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.

1 answer

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:

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