A large tank of fish from hatchery is being delivered to a lake. The hatchery claims that the mean length of fish in the tank is 15 inches, and the standard deviation is 2 inches. A random sample of 48 fish is taken from the tank. Let x bar be the mean sample length of these fish. What is the probability that x bar is within 0.5 inch of claimed population mean? (Round your answer to four decimal places.)

Step 1 ：Given the population mean (\(\mu\)) is 15 inches, the population standard deviation (\(\sigma\)) is 2 inches, and the sample size (\(n\)) is 48 fish. We want to find the probability that the sample mean (\(\bar{x}\)) is within 0.5 inch of the population mean.

Step 2 ：First, calculate the standard deviation of the sample mean, also known as the standard error (SE). The formula for the standard error is: \(SE = \frac{\sigma}{\sqrt{n}}\)

Step 3 ：Substitute the given values into the formula: \(SE = \frac{2}{\sqrt{48}} \approx 0.289\)

Step 4 ：Next, calculate the z-scores for the boundaries of the interval around the population mean. The formula for a z-score is: \(z = \frac{\bar{x} - \mu}{SE}\)

Step 5 ：For the lower boundary (14.5 inches), calculate the z-score: \(z_1 = \frac{14.5 - 15}{0.289} \approx -1.73\)

Step 6 ：For the upper boundary (15.5 inches), calculate the z-score: \(z_2 = \frac{15.5 - 15}{0.289} \approx 1.73\)

Step 7 ：Finally, find the probability that the z-score is between -1.73 and 1.73. This can be found using a standard normal distribution table or a calculator with a normal distribution function. The probability corresponds to the area under the standard normal curve between these two z-scores.

Step 8 ：The probability is: \(P(-1.73 < Z < 1.73) \approx 0.9164\)

Step 9 ：So, the probability that the sample mean is within 0.5 inch of the population mean is approximately 0.9164, or 91.64% when rounded to four decimal places.

Step 10 ：The final answer is \(\boxed{0.9164}\)