Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 10.5 years, and standard deviation of 1.6 years.

The 2% of items with the shortest lifespan will last less than how many years?

Give your answer to one decimal place.

Answer 1

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:

\[
\boxed{7.2} \text{ years}
\]