Questions: Wristwatch Lifetimes The mean lifetime of a certain brand of wristwatch is 40 months, with a standard deviation of 4 months. If the distribution is normal, for how many months should a guarantee be made if the manufacturer does not want to exchange more than 10% of the watches? Assume the variable is normally distributed. Use The Standard Normal Distribution Table. Round the final answer to at least one decimal place and intermediate z value calculations to 2 decimal places. The guarantee should be offered for months.

Solution Steps

To determine the number of months for the guarantee, we need to find the value of the wristwatch lifetime that corresponds to the 90th percentile of a normal distribution with a mean of 40 months and a standard deviation of 4 months. This involves finding the z-score that corresponds to the 90th percentile and then using the z-score formula to calculate the corresponding lifetime in months.

We are given that the mean lifetime of the wristwatch is \( \mu = 40 \) months and the standard deviation is \( \sigma = 4 \) months. We need to find the guarantee period such that no more than \( 10\% \) of the watches will need to be exchanged.

To find the guarantee period, we first determine the z-score that corresponds to the \( 90\% \) percentile of the standard normal distribution. The z-score for the \( 90\% \) percentile is approximately \( z = 1.2816 \).

Using the z-score, we can calculate the guarantee period \( x \) using the formula: \[ x = \mu + z \cdot \sigma \] Substituting the known values: \[ x = 40 + 1.2816 \cdot 4 \] Calculating this gives: \[ x = 40 + 5.1264 = 45.1264 \] Rounding to one decimal place, we find: \[ x \approx 45.1 \]

Final Answer