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

3000

miles. What warranty should the company use if they

want 96% of the tires to outlast the warranty

Answer 1

µ = 60000

sigma = 3000

$or,P(\frac{X-\mu}{\sigma}>\frac{x-\mu}{\sigma}) = 0.96$

$or,P(Z>\frac{x-60000}{3000}) = 0.96$

$or,1-P(Z<\frac{x-60000}{3000}) = 0.96$

$or,P(Z<\frac{x-60000}{3000}) = 0.04$

$or,\frac{x-60000}{3000} = -1.75$

or, x = 60000 - 1.75 * 3000

or, x = 54750 (ans)