Answer to Question #153412 in Statistics and Probability for EUGINE HAWEZA

Question

Question 3. A tire manufacturer warranties its tires to last at least 20,000 miles or “you get a new set of tires.” In its experience, a set of these tires lasts on average 26,000 miles with a standard deviation of 5,000 miles. Assume that the wear is normally distributed. The manufacturer profits $200 on each set sold, and replacing a set costs the manufacturer $400.

(a) What is the probability that a set of tires wears out before 20,000 miles?

(b) What is the probability that the manufacturer turns a profit on selling a set to one customer?

(c) If the manufacturer sells 500 sets of tires, what is the probability that it earns a profit after paying for any replacements? Assume that the purchases are made around the country and that the drivers experience independent amounts of wear.

Answer 1

Let $X=$ the tire life in miles: $X\sim N(\mu, \sigma^2)$

Given μ = 26000 miles, σ = 5000 miles

(a)

$$P(X<20000)=P(Z<\dfrac{20000-26000}{5000})$$

$$=P(Z<-1.2)\approx 0.11507$$

The probability that a set of tires wears out before 20,000 miles is $0.11507.$

(b)

$$P(X\geq20000)=1-P(X<20000)$$

$$\approx1-0.11507=0.88493$$

The probability that the manufacturer turns a profit on selling a set to one customer is $0.88493.$

(c) $X\sim N(\mu,\sigma^2/n)$

$$P(X\geq20000)=1-P(X<20000)$$

$$=1-P(Z<\dfrac{20000-26000}{5000/\sqrt{500}})$$

$$\approx 1-P(Z<-26.8328)\approx1-0=1$$

The probability that it earns a profit after paying for any replacements is 1.