The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 44,663 miles, with a standard deviation of 2594 miles.
What is the probability that the sample mean would differ from the population mean by less than 199 miles in a sample of 209 tires if the manager is correct? Round your answer to four decimal places.
Answer
So, the final answer is \(\boxed{0.7338}\) or 73.38% when rounded to four decimal places. This means that if the manager is correct, there is a 73.38% chance that the sample mean would differ from the population mean by less than 199 miles in a sample of 209 tires.
Step 1 ：Given the population mean \(\mu = 44663\) miles, standard deviation \(\sigma = 2594\) miles, sample size \(n = 209\) tires, and difference from the mean \(d = 199\) miles.
Step 2 ：We are asked to find the probability that the sample mean differs from the population mean by less than 199 miles. This means we are looking for \(P(-199 < X - \mu < 199)\).
Step 3 ：First, we calculate the standard error (SE), which is the standard deviation of the sampling distribution of the sample mean. The formula for standard error is \(SE = \sigma / \sqrt{n}\).
Step 4 ：Substituting the given values, we get \(SE = 2594 / \sqrt{209} \approx 179.32\).
Step 5 ：Next, we convert the difference from the mean (199 miles) to a z-score. The z-score is a measure of how many standard deviations an element is from the mean. The formula for the z-score is \(Z = (X - \mu) / SE\).
Step 6 ：Substituting the values for the lower limit (-199), we get \(Z1 = (-199 - 0) / 179.32 \approx -1.11\).
Step 7 ：And for the upper limit (199), we get \(Z2 = (199 - 0) / 179.32 \approx 1.11\).
Step 8 ：Now, we find the probability that the z-score is between -1.11 and 1.11. We can look up these values in the z-table (also known as the standard normal distribution table).
Step 9 ：The value for \(Z1 = -1.11\) is 0.1331 and for \(Z2 = 1.11\) is 0.8669.
Step 10 ：So, the probability that the sample mean would differ from the population mean by less than 199 miles is \(P(-1.11 < Z < 1.11) = P(Z < 1.11) - P(Z < -1.11) = 0.8669 - 0.1331 = 0.7338\).
Step 11 ：So, the final answer is \(\boxed{0.7338}\) or 73.38% when rounded to four decimal places. This means that if the manager is correct, there is a 73.38% chance that the sample mean would differ from the population mean by less than 199 miles in a sample of 209 tires.
