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.

**Step 1. Define the problem parameters** Let \( X \) denote the lifetime (in years) of the MP3 players. We know that \[ X \sim N(\mu, \sigma) \] where \[ \mu = 3.8 \quad \text{years}, \quad \sigma = 1.1 \quad \text{years}. \] The company wants the warranty to be such that only \( 0.6\% \) of the players fail before the warranty expires, which translates to: \[ P(X < t) = 0.006. \] **Step 2. Standardize the variable** Standardize \( X \) to a standard normal variable \( Z \) as follows: \[ Z = \frac{X - \mu}{\sigma}. \] Thus, the probability statement becomes: \[ P\left(Z < \frac{t - 3.8}{1.1}\right) = 0.006. \] Let \[ z_0 = \frac{t - 3.8}{1.1}. \] Then, we have: \[ P(Z < z_0) = 0.006. \] **Step 3. Find the corresponding \( z \)-score** We need to determine the \( z \)-score that corresponds to a cumulative probability of \( 0.006 \). Using the standard normal distribution table or calculator, we find: \[ z_0 \approx -2.50. \] **Step 4. Solve for the warranty time \( t \)** Now, substitute \( z_0 \) back into the standardization equation: \[ \frac{t - 3.8}{1.1} = -2.50. \] Multiply both sides by \( 1.1 \): \[ t - 3.8 = -2.50 \times 1.1 = -2.75. \] Add \( 3.8 \) to both sides: \[ t = 3.8 - 2.75 = 1.05. \] **Step 5. Round the answer to 1 decimal place** Rounding \( 1.05 \) to 1 decimal place gives: \[ t \approx 1.1 \quad \text{years}. \] The time length of the warranty is \( \boxed{1.1} \) years.