Meredith is a general surgeon who performs surgeries such as appendectomies and laparoscopic cholecystectomies. The average number of sutures that Meredith uses to close a patient is 37, and the standard deviation is 8. The distribution of number of sutures is right skewed.

32 files are randomly drawn from Meredith's patient population, and the number of sutures used to close each patient is noted. What is the probability that the mean number of sutures is more than 40?

The answer should be typed as a decimal with 4 decimal places as needed.

Answer:

Step-by-step explanation:

We can use the central limit theorem to approximate the distribution of the sample mean of sutures used by Meredith's patients.

According to the central limit theorem, the sample mean follows a normal distribution with a mean of the population mean (μ = 37) and a standard deviation of the population standard deviation divided by the square root of the sample size (σ/√n = 8/√32 = 1.414).

So, we need to calculate the probability of getting a sample mean of more than 40.

Z-score = (sample mean - population mean) / (population standard deviation / √sample size)

Z-score = (40 - 37) / (8 / √32) = 3.19

Using a standard normal distribution table or calculator, we find that the probability of getting a Z-score of 3.19 or more is approximately 0.0007.

Therefore, the probability that the mean number of sutures is more than 40 is 0.0007 (rounded to 4 decimal places).