The amount of detergent dispensed into bottles of liquid laundry detergent bottles for a particular brand is normally distributed with a mean of 81.5 ounces with a standard deviation of 2.4 ounces. If eighteen bottles are randomly chosen from the factory, what is the probability that the mean fill is more than 82.4 ounces?

Round your answer to at least three decimal places.

If eighteen randomly selected bottles had a mean of $82.4 \mathrm{oz}$, should the machine dispensing the detergent be recalibrated? In other words, is this an unusual event? If it is unusual, then the machine should be recalibrated.

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.