Question

the lifetime in miles for a certain brand of tire is normally distributed with a mean of 22,000 miles and a standard deviation of 3,100 miles the tire manufacturer wants to offer a money - back guarantee so that no more than 3% of tires will qualify for a refund. what is the minimum number of miles the manufacturer should guarantee that the tires will last? a. 27,828 miles b. 16,172 miles c. 15,800 miles d. 25,007 miles

Explanation:

Step1: Find the z - score

We want the area to the left of the z - score to be $1 - 0.03=0.97$. Looking up in the standard normal distribution table (z - table), the z - score corresponding to an area of 0.97 is approximately $z = 1.88$.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value we want to find, $\mu$ is the mean, and $\sigma$ is the standard deviation. We know $\mu = 22000$, $\sigma=3100$, and $z = 1.88$. Rearranging the formula for $x$ gives $x=\mu+z\sigma$.

Step3: Calculate the value of x

Substitute the values into the formula: $x = 22000+1.88\times3100$. First, calculate $1.88\times3100 = 1.88\times31\times100=(1.88\times31)\times100=(1.88\times(30 + 1))\times100=(1.88\times30+1.88\times1)\times100=(56.4+1.88)\times100 = 5828$. Then $x=22000 + 5828=16172$.

Answer:

B. 16,172 miles