The probability that the sample mean score is greater than 550 is 0.002. The 80% percentile of the sample mean is approximately 525.4.
To answer part A of the question, we need to calculate the z-score for the sample mean score of 550. The formula for the z-score is given by:
z = (x - μ) / (σ / √n)
where x is the sample mean score, μ is the population mean score, σ is the population standard deviation, and n is the sample size. Plugging in the values, we have:
z = (550 - 514) / (118 / √65) = 2.797
We can then use a standard normal distribution table or a calculator to find the probability associated with a z-score of 2.797, which is approximately 0.002. Therefore, the probability that the sample mean score is greater than 550 is 0.002.
For part B of the question, we need to find the 80% percentile of the sample mean. The 80% percentile represents the score below which 80% of the sample means fall. Since the sample means follow a normal distribution, we can use the z-score to find the corresponding score. Using a z-score table or calculator, we find that a z-score of 0.842 corresponds to the 80% percentile. The score associated with this z-score can be calculated using the formula:
x = μ + (z * (σ / √n))
Plugging in the values, we have:
x = 514 + (0.842 * (118 / √65)) ≈ 525.4
Therefore, the 80% percentile of the sample mean is approximately 525.4.