We are given the following information in this problem:

• The lifespan of items is normally distributed with a mean ($\mu$) of 10.7 years and a standard deviation ($\sigma$) of 1.3 years.
• You purchase 6 items, and the goal is to find the probability that their mean life will be longer than 11 years.

Step-by-Step Approach:

1. Find the standard error of the mean (SEM) for the sample: $\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{1.3}{\sqrt{6}}$ where $n = 6$ is the number of items purchased.

2. Find the z-score for the sample mean of 11 years: $z = \frac{\text{sample mean} - \mu}{\text{SEM}} = \frac{11 - 10.7}{\text{SEM}}$

3. Use the z-score to find the probability from the standard normal distribution. This will give the probability that the sample mean is less than 11 years, and subtracting that value from 1 will give the probability that the sample mean is more than 11 years.

Let me calculate the exact probability for you.The z-score for a sample mean of 11 years is approximately 0.57. Using this z-score, the probability that the sample mean lifespan will be longer than 11 years is approximately 0.286 (or 28.6%).

Would you like any further details or have questions?

Here are 5 related questions for further exploration:

1. How would the probability change if the sample size increased to 10?
2. What if the standard deviation were smaller, say 1.0 years?
3. How can we interpret the z-score in this context?
4. What is the probability that the mean lifespan is less than 10 years?
5. How does the Central Limit Theorem apply to this problem?

Tip: When working with the normal distribution, increasing the sample size reduces the standard error, making estimates more precise.