a tire company finds the lifespan for one brand of its tires is normally distributed with a mean of 46,900…

a tire company finds the lifespan for one brand of its tires is normally distributed with a mean of 46,900 miles and a standard deviation of 6,000 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? click to view page 1 of the table. click to view page 2 of the table. \n\na. 54,580\nb. 37,030\nc. 56,770\nd. 39,220

a tire company finds the lifespan for one brand of its tires is normally distributed with a mean of 46,900 miles and a standard deviation of 6,000 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? click to view page 1 of the table. click to view page 2 of the table. \n\na. 54,580\nb. 37,030\nc. 56,770\nd. 39,220

Answer

Explanation:

Step1: Find the z - score

We want to find the value such that the area to the left of it under the normal - distribution curve is 0.10. Looking up in the standard normal distribution table (z - table), the z - score corresponding to an area of 0.10 is approximately $z=- 1.28$.

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 that $\mu = 46900$, $\sigma=6000$, and $z=-1.28$. Rearranging the formula for $x$ gives $x=\mu + z\sigma$.

Step3: Calculate the value of x

Substitute the values into the formula: $x = 46900+( - 1.28)\times6000=46900-7680 = 39220$.

Answer:

D. 39,220