# Normal distribution: tire durability

• k77i
In summary, the top-selling Red and Voss tire is rated at 60000 miles, but the actual mileage it can run before wearing out follows a normal distribution with a mean of 75000 miles and a standard deviation of 5000 miles. The probability that the tire wears out before 60000 miles is 0.13%, while the probability that it lasts more than 85000 miles is 2.28%.
k77i

## Homework Statement

The top-selling Red and Voss tire is rated 60000 miles, which means nothing. In fact, the distance the tires can run until wear-out is a normally distributed random variable with a mean of 75000 miles and a standard deviation of 5000 miles.

A. What is the probability that the tire wears out before 60000 miles?

B. What is the probability that a tire lasts more than 85000 miles?

## Homework Equations

z = [X - (mean of X)]/Standard deviation

## The Attempt at a Solution

This is another simple question that I can't seem to be getting correct.

For part A, I did z = (60000 - 75000)/5000 = -3
Then P(z < -3) = 0.0013 according to the chart.. well 0.0013 is P(z= -3.0).. I tried entering the answer for P( z= -2.99) = 0.0014 since it's the last value on the chart lower than -3 also but that was also incorrect.

For part B, z = [85000 - 75000]/5000 = 2
Then P(z > 2) = 1- P(z = 2) = 1 - 0.9772 = 0.0288
Also incorrect..

k77i said:

## The Attempt at a Solution

This is another simple question that I can't seem to be getting correct.

For part A, I did z = (60000 - 75000)/5000 = -3
Then P(z < -3) = 0.0013 according to the chart.. well 0.0013 is P(z= -3.0).. I tried entering the answer for P( z= -2.99) = 0.0014 since it's the last value on the chart lower than -3 also but that was also incorrect.

What region does you chart provide z-values for? Some charts will directly give you the value of P(z<a).

k77i said:
For part B, z = [85000 - 75000]/5000 = 2
Then P(z > 2) = 1- P(z = 2) = 1 - 0.9772 = 0.0288
Also incorrect..

P(z>2) = 1-P(z≤2) = 1-P(z=2)-p(z<2).

im not sure what u mean by which region but my hart goes to a point z?

Forget my initial posts with P(z=a), you can't have P(z=a) for a normal distribution. I would think your answers are correct, I am not sure why exactly it is saying is incorrect.

Perhaps you need to find a certain input format?
Like present it as a percentage, or with a different number of significant digits?

You made a calculation error in the second. It should be 0.0228 or 2.28%.

## 1. What is a normal distribution?

A normal distribution is a probability distribution that is commonly seen in nature and is symmetrical around the mean. It is also known as a bell curve due to its shape. In a normal distribution, the majority of the data falls within one standard deviation of the mean, with a smaller proportion falling further away from the mean.

## 2. How is tire durability related to normal distribution?

Tire durability can be represented by a normal distribution as it follows a similar pattern as other natural phenomena. The majority of tires will have an average durability, with a smaller proportion having significantly lower or higher durability. This distribution can help predict the likelihood of a tire lasting for a certain number of miles.

## 3. How is the normal distribution used in predicting tire durability?

By using the normal distribution, scientists can estimate the average tire durability and the likelihood of a tire lasting a certain number of miles. This information can be helpful for manufacturers in determining the quality of their tires and for consumers in selecting the most durable tires for their vehicles.

## 4. What factors can affect the normal distribution of tire durability?

Some factors that can affect the normal distribution of tire durability include the quality of materials used, the manufacturing process, and the usage conditions of the tires. These factors can cause the distribution to shift, with some tires having higher or lower durability than expected based on the normal distribution.

## 5. Can the normal distribution accurately predict the durability of all tires?

No, the normal distribution is a statistical model and cannot predict the exact durability of individual tires. It provides an estimate of the average durability and the likelihood of a tire falling within a certain range of durability. Other factors, such as external damage or defects, can also affect the durability of a tire.

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