a tire manufacturer believes that the tread life of its tires are normally distributed and will lst an average of 60,000 miles with a standard deviation of 3,000 miles. sixty-four randomly selected tires were tested and the average miles where the tire failed were recorded. what is the probability that the mean miles recorded when the tires failed will be more than 59,500 miles?

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)(To make things simpler, you could use n.)

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to that Z score.

r4err

To find the probability that the mean miles recorded when the tires fail will be more than 59,500 miles, we need to use the concept of the sampling distribution of the sample mean.

Given:
- The average tread life of the tires is normally distributed with a mean (μ) of 60,000 miles and a standard deviation (σ) of 3,000 miles.
- We have a sample size (n) of 64 tires.

The sampling distribution of the sample mean, also known as the distribution of the sample means, can be approximated as a normal distribution when the sample size is large enough (usually n ≥ 30) due to the Central Limit Theorem.

Here's how we can calculate the probability using the sampling distribution:

1. Calculate the standard deviation of the sampling distribution.
The standard deviation of the sampling distribution is also known as the standard error (SE) and can be calculated using the formula:
SE = σ / sqrt(n)
where σ is the standard deviation of the population and n is the sample size.

In this case, SE = 3000 / sqrt(64) = 375.

2. Calculate the z-score for the given value of 59,500 miles.
The z-score measures the number of standard deviations a particular value is away from the mean. It can be calculated using the formula:
z = (x - μ) / SE
where x is the given value and μ is the mean.

In this case, z = (59500 - 60000) / 375 = -0.1333 (rounded to four decimal places).

3. Find the probability using the standard normal (z) distribution table.
We want to find the probability that the mean miles recorded when the tires fail is more than 59,500 miles, which corresponds to the area under the curve to the right of the z-score.

By looking up the z-score (-0.1333) in the z-table, we find that the probability is 0.552 (rounded to three decimal places).

Therefore, the probability that the mean miles recorded when the tires fail will be more than 59,500 miles is approximately 0.552 or 55.2%.