The mean cost of a five pound bag of shrimp is 40 dollars with a standard deviation of 8 dollars.
If a sample of 51 bags of shrimp is randomly selected, what is the probability that the sample mean would be less than 42.7 dollars? Round your answer to four decimal places.

Answer How to enter your answer (opens in new window) 2 Points

Step 1 ：We are given a problem of probability involving the normal distribution. The Central Limit Theorem tells us that the distribution of sample means will be approximately normally distributed if the sample size is large enough (usually n > 30).

Step 2 ：We are given the population mean (\(\mu = 40\)), the population standard deviation (\(\sigma = 8\)), the sample size (\(n = 51\)), and we are asked to find the probability that the sample mean (\(\bar{x}\)) is less than 42.7.

Step 3 ：To solve this, we need to standardize the sample mean using the Z-score formula: \(Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\)

Step 4 ：Substituting the given values into the formula, we get \(Z = \frac{42.7 - 40}{8 / \sqrt{51}} = 2.4102320946332147\)

Step 5 ：Then, we can use the cumulative distribution function (CDF) of the standard normal distribution to find the probability that Z is less than the calculated Z-score. The probability is 0.992

Step 6 ：Final Answer: The probability that the sample mean would be less than 42.7 dollars is \(\boxed{0.992}\)