Question

The mean cost of a five pound bag of shrimp is 40 dollars with a standard deviation of 8 dollars.

If a sample of 49 bags of shrimp is randomly selected, what is the probability that the sample mean would be less than 37.4 dollars? Round your answer to four decimal places.

Answer 1

The mean of the sample distribution of the sample mean is the same as the population mean, which is 40 dollars. The standard deviation of the sample distribution of the sample mean (also called the standard error) is given by:

standard error = standard deviation / sqrt(sample size) = 8 / sqrt(49) = 8 / 7

To find the probability that the sample mean would be less than 37.4 dollars, we need to standardize the sample mean using the standard error and then look up the probability from a standard normal distribution table. The z-score for a sample mean of 37.4 dollars is:

z = (37.4 - 40) / (8 / 7) = -1.225

Looking up this z-score in a standard normal distribution table, we find that the probability of getting a sample mean less than 37.4 dollars is 0.1103 (rounded to four decimal places). Therefore, the probability that the sample mean would be less than 37.4 dollars is 0.1103.

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Explanation: