According to internal testing done by the Get-A-Grip tire company, the mean lifetime of tires sold on new cars is 23,000 miles, with a standard deviation of 2500 miles.

a) If the claim by Get-A-Grip is true, what is the mean of the sampling distribution of for samples of size ? (2 pts)

b) If the claim by Get-A-Grip is true, what is the standard deviation of the sampling distribution of for samples of size ? (4 pts)

c) If the distribution of tire life is approximately normal, what is the probability that the mean of a random sample of size n = 4 of tire lifetimes will be less than 20,000 miles? (6 pts)

a) The mean of the sampling distribution of tire lifetimes for samples of size n is 23,000 miles.

b) The standard deviation of the sampling distribution of tire lifetimes for samples of size n is 2500/√n miles.

c) The probability that the mean of a random sample of size n = 4 of tire lifetimes will be less than 20,000 miles is 0.0668.

a) Well, if Get-A-Grip's claim is true, then the mean of the sampling distribution of X-bar for samples of size n would also be 23,000 miles. Why? Because the mean of the sampling distribution is equal to the population mean.

b) If the claim is true, then the standard deviation of the sampling distribution of X-bar for samples of size n would be 2500 divided by the square root of n. This is because the standard deviation of the sampling distribution is equal to the population standard deviation divided by the square root of the sample size. So, in this case, it would be 2500 divided by the square root of n.

c) To calculate the probability that the mean of a random sample of size n = 4 of tire lifetimes will be less than 20,000 miles, we need to use the sampling distribution of X-bar. Since the distribution of tire life is approximately normal, we can use the properties of the normal distribution to calculate this probability. We can standardize the sample mean using the formula (X-bar - mean of sampling distribution) / (standard deviation of sampling distribution). In this case, the mean of the sampling distribution is 23,000 miles, and the standard deviation of the sampling distribution is 2500 / sqrt(4). So, the standardized value is (20,000 - 23,000) / (2500 / sqrt(4)). Once we have the standardized value, we can use a standard normal distribution table or calculator to find the probability associated with that value.

a) The mean of the sampling distribution of the sample mean (x̄) is equal to the population mean (μ). Therefore, the mean of the sampling distribution of x̄ for samples of size n is also 23,000 miles.

b) The standard deviation of the sampling distribution of the sample mean (x̄) can be calculated using the formula:

σx̄ = σ / √n

where σ is the population standard deviation and n is the sample size.

In this case, σ = 2500 miles and n = ?

Since the value of n is not provided in the question, we cannot calculate the standard deviation of the sampling distribution without knowing the sample size.

c) To find the probability that the mean of a random sample of size n = 4 of tire lifetimes will be less than 20,000 miles, we need to use the sampling distribution of the sample mean.

First, we need to calculate the standard error of the mean (SE) using the formula:

SE = σ / √n

where σ is the population standard deviation and n is the sample size.

In this case, σ = 2500 miles and n = 4.

SE = 2500 / √4 = 2500 / 2 = 1250 miles

Next, we need to standardize the value 20,000 miles using the sampling distribution by calculating the z-score:

z = (x - μ) / SE

where x is the value we want to find the probability for, μ is the population mean, and SE is the standard error of the mean.

In this case, x = 20,000 miles, μ = 23,000 miles, and SE = 1250 miles.

z = (20,000 - 23,000) / 1250 = -24000 / 1250 = -19.2

Using a standard normal distribution table or a calculator, we can find the probability that the z-score is less than -19.2, which is extremely close to 0. Therefore, the probability that the mean of a random sample of size n = 4 of tire lifetimes will be less than 20,000 miles is almost 0.

In order to answer these questions, we need to apply statistical concepts and formulas. Let's break down each question step by step:

a) To find the mean of the sampling distribution of the sample mean (X̄), we use the same mean as the population, which is 23,000 miles. Therefore, the mean of X̄ is also 23,000 miles.

b) The standard deviation of the sampling distribution of X̄ (σX̄) can be calculated using the formula: σX̄ = σ / √n, where σ is the standard deviation of the population (2500 miles) and n is the sample size.

Given that n is not specified, we can't calculate the exact value of σX̄. However, assuming n = 4 as mentioned in part c), we can calculate the standard deviation of the sampling distribution:
σX̄ = 2500 / √4 = 2500 / 2 = 1250 miles.

c) To find the probability that the mean of a random sample of size n = 4 will be less than 20,000 miles, we need to standardize the distribution using the Z-score formula.

The formula to calculate the Z-score is: Z = (X - μ) / (σ / √n), where X is the value we are interested in (20,000 miles), μ is the population mean (23,000 miles), σ is the population standard deviation (2500 miles), and n is the sample size (4).

Plugging in the values:
Z = (20,000 - 23,000) / (2500 / √4) = -3000 / 1250 = -2.4

To find the probability associated with a Z-score of -2.4, we can refer to a Z-table or use a statistical calculator. Using a Z-table, we find that the probability is approximately 0.0082, or 0.82%.

Therefore, the probability that the mean of a random sample of size n = 4 of tire lifetimes will be less than 20,000 miles is approximately 0.0082, or 0.82%.