Question

The thread life of a particular brand of tire is a random variable best deacribed bu a normal distribution with a mean of 68,000 miles and standard deviation of 1500 miles.

A. Find the probability that randomly selcted tire feom this particular brand has a thread life of at least 75,000 miles.

B. Find the probability that a randomly selected tire from this particular brand has thread life less than 66,000 miles?

C. If manufacturer wants to set warranty of the yhread life of the tire so that only 2% of all tires will outlast the warranty, what should they set the millage at? Round answer to the nearest hundred miles.

Answer 1

Given
X̅ = 68000           ....... Mean
σ = 1500             ....... Standard Deviation

A) To find P( randomly selcted tire feom this particular brand has a thread life of at least 75,000 miles )

that is to find P(X ≥ 75000)

P(X ≥ 75000) = 1 - P(X < 75000)

Using standard normal tables Or Using Excel Function "NORM.DIST", we get

P(X ≥ 75000) = 1 - NORM.DIST(75000, 68000, 1500, TRUE)

                      = 1 - 1

                         = 0

P( randomly selcted tire feom this particular brand has a thread life of at least 75,000 miles ) = 0

B)

To find P( randomly selcted tire feom this particular brand has a thread life less than 66000 miles )

that is to find P(X 66000)

Using standard normal tables Or Using Excel Function "NORM.DIST", we get

P(X 66000) = NORM.DIST(66000, 68000, 1500, TRUE)

                      = 0.0912

P( randomly selcted tire feom this particular brand has a thread life less than 66000 miles) = 0.0912

C) Let the mileage to be set for warranty be X'

It is required that only 2% of all tires will outlast the warranty

Thus,

P(X > X') = 0.02

1 - P(X ≤ X') = 0.02

P(X ≤ X') = 0.98

Using standard normal tables Or Using Excel Function "NORM.INV", we get

X' = NORM.INV(0.98, TRUE)

X' = 71080.62 miles

Thus, the mileage to be set for warranty = 71100 miles          (Rounded to the nearest 100)