Step 1 :Given that the sample mean (X) is 82.4 ounces, the population mean (μ) is 81.5 ounces, the standard deviation of the population (σ) is 2.4 ounces, and the size of the sample (n) is 18.
Step 2 :We can calculate the z-score using the formula: \(Z = \frac{X - μ}{σ / \sqrt{n}}\)
Step 3 :Substituting the given values into the formula, we get: \(Z = \frac{82.4 - 81.5}{2.4 / \sqrt{18}} = 2.25\)
Step 4 :The z-score of 2.25 means that the sample mean is 2.25 standard deviations above the population mean.
Step 5 :To find the probability that the mean fill is more than 82.4 ounces, we need to find the area to the right of the z-score of 2.25 in the standard normal distribution. This is also known as the p-value.
Step 6 :Using a standard normal distribution table or a calculator, we find that the area to the left of z = 2.25 is approximately 0.9878.
Step 7 :Since we want the area to the right, we subtract this value from 1: \(1 - 0.9878 = 0.0122\)
Step 8 :\(\boxed{0.0122}\) or 1.2% is the probability that the mean fill is more than 82.4 ounces.
Step 9 :A common rule of thumb is that an event is considered unusual if its probability is less than 5%. Since 1.2% is less than 5%, this would be considered an unusual event, and it would be recommended to recalibrate the machine.