are too small or too large, customers can get very frustrated, especially if they end up spilling their drinks. At one particular restaurant, large drink cups require lids with a 鈥渄iameter鈥� of between 3.95 and 4.05 inches. The restaurant鈥檚 lid supplier claims that the mean diameter of their large lids is 3.98 inches with a standard deviation of 0.02 inches. Assume that the supplier鈥檚 claim is true. (a) What percent of large lids are too small to fit? Show your method. (b) What percent of large lids are too big to fit? Show your method. (c) Compare your answers to (a) and (b). Does it make sense for the lid manufacturer to try to make one of these values larger than the other? Why or why not?

Step by step solution

01

Understand the Problem

We are dealing with a normal distribution with a mean diameter of 3.98 inches and a standard deviation of 0.02 inches. We need to find the percentage of lids that don't fit for both being too small (<3.95 inches) and too large (>4.05 inches).

02

Calculate Z-score for Too Small Lids

The Z-score formula is \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. Plug in the values: \( X = 3.95 \), \( \mu = 3.98 \), \( \sigma = 0.02 \). Thus: \[Z = \frac{3.95 - 3.98}{0.02} = -1.5\]

03

Find the Percentage for Too Small Lids

Use the standard normal distribution table to find the proportion of the distribution that falls below a Z-score of -1.5. This value is approximately 0.0668, meaning about 6.68% of lids are too small.

04

Calculate Z-score for Too Large Lids

Similar to Step 2, calculate for lids being too large with \( X = 4.05 \): \[Z = \frac{4.05 - 3.98}{0.02} = 3.5\]

05

Find the Percentage for Too Large Lids

Use the standard normal distribution table to find the upper tail probability for a Z-score of 3.5. This value is approximately 0.0002, meaning about 0.02% of lids are too large.

06

Compare the Percentages

From the calculations, 6.68% of lids are too small and 0.02% are too large. It is less balanced, with far more lids being too small.鈥攊t doesn't make sense to have one error much larger than the other, but addressing the too-small lids (since they are more prevalent) might be more beneficial.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Z-score Calculation

To determine how many lids are either too small or too large to fit properly, we use a statistical measurement called "Z-score." This helps us understand how far away a particular lid size is from the average (mean) size, in terms of standard deviations.
The Z-score is calculated using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where:

• \( Z \) is the Z-score
• \( X \) is the observed value
• \( \mu \) is the mean of the distribution
• \( \sigma \) is the standard deviation

For lids that are too small, we set \( X \) to 3.95 inches, the minimum fitting size. Plugging in the numbers: \[ Z = \frac{3.95 - 3.98}{0.02} = -1.5 \] This indicates that these smaller lids are 1.5 standard deviations below the mean.
Now, for lids that are too large, we use 4.05 inches鈥� the maximum acceptable lid size. Our formula looks like this: \[ Z = \frac{4.05 - 3.98}{0.02} = 3.5 \] This tells us that lids exceeding 4.05 inches are 3.5 standard deviations above the mean. Utilizing the Z-score can help manufacturers adjust production to reduce the number of lids that do not meet the size criteria.

Standard Deviation

Standard deviation plays a critical role in understanding the spread of measurements around an average value (mean). It quantifies how much the individual pieces of data (in this case, lid diameters) are expected to deviate from the mean of the entire data set.
A smaller standard deviation, like the 0.02 inches in our lid problem, means that most lid sizes are very close to 3.98 inches. This tight clustering is ideal for product consistency.
In practical terms for the restaurant, if the standard deviation were larger, we would expect to see more lids that are either too large or too small as they would spread further from the mean. Hence, maintaining a low standard deviation ensures that nearly all lids fall within the acceptable range of 3.95 to 4.05 inches.
Maintaining a minimal standard deviation not only supports the reliability of the lids fitting consistently but also plays a crucial role in minimizing customer dissatisfaction stemming from poor fitting lids.

Proportion Calculation

After determining the respective Z-scores, we find the proportions of lids that fall outside the acceptable range using the standard normal distribution table. This table provides the probability or proportion of values that lie below or above a particular Z-score.
For lids that are too small, a Z-score of -1.5 corresponds to a proportion of approximately 0.0668, or 6.68%. This tells us that about 6.68% of lids are too small to properly fit the cups.
For lids that exceed the acceptable range with a Z-score of 3.5, the table tells us that only 0.02% of lids are too big.
It's crucial for manufacturers to focus on these proportions as they inform quality control processes. If a significant portion of lids is too small or too large, adjustments in the production line can be made to address these discrepancies. This ultimately aids in ensuring customer satisfaction and maintaining product consistency in the marketplace.