Question

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 47,225 miles, with a standard deviation of 3178 miles. What is the probability that the sample mean would be less than 47,050 miles in a sample of 208 tires if the manager is correct? Round your answer to four decimal places.

Answer 1

The probability that the sample mean of tire mileage is less than 47,050 miles in a sample of 208 tires is 0.2137, or 21.37% when rounded to four decimal places.

Step-by-step explanation:

To find the probability that the sample mean is less than 47,050 miles for a sample of 208 tires, given a population mean (μ) of 47,225 miles and a population standard deviation (σ) of 3,178 miles, we can use the Central Limit Theorem. This theorem allows us to treat the distribution of the sample mean as a normal distribution because the sample size is large (n ≥ 30). The standard error of the mean (SEM) is calculated using the formula SEM = σ / √n, where n is the sample size.

First, calculate the SEM:

1. 
   
2. SEM = 3,178 / √208
3. 
   
4. SEM ≈ 3,178 / 14.422
5. 
   
6. SEM ≈ 220.4 miles

Next, calculate the Z-score for 47,050 using the formula Z = (X - μ) / SEM, where X is the sample mean:

1. 
   
2. Z = (47,050 - 47,225) / 220.4
3. 
   
4. Z ≈ -175 / 220.4
5. 
   
6. Z ≈ -0.794

Using a Z-table, we find the probability corresponding to a Z-score of -0.794.

The probability (P) that the sample mean is less than 47,050 is the area under the standard normal curve to the left of Z = -0.794. We find this probability to be approximately 0.2137, which is 21.37% when converted to a percentage.

Therefore, the probability that the sample mean of tire mileage will be less than 47,050 miles in a sample of 208 tires is 0.2137, or 21.37% when rounded to four decimal places.