To solve the problem shown in the image, we need to determine the minimum sample size required for estimating the mean completion time of an assembly line operation.

Given Information:

• Confidence level = 95%
• Population standard deviation ($\sigma$) = 10.2 minutes
• Desired margin of error (E) = 1.7 minutes
• Z-score for 95% confidence level = 1.96 (from the standard normal distribution)

Formula for Sample Size (n):

$n = \left(\frac{Z \cdot \sigma}{E}\right)^2$ Where:

• $Z$ is the Z-score corresponding to the confidence level.
• $\sigma$ is the population standard deviation.
• $E$ is the margin of error.

Substituting the values:

$n = \left(\frac{1.96 \times 10.2}{1.7}\right)^2$ Now, let's compute this step by step:

$\frac{1.96 \times 10.2}{1.7} = \frac{19.992}{1.7} \approx 11.76$ $n = 11.76^2 \approx 138.36$

Since the sample size must be a whole number, we round up to the next whole number:

$n = 139$

Final Answer:

The minimum sample size needed is 139.

Let me know if you'd like more details or have any questions.

Here are 5 related questions to deepen understanding:

1. How would the required sample size change if the margin of error was 1 minute instead of 1.7 minutes?
2. What is the effect of increasing the confidence level to 99% on the sample size?
3. How does a larger population standard deviation affect the required sample size?
4. Can you explain the relationship between sample size and margin of error?
5. What happens to the required sample size if we halve the standard deviation?

Tip: Always round up the sample size to ensure the desired margin of error is met.