Question

A tire manufacturer believes that the tread life of its snow tires can be described by a normal model with a mean of 32,000 miles and a standard deviation of 2500 miles.
a) if you buy one of these tires, would it be reasonable for you to hope it will last 40,000 miles? explain
b) approx. what fraction of these tires can be expected to last less than 30,000 miles?
c) approx. what fraction of these tires can be expected to last between 30,000 and 35,000 miles?

Answer 1

Answer on Question #54884 – Math – Statistics and Probability

A tire manufacturer believes that the tread life of its snow tires can be described by a normal model with a mean of 32,000 miles and a standard deviation of 2500 miles.

a) if you buy one of these tires, would it be reasonable for you to hope it will last 40,000 miles? explain

b) approx. what fraction of these tires can be expected to last less than 30,000 miles?

c) approx. what fraction of these tires can be expected to last between 30,000 and 35,000 miles?

Solution

a) Mean = 32,000 miles

SD = 2,500 miles

x = 40,000 miles

$$z = \frac{x - \text{mean}}{\text{SD}} = \frac{40,000 - 32,000}{2,500} = 3.2.$$

From a normal distribution table

$$P(z < 3.2) = 0.9993$$

so

$$P(3.2 < z) = 1 - 0.9993 = 0.0007 = 0.07\%$$

Thus, it is not reasonable to hope for the set of these tires last for 40,000 miles (or more). The probability for this to occur is just too small.

b) x = 30,000 miles

$$z = \frac{x - \text{mean}}{\text{SD}} = \frac{30,000 - 32,000}{2,500} = -0.8.$$ $$P(z < -0.8) = 0.2119$$

Thus, approximately 21.19% of these tires can be expected to last less than 30,000 miles.

c) We have already calculated $P(z < -0.8) = 0.2119$, thus let's focus on $x = 35,000$ miles:

$$z = \frac{(35,000 - 32,000)}{2,500} = 1.2$$ $$P(z < 1.2) = 0.8849.$$

So,

$$P(-0.8 < z < 1.2) = P(z < 1.2) - P(z < -0.8) = 0.8849 - 0.2119 = 0.6730.$$

Thus, approximately 67.3% of these tires are expected to last between 30,000 and 35,000 miles.

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