Step 1 :We are given that the population mean (\(\mu\)) is 133 liters, the population variance (\(\sigma^2\)) is 576 liters^2, and the sample size (n) is 195. This implies that the standard deviation (\(\sigma\)) is \(\sqrt{576}\) = 24 liters.
Step 2 :We are asked to find the probability that the sample mean differs from the true mean by less than 3.62 liters. This means we want to find P(129.38 < \(\bar{X}\) < 136.62).
Step 3 :The standard deviation of the sampling distribution (standard error) is given by \(\sigma/\sqrt{n}\).
Step 4 :We can standardize the sample means to z-scores using the formula z = (\(\bar{X}\) - \(\mu\)) / (\(\sigma/\sqrt{n}\)), and then use the standard normal distribution (z-distribution) to find the probabilities.
Step 5 :Calculating the z-scores for 129.38 and 136.62, we get approximately -2.106 and 2.106 respectively.
Step 6 :Using the standard normal distribution, we find that the probabilities corresponding to these z-scores are approximately 0.0176 and 0.9824 respectively.
Step 7 :The probability that the sample mean would differ from the true mean by less than 3.62 liters is the difference of these two probabilities, which is approximately 0.9648.
Step 8 :Final Answer: The probability that the sample mean would differ from the true mean by less than 3.62 liters is approximately \(\boxed{0.9648}\).