Company XYZ know that replacement times for the portable MP3 players it produces are normally distributed with a mean of 3.8 years and a standard deviation of 1.1 years. If the company wants to provide a warranty so that only \( 0.6 \% \) of the portable MP3 players 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 find the warranty period such that only 0.6% of the portable MP3 players will be replaced before the warranty expires, we need to find the \( z \)-score corresponding to the 0.6% in the lower tail of the normal distribution. Using standard normal distribution tables or a calculator, the \( z \)-score that corresponds to the cumulative probability of 0.006 is approximately -2.54. Now, we can use the \( z \)-score formula to find the warranty time length: \[ z = \frac{X - \mu}{\sigma} \] Where \( X \) is the time length we want to find, \( \mu \) is the mean (3.8 years), and \( \sigma \) is the standard deviation (1.1 years). Rearranging the formula: \[ X = z \cdot \sigma + \mu \] Substituting the values we have: \[ X = (-2.54) \cdot (1.1) + 3.8 \] Calculating \( X \): \[ X = -2.794 + 3.8 = 1.006 \] Rounding to one decimal place, the warranty length is approximately: \[ \text{warranty} = 1.0 \text{ years} \]