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. 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. What should 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?

Answer 1

The tire company should offer a guarantee of 29,032 miles to ensure that no more than 2 percent of the tires fail to meet the guaranteed mileage. This is calculated using the 2nd percentile of a normal distribution with a mean of 37,000 miles and a standard deviation of 3,887 miles.

Step-by-step explanation:

To determine the guaranteed mileage the tire company should offer, we must find the value of mileage such that no more than 2 percent of the tires will fail to exceed it. This involves finding the 2nd percentile of a normal distribution with a mean of 37,000 miles and a standard deviation of 3,887 miles.

Steps to find the 2nd percentile:

1. Convert the desired tail probability to a z-score. For the 2nd percentile, the z-score corresponds to the value such that 2% of the area under the normal curve falls to its left.
2. Consult a standard normal (Z) distribution table, or use a statistical calculator or software, to find the z-score that corresponds to the cumulative probability of 0.02.
3. Once the z-score is found, use the formula for a value X on a normal distribution X = μ + z* σ, where μ is the mean and σ is the standard deviation.

The z-score corresponding to the 2nd percentile is approximately -2.05. Therefore, the guaranteed mileage X can be calculated as follows:

X = 37,000 + (-2.05)(3,887) = 37,000 - 7,968.35 ≈ 29,031.65 miles

Therefore, the tire company should set the guaranteed mileage at 29,032 miles (rounded to the nearest mile) to ensure that no more than 2 percent of the tires will fail to meet the guaranteed mileage.

Answer 2

Step-by-step explanation:

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 ( (X-\mu )/(\sigma )\leq (k-\mu )/(\sigma ) \right )=0.02 \\\\P\left ( Z \leq (k-37,000 )/(3,887 ) \right )=0.02`

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

`(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

`(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: