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The mean per capita consumption of milk per year is 133 liters with a variance of 576 .
If a sample of 195 people is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 3.62 liters? Round your answer to four decimal places.

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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 \(\boxed{0.9648}\).