Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average (mean) of 32 grams of fat per pound, with a standard deviation of 8 grams of fat per pound. A random sample of 36 farm-raised trout is selected. The mean fat content for the sample is 30.7 grams per pound. Find the probability of observing a sample mean of 30.7 grams of fat per pound or less in a random sample of 36 farm-raised trout. Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.

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Accepted Answer

Step:1 ：This problem involves the concept of sampling distributions and the Central Limit Theorem. The Central Limit Theorem states that if you have a population with mean (mu) and standard deviation (sigma) and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large (usually n > 30).Step:2 ：In this case, we are given the population mean ((mu = 32) grams), the population standard deviation ((sigma = 8) grams), the sample size (n = 36), and the sample mean ((bar{x} = 30.7) grams). We are asked to find the probability of observing a sample mean of 30.7 grams or less.Step:3 ：To solve this, we can use the formula for the z-score, which is ((bar{x} - mu) / (sigma / sqrt{n})). The z-score tells us how many standard deviations an element is from the mean. Once we have the z-score, we can look up the corresponding probability in the standard normal distribution table or use a function to calculate it.Step:4 ：Calculating the z-score gives us -0.975. This indicates that the sample mean of 30.7 grams is approximately 0.975 standard deviations below the population mean.Step:5 ：The probability of observing a sample mean of 30.7 grams or less is approximately 0.165 (or 16.5% when expressed as a percentage).Step:6 ：Final Answer: The probability of observing a sample mean of 30.7 grams of fat per pound or less in a random sample of 36 farm-raised trout is approximately (boxed{0.165}).

Answer

Step: 1 ：This problem involves the concept of sampling distributions and the Central Limit Theorem. The Central Limit Theorem states that if you have a population with mean (mu) and standard deviation (sigma) and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large (usually n > 30).Step: 2 ：In this case, we are given the population mean ((mu = 32) grams), the population standard deviation ((sigma = 8) grams), the sample size (n = 36), and the sample mean ((bar{x} = 30.7) grams). We are asked to find the probability of observing a sample mean of 30.7 grams or less.Step: 3 ：To solve this, we can use the formula for the z-score, which is ((bar{x} - mu) / (sigma / sqrt{n})). The z-score tells us how many standard deviations an element is from the mean. Once we have the z-score, we can look up the corresponding probability in the standard normal distribution table or use a function to calculate it.Step: 4 ：Calculating the z-score gives us -0.975. This indicates that the sample mean of 30.7 grams is approximately 0.975 standard deviations below the population mean.Step: 5 ：The probability of observing a sample mean of 30.7 grams or less is approximately 0.165 (or 16.5% when expressed as a percentage).Step: 6 ：Final Answer: The probability of observing a sample mean of 30.7 grams of fat per pound or less in a random sample of 36 farm-raised trout is approximately (boxed{0.165}).

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