Question

The tread life of a particular brand of tire is a random variable best described by a normal distribution. With a mean of 60,000 miles and a standard deviation of 2,100 miles, what is the probability that a particular tire of this brand will last for longer than 57,900 miles?

Answer 1

Given information about the normal distribution of trend lift of a particular brand of tire:

μ = 60,000

s = 2,100

We need to find: P(X > 57,900)

z₅₇₉₀₀ = (x - μ)/s = (57900 - 60000)/2100 = -1

=> P(X > 57900) = P(z > z₅₇₉₀₀) = P(z > -1) = 1 - P(z <= -1)

From the z-table, we can find out P(z <= -1) which is 0.1587

P(X > 57,900) = 1 - 0.1587 = 0.8413

Hence, the probability that a particular tire of this brand will last for longer than 57,900 miles is 0.8413.