Lifetred Inc makes automobile tires that have a mean life of

Lifetred, Inc., makes automobile tires that have a mean life of 60,000 miles with a standard deviation of 2500 miles. a. What fraction of tires s expected to survive beyond 66,250 miles? b. What fraction will survive fewer than 57,000 miles? c. ---- (skip this) d. What length of warranty is needed so that no more than 10 percent of the tires will be expected to fail during the warranty period?

Solution

P(X < x) = P(Z < x - mean/std)

a)

P(X >66250) = P(Z > 66250 - 60000/2500)

= P(Z > 2.5)

= 0.0062

b)

P(X < 57000)

= P(Z < 57000-60000/2500)

= P(Z < -1.2)

= 0.1151

d)

P(X <x) = 0.1

=>

P(Z < x-mean/std) = 0.1

=>

x-mean/std = invnorm(0.1) = -1.282

=>

x-60000/2500 = -1.282

x = 56795