To find the lifespan corresponding to the shortest 2% of items, we need to determine the z-score that corresponds to the 2nd percentile of the standard normal distribution.
Using a standard normal distribution table or a calculator, we find that the z-score for the 2nd percentile is approximately \(-2.054\).
Now, we can use the z-score formula to convert this back to the original lifespan distribution:
\[ X = \mu + z \cdot \sigma \]
Where:
- \(X\) is the lifespan we want to find.
- \(\mu\) is the mean (10.5 years).
- \(z\) is the z-score (\(-2.054\)).
- \(\sigma\) is the standard deviation (1.6 years).
Substituting the values into the formula, we get:
\[ X = 10.5 + (-2.054) \cdot 1.6 \] \[ X = 10.5 - 3.2864 \] \[ X \approx 7.2136 \]
Rounding this to one decimal place, the lifespan corresponding to the shortest 2% of items is approximately:
\[ \boxed{7.2} \text{ years} \]