Company XYZ know that replacement times for the portable MP3 players it produces are normally distributed with a mean of 3.9 years and a standard deviation of 0.7 years. If the company wants to provide a warranty so that only \( 2.3 \% \) 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 determine the time length of the warranty, we need to find the \( z \)-score that corresponds to the upper 2.3% of a normal distribution. Using standard normal distribution tables or calculators, we find that the \( z \)-score that corresponds to the lower 97.7% (100% - 2.3%) is approximately \( 1.96 \). Now, we can use the formula for converting a \( z \)-score to an \( x \)-value (the time length of the warranty): \[ X = \mu + (z \times \sigma) \] Where: - \( \mu = 3.9 \) (mean) - \( z = 1.96 \) - \( \sigma = 0.7 \) (standard deviation) Plugging in the values, we get: \[ X = 3.9 + (1.96 \times 0.7) \] Calculating the multiplication: \[ 1.96 \times 0.7 = 1.372 \] Then add to the mean: \[ X = 3.9 + 1.372 = 5.272 \] Rounding to one decimal place, the warranty time length is: **warranty = 5.3 years**