A tire manufacturer warranties its tires to last at least 20,000 miles orâ "you get a new set ofâ tires." In itsâ experience, a set of these tires last on average 26,000 miles with SD 5,000 miles. Assume that the wear is normally distributed. The manufacturer profits â$200 on each setâ sold, and replacing a set costs the manufacturer â$400. Complete parts a through c.
(a) What is the probability that a set of tires wears out before 20,000 miles?

We are given that a tire manufacturer warranties its tires to last at least 20,000 miles or "you get a new set of tires." In its past experience, a set of these tires last on average 26,000 miles with S.D. 5,000 miles. Assume that the wear is normally distributed.

Let X = wearing of tires

So, X ~ N([tex]\mu=26,000,\sigma^{2}=5,000^{2}[/tex])

Now, the z score probability distribution is given by;

Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = average lasting of tires = 26,000 miles

[tex]\sigma[/tex] = standard deviation = 5,000 miles

So, probability that a set of tires wears out before 20,000 miles is given by = P(X < 20,000 miles)

P(X < 20,000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{20,000-26,000}{5,000}[/tex] ) = P(Z < -1.2) = 1 - P(Z [tex]\leq[/tex] 1.2)

= 1 - 0.88493 = 0.1151

Therefore, probability that a set of tires wears out before 20,000 miles is 0.1151.