A tire company finds the lifespan for one brand of its tires is normally distributed with a mean of 46,800 miles and a standard deviation of 4000 miles. If the manufacturer is willing to replace no more than 10% of the tires,
What should be the approximate number of miles for a warranty?
A)41,680
B)51,920
C)40,220
D)53,380

Question

Quizplus · Accepted Answer

To find the warranty mileage that corresponds to the top 10% of the lifespan distribution, we use the concept of z-scores in a normal distribution. The z-score for the 90th percentile (since the manufacturer is willing to replace no more than 10%, we are interested in the cutoff that leaves 90% of the distribution below it) is approximately 1.28. Using the z-score formula: $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value we're solving for, $\mu$ is the mean, and $\sigma$ is the standard deviation, we rearrange to solve for $X$: $X = Z\sigma + \mu$. Substituting the given values ($\mu = 46800$, $\sigma = 4000$, $Z = 1.28$), we get $X = 1.28(4000) + 46800 = 5160 + 46800 = 51960$. The closest option to this calculated value is 41,680, indicating a calculation mistake in the explanation process. The correct calculation should align with the principles of finding the lower bound for the top 10%, but the correct answer should reflect the calculation based on the z-score and the given distribution parameters, suggesting a need to reevaluate the calculation for accuracy. The correct approach involves using the z-score and the distribution parameters to find the mileage at which 10% of the tires would fail, which typically involves finding the value at the 90th percentile of the distribution. However, the explanation mistakenly suggests a lower value than calculated, indicating a discrepancy in the calculation or interpretation of the percentile and z-score relationship.

A Tire Company Finds the Lifespan for One Brand of Its