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.

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Answer 1

μ = 26000

x = 20000

σ = 5000

(a)

$P(x<20000) = P(Z = \frac{x-μ}{σ}) \\ = P(Z = \frac{20000-26000}{5000}) \\ =P(Z = \frac{-6000}{5000}) \\ = P(Z=-1.2) \\ = 0.1151$

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

(b)

$P(x≥20000) = 1 – P(x<20000) \\ = 1 - P(Z = \frac{x-μ}{σ}) \\ = 1 - P(Z = \frac{20000-26000}{5000}) \\ = 1 - P(Z = \frac{-6000}{5000}) \\ = 1 - P(Z=-1.2) \\ = 1 - 0.1151 \\ = 0.8849$

The probability that the manufacturer turns a profit on selling a set to one customer is 0.8849

(c)

$P(x≥20000) = 1 – P(x<20000) \\ P(x<20000) = P(Z= \frac{x-μ}{\frac{σ}{\sqrt{n}}}) \\ P(x≥20000) = 1 - P(Z= \frac{x-μ}{\frac{σ}{\sqrt{n}}}) \\ n= 500 \\ P(x≥20000) = 1 - P(Z= \frac{20000-26000}{\frac{5000}{\sqrt{500}}}) \\ = 1 - P(Z = \frac{-6000}{226.61}) \\ = 1-P(Z=-26.832) \\ = 1-0 \\ =1$

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