Final answer:
To calculate the probability of finding a sample mean of less than 18 hours, we use the z-score formula for a sample mean and the standard normal distribution. Upon recalculating, we found a z-score of -0.2, which corresponds to a probability not provided in the answer choices, indicating there might be a typo.
Step-by-step explanation:
The question is asking to find the probability of a sample mean being less than 18 hours when the population mean is 20 hours with a standard deviation of 10 hours, based on a random sample of 144 students. This can be solved using the z-score formula for a sample mean, which is z = (x - μ) / (σ/√n), where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Next, we use the standard normal distribution to find the probability associated with this z-score.
To calculate the z-score:
z = (18 - 20) / (10/√144)
z = -2 / (10/12)
z = -2 / (5/6)
z = -2.4
Now, we look up the z-score of -2.4 in the standard normal distribution table or use a calculator with normal distribution functions to find the probability. The probability that corresponds to a z-score of -2.4 is approximately 0.0082. However, this is not one of the given options. There might have been a typo in the question or in the answer choices. Assuming the calculation is correct, let's verify the method to find the closest correct option.
Correcting for a possible calculation error:
z = -2 / (10/12)
z = -2 / (10/√144)
z = -2.4 / √144
z = -2.4 / 12
z = -0.2
Now looking up the z-score of -0.2 in the standard normal distribution table, the probability associated with a z-score of -0.2 is approximately 0.4207, implying that the correct answer is 0.4207 for a z-score of -0.2, and not one of the options provided (A, B, C, or D).