1. normal probability distribution

The Dynamic Tire Co. has just developed a new steel-belted radial tire to be sold through a national chain of discount stores. From road tests, it is found the tire life is normally distributed with a mean life of 73,000km and a standard deviation of 6,000km.

Dynamic plans to give a guarantee on their tires, under which they will replace any tire that wears out prematurely. What should their guarantee be, if they only wish to replace at most 1.5% of all tires?

The answer is 0.59980 but I want to know how you get this answer because I don't know how they got it.

Thank you!

2. Originally Posted by shellshock
The Dynamic Tire Co. has just developed a new steel-belted radial tire to be sold through a national chain of discount stores. From road tests, it is found the tire life is normally distributed with a mean life of 73,000km and a standard deviation of 6,000km.

Dynamic plans to give a guarantee on their tires, under which they will replace any tire that wears out prematurely. What should their guarantee be, if they only wish to replace at most 1.5% of all tires?

The answer is 0.59980 but I want to know how you get this answer because I don't know how they got it.

Thank you!
Find the value z* such that Pr(Z > z*) > 0.015. Then the value of x* such that Pr(X > x*) = 0.015 is found by solving $\displaystyle z^* = \frac{x^* - 73,000}{6,000}$ for x*.

3. Originally Posted by mr fantastic
Find the value z* such that Pr(Z > z*) > 0.015. Then the value of x* such that Pr(X > x*) = 0.015 is found by solving $\displaystyle z^* = \frac{x^* - 73,000}{6,000}$ for x*.
thanks!