Question #153412

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.


1
Expert's answer
2021-01-03T14:27:50-0500

Let X=X= the tire life in miles: XN(μ,σ2)X\sim N(\mu, \sigma^2)

Given μ = 26000 miles, σ = 5000 miles

(a)

P(X<20000)=P(Z<20000260005000)P(X<20000)=P(Z<\dfrac{20000-26000}{5000})

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


The probability that a set of tires wears out before 20,000 miles is 0.11507.0.11507.


(b)


P(X20000)=1P(X<20000)P(X\geq20000)=1-P(X<20000)

10.11507=0.88493\approx1-0.11507=0.88493

The probability that the manufacturer turns a profit on selling a set to one customer is 0.88493.0.88493.


(c) XN(μ,σ2/n)X\sim N(\mu,\sigma^2/n)


P(X20000)=1P(X<20000)P(X\geq20000)=1-P(X<20000)

=1P(Z<20000260005000/500)=1-P(Z<\dfrac{20000-26000}{5000/\sqrt{500}})

1P(Z<26.8328)10=1\approx 1-P(Z<-26.8328)\approx1-0=1

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



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