We are given that the lifespans of the items are normally distributed with a mean of 10.2 years and a standard deviation of 1.3 years. We need to find the threshold mean lifespan for 6 randomly chosen items, below which 10% of the samples fall.

Steps:

1. Population Mean ($\mu$): 10.2 years
2. Population Standard Deviation ($\sigma$): 1.3 years
3. Sample Size ($n$): 6 items
4. Percentage for the threshold: 10%, which corresponds to a z-score in the lower 10% of the normal distribution.

We will use the formula for the distribution of the sample mean:

$\text{Standard error of the mean} = \frac{\sigma}{\sqrt{n}}$

First, let's calculate the standard error of the mean.

$\text{Standard error} = \frac{1.3}{\sqrt{6}}$

Next, we look up the z-score corresponding to the 10th percentile in the standard normal distribution, which is approximately $-1.2816$.

Finally, we can use the formula for the sample mean distribution:

$X_{\text{mean}} = \mu + (z \times \text{Standard error})$

Let me calculate this.The mean life below which 10% of the samples of 6 randomly picked items will fall is approximately 9.5 years.

Would you like more details on this process or have any other questions?

Here are 5 related questions you might find useful:

1. How do we calculate the z-score for any given percentage?
2. How would the threshold change if the sample size were larger or smaller?
3. What does the standard error represent in this context?
4. How is the normal distribution applied in quality control?
5. Can the same approach be used for non-normally distributed data?

Tip: The standard error decreases as the sample size increases, making the sample mean more closely aligned with the population mean.