According to the given question, a tire company has developed a new type of steel-belted radial tire. Extensive testing indicates the population of mileages obtained by all tires of this new type is normally distributed with a mean of 37,000 miles and a standard deviation of 3,887 miles.

Let us define X be the random variable shows that the mileages tires normally distributed with

mean

μ = 37000

standard deviation

σ =3, 887

Therefore

X ~ (μ = 37000, σ =3,887)

The company wishes to offer a guarantee providing a discount on a new set of tires if the original tires purchased do not exceed the mileage stated in the guarantee. Therefore the guaranteed mileage be if the tire company desires that no more than 2 percent of the tires will fail to meet the guaranteed mileage is determined as:

P(X < k) = 0.02

$\Rightarrow P\left ( \frac{X-\mu }{\sigma }\leq \frac{k-\mu }{\sigma } \right )=0.02
\\\\P\left ( Z \leq \frac{k-37,000 }{3,887 } \right )=0.02$

From the standard normal curve 2% area is determined as -2.0537 and hence

$\frac{k-37,000}{3,887}=-2.0537\\\\k=37000-7982.7319\\\\k=29017.2681\\\\\Rightarrow k=29018$

If we consider z value at two decimal places then

$\frac{k-37,000}{3,887} =-2.05\\\\\Rightarrow k=29032
$

Therefore the guaranteed 29032 mileage be if the tire company desires that no more than 2 percent of the tires will fail to meet the guaranteed mileage.

The area under the standard normal curve is determined as: