Company XYZ know that replacement times for the quartz time pieces it produces are normally distributed with a mean of 16.6 years and a standard deviation of 2.1 years. If the company wants to provide a warranty so that only \( 4.5 \% \) of the quartz time pieces will be replaced before the warranty expires, what is the time length of the warranty? warranty = years Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact \( z \) - scores or \( z \)-scores rounded to 3 decimal places are accepted.

To determine the time length of the warranty, we need to find the z-score that corresponds to the lower 4.5% of the standard normal distribution. From z-tables or using a calculator, we find that the z-score for 4.5% is approximately -1.645. We can use this z-score in the formula for z-scores: \[ z = \frac{X - \mu}{\sigma} \] Where: - \( z \) is the z-score (-1.645) - \( X \) is the warranty time we want to find - \( \mu \) is the mean (16.6 years) - \( \sigma \) is the standard deviation (2.1 years) Rearranging the formula to solve for \( X \): \[ X = \mu + z \cdot \sigma \] Substituting the values: \[ X = 16.6 + (-1.645) \cdot 2.1 \] \[ X = 16.6 - 3.4485 \] \[ X \approx 13.1 \text{ years} \] Thus, the time length of the warranty should be approximately **13.1 years**.