A manufacturer claims that the life span of its tires is 50,000 miles. You work for a consumer protection agency and you are testing this manufacturer's tires. Assume the life spans of the tires are normally distributed. You select 100 tires at random and test them. THe mean life span is 49,271 miles. Assume ó = 800 miles. Assuming the manufacturer's claim is correct, what is the probability the mean of the sample is 49,271 miles or less.

- I used the central limit theorem to answer the question..I'm not sure of my answer. I just wanted to check it with somebody else's answer to make sure I did it right ...Thank you for taking time to read this.

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m>n

40

20

huh?

To find the probability that the mean of the sample is 49,271 miles or less, we can use the standard normal distribution and the Z-score.

The formula for the Z-score is Z = (X - μ) / (σ / √n), where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

In this case, X (sample mean) is 49,271 miles, μ (population mean) is 50,000 miles, σ (population standard deviation) is 800 miles, and n (sample size) is 100.

First, let's calculate the Z-score:

Z = (X - μ) / (σ / √n)
= (49,271 - 50,000) / (800 / √100)
= -729 / 80
= -9.1125

Now, we need to find the probability corresponding to this Z-score using a standard normal distribution table or a statistical software.

Looking up the Z-score of -9.1125 in a standard normal distribution table, we find that the probability is essentially zero, as the Z-score is extremely low.

Therefore, the probability that the mean of the sample is 49,271 miles or less, assuming the manufacturer's claim is correct, is nearly zero.

It is worth noting that this probability indicates that it is highly unlikely for the sample mean to be at or below 49,271 miles if the manufacturer's claim of a 50,000-mile life span is true.