Find the specified probability.The number of hours per week that high school seniors spend on homework is normally distributed, with a mean of 10 hours and a standard deviation of 3 hours. 60 students are chosen at random.Find the probability that the mean number of hours spent on homework for this sample group is between 9.8 and 10.4.

To solve this problem, we will use the Central Limit Theorem which states that the sampling distribution of the sample means approaches a normal distribution as the sample size gets larger, regardless of the shape of the population distribution.

Step 1: Identify the given information. The population mean (μ) is 10 hours. The population standard deviation (σ) is 3 hours. The sample size (n) is 60 students.

Step 2: Calculate the standard error. The standard error (SE) is the standard deviation of the sampling distribution of the sample mean. It is calculated as σ/√n. So, SE = 3/√60 ≈ 0.3873.

Step 3: Convert the range of sample means to z-scores. A z-score is a measure of how many standard deviations an element is from the mean. For 9.8 hours, Z = (9.8 - 10) / 0.3873 ≈ -0.5164. For 10.4 hours, Z = (10.4 - 10) / 0.3873 ≈ 1.0328.

Step 4: Find the probability that the sample mean is between these z-scores. You can use a standard normal distribution table or a calculator to find the probabilities associated with these z-scores. The probability for Z = -0.5164 is 0.3023 (this is the probability that the sample mean is less than 9.8 hours). The probability for Z = 1.0328 is 0.8508 (this is the probability that the sample mean is less than 10.4 hours).

Step 5: Subtract the two probabilities to find the probability that the sample mean is between 9.8 and 10.4 hours. P(9.8 < X < 10.4) = P(X < 10.4) - P(X < 9.8) = 0.8508 - 0.3023 = 0.5485.

So, the probability that the mean number of hours spent on homework for this sample group is between 9.8 and 10.4 is approximately 0.5485 or 54.85%.