Questions: Suppose that the lifetimes of tires of a certain brand are normally distributed with a mean of 75,000 miles and a standard deviation of o miles. These tires come with a 60,000-mile warranty. The manufacturer of the tires can adjust σ during the production process, but the adjustment of σ is quite costly. The manufacturer wants to set σ once and for all so that only 2% of the tires will fail before the warranty expires. Find the standard deviation to be set. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.

Solution Steps

To find the initial z-score for the warranty miles, we use the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \( X = 60000 \) (warranty miles),
• \( \mu = 75000 \) (mean lifetime of tires),
• \( \sigma = 10000 \) (initial guess for standard deviation).

Substituting the values, we have:

\[ z = \frac{60000 - 75000}{10000} = \frac{-15000}{10000} = -1.5 \]

Thus, the initial z-score for warranty miles is \( z = -1.5 \).

To ensure that only \( 2\% \) of the tires fail before the warranty expires, we need to find the standard deviation \( \sigma \) such that the z-score corresponds to the \( 2\% \) failure rate. The z-score for the \( 2\% \) left tail is approximately \( -2.054 \).

Using the rearranged z-score formula:

\[ \sigma = \frac{X - \mu}{z} \]

we substitute the values:

\[ \sigma = \frac{60000 - 75000}{-2.054} = \frac{-15000}{-2.054} \approx 7302.8 \]

Thus, the required standard deviation to achieve a \( 2\% \) failure rate is approximately \( 7302.8 \).

Final Answer