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a survey of 250 lobster fishermen found that they catch an average of 32.0 pounds of lobster per day with a standard deviation of 4.0 pounds. if a random sample of 36 lobster fishermen is selected, what is the probability that their average catch is less than 31.5 pounds? assume the distribution of the weights of lobster is normal. group of answer choices

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Answer: The probability that a random sample of 36 lobster fishermen will have an average catch of less than 31.5 pounds is approximately 0.2266 or 22.66%.

Explanation:

We can use the central limit theorem to approximate the sampling distribution of the sample mean. The mean of the sample means is equal to the population mean, which is 32.0 pounds, and the standard deviation of the sample means is equal to the population standard deviation divided by the square root of the sample size, which is 4.0 pounds divided by the square root of 36, or 0.67 pounds.

To find the probability that the average catch of a random sample of 36 lobster fishermen is less than 31.5 pounds, we need to convert this value to a z-score using the formula:

z = (x - mu) / (sigma / sqrt(n))

where x is the sample mean, mu is the population mean, sigma is the population standard deviation, and n is the sample size.

Substituting the values we get:

z = (31.5 - 32.0) / (4.0 / sqrt(36))

z = -0.75

Using a standard normal distribution table or calculator, we can find that the probability of getting a z-score of -0.75 or less is 0.2266.

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