Model Answer
0 min readIntroduction
In the realm of quality control and operations management, understanding process variability is crucial for manufacturers. Statistical process control relies heavily on sampling and inference to assess whether a process is operating as intended. The Central Limit Theorem states that the distribution of sample means will approximate a normal distribution, regardless of the underlying population distribution, provided the sample size is sufficiently large. This question assesses the probability of observing a sample mean exceeding a specified value, given the manufacturer’s claims about the assembly time of their electric saw.
Understanding the Problem
We are given the following information:
- Population mean (μ) = 80 minutes
- Population standard deviation (σ) = 40 minutes
- Sample size (n) = 64
- We want to find the probability that the sample mean (x̄) > 88 minutes
Formulating Hypotheses
While not explicitly asked for, framing the hypotheses clarifies the problem.
- Null Hypothesis (H0): μ = 80 minutes
- Alternative Hypothesis (H1): μ > 80 minutes
Calculating the Standard Error
The standard error (SE) of the sample mean is calculated as:
SE = σ / √n = 40 / √64 = 40 / 8 = 5 minutes
Calculating the Z-Score
The z-score measures how many standard errors the sample mean is away from the population mean. It is calculated as:
z = (x̄ - μ) / SE = (88 - 80) / 5 = 8 / 5 = 1.6
Finding the Probability
We want to find P(x̄ > 88), which is equivalent to finding P(z > 1.6). Using a standard normal distribution table or a statistical calculator, we find the area to the left of z = 1.6 is approximately 0.9452. Therefore, the area to the right (P(z > 1.6)) is:
P(z > 1.6) = 1 - P(z ≤ 1.6) = 1 - 0.9452 = 0.0548
Conclusion based on the Calculation
Therefore, the probability that the sample mean will be greater than 88 minutes is approximately 0.0548 or 5.48%. This suggests that observing a sample mean of 88 minutes or higher is relatively unlikely if the manufacturer's claim of a mean assembly time of 80 minutes is true.
Table Summarizing the Calculations
| Parameter | Value |
|---|---|
| Population Mean (μ) | 80 minutes |
| Population Standard Deviation (σ) | 40 minutes |
| Sample Size (n) | 64 |
| Sample Mean (x̄) | 88 minutes |
| Standard Error (SE) | 5 minutes |
| Z-Score (z) | 1.6 |
| Probability (P(x̄ > 88)) | 0.0548 |
Conclusion
In conclusion, the probability of observing a sample mean assembly time greater than 88 minutes, given the manufacturer’s claims, is approximately 5.48%. This relatively low probability suggests that the observed sample mean is somewhat unusual and might warrant further investigation into the manufacturing process. The application of the Central Limit Theorem and z-score calculation provides a robust framework for assessing such claims in operations management and quality control.
Answer Length
This is a comprehensive model answer for learning purposes and may exceed the word limit. In the exam, always adhere to the prescribed word count.