Question

3) A tire manufacturer believes that the treadlife of its snow tires can be described by a normal model with a mean of 32000 miles and a standard deviation of 2500 miles. What percentage of these tires can be expected to last for more than 40,000 miles? What number of miles represents the 90th percentile?

Answer 1

Solution:

Given that,

mean = = 3200

standard deviation = = 2500

p ( x > 40,000 )

= 1 - p (x < 40,000 )

= 1 - p ( x -  / ) < ( 40,000 - 32,000 / 2500 )

= 1 - p ( z < 8000 / 2500 )

= 1 - p ( z < 3.2)

Using z table

= 1 - 0.9993

= 0.0007

Probability = 0.0007 =0.07%

Using standard normal table,

P(Z < z) = 90%

P(Z < z) = 0.90

P(Z < 1.28) = 0.90

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 + 2500 + 32000

x = 35200 miles

The 90th percentile is 35200