Final answer:
Using the normal distribution and z-scores, we find the probability that the average hospital stay is no more than 6.5 days for a sample of 50 patients. For a smaller sample of 8 patients, the population must be assumed to be approximately normal to apply the same method, and the probability would decrease due to an increased standard error.
Step-by-step explanation:
To calculate the probability that the average length of stay for a sample of 50 patients is no more than 6.5 days, we use the normal distribution and z-scores. Given the population mean (μ) is 5.5 days and the population standard deviation (σ) is 2.5 days, we calculate the z-score for 6.5 days using the formula Z = (X - μ) / (σ / √ n), where X is the sample mean and n is the sample size.
For n = 50, Z = (6.5 - 5.5) / (2.5 / √ 50) = 1 / (2.5 / 7.071) ≈ 2.828. We can then look up this z-score in the standard normal distribution table or use a calculator to find the probability associated with this z-score, which gives us the probability of the average length of stay being no more than 6.5 days.
If the sample size were decreased to 8 patients, we would need to assume that the population distribution is approximately normal in order to use the normal distribution as a model, because the central limit theorem is not as robust for smaller samples. This assumption allows us to use the sample mean as a normally distributed variable even with a small sample size. The relevant options for the change in probability with a smaller sample size would be a decrease (option b), as the standard error increases with smaller sample sizes, making it more difficult for the sample mean to be far from the population mean.