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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!

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 7.6 years and a standard deviation of 0.9 years. The 1% of items with the shortest lifespan will last less than how many years?

Learn how to calculate the 1% shortest lifespan for a normally distributed dataset with a mean of 7.6 years and a standard deviation of 0.9 years. This problem involves understanding normal distribution, percentiles, and z-scores. Suitable for Grades 10-12.

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