Thread: 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!

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 for x*.

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 for x*.

thanks!