To solve this problem, we need to calculate the value below which the shortest 1% of the normally distributed lifespans fall. The parameters given are:

• Mean $\mu = 7.6$ years
• Standard deviation $\sigma = 0.9$ years
• We are looking for the 1st percentile, which means the point at which 1% of the distribution lies below it.

This can be solved using the formula for a value in a normal distribution:

$X = \mu + Z \cdot \sigma$

Where $Z$ is the z-score corresponding to the 1st percentile in a standard normal distribution. We can find $Z$ from standard z-tables or a calculator. For the 1st percentile, $Z \approx -2.326$.

Now, we substitute the values into the formula:

$X = 7.6 + (-2.326) \cdot 0.9$

Let me calculate that for you.The lifespan corresponding to the 1% of items with the shortest lifespan is approximately 5.5 years.

Do you need further details or have any questions?

Here are some related questions to explore:

1. What is the z-score for the 10th percentile in a normal distribution?
2. How would the lifespan change if the standard deviation increased?
3. What if the mean lifespan was 8 years? How would the 1% lifespan change?
4. How can you use a z-table to find percentiles in a normal distribution?
5. How does the concept of z-scores apply in quality control?

Tip: The smaller the z-score, the further away you are from the mean on the lower side of the distribution!