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matrenka [14]
3 years ago
11

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?
Mathematics
1 answer:
Y_Kistochka [10]3 years ago
4 0

Answer:

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

Step-by-step explanation:

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.

<em>Let X = wearing of tires</em>

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

Now, the z score probability distribution is given by;

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

where, \mu = average lasting of tires = 26,000 miles

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

                                                                    = 1 - 0.88493 = 0.1151

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

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2(4+x)+(13+x)

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------------

So the original equation

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