Final answer:
A. 505 represents the 33rd percentile. B. The probability that a randomly selected student scored between 500 and 524 on the test is 73.89%. C. The probability that the mean score for a sample of 16 students is between 500 and 524 on the test is 39.28%.
Step-by-step explanation:
A. To find the score that represents the 33rd percentile on the test, we need to find the z-score corresponding to that percentile and then convert it back to the original score using the formula z = (x - mean) / standard deviation. Using the z-table, we find the z-score corresponding to the 33rd percentile is approximately -0.44. Plugging this value into the formula, we get -0.44 = (x - 512) / 15.6. Solving for x, we find x = -0.44 * 15.6 + 512 ≈ 505.36. Therefore, a score of approximately 505 represents the 33rd percentile on the test.
B. To find the probability that a randomly selected student scored between 500 and 524 on the test, we need to find the z-scores corresponding to these scores and then find the area under the normal distribution curve between those z-scores. Using the formula z = (x - mean) / standard deviation, the z-score corresponding to 500 is (500 - 512) / 15.6 ≈ -0.77, and the z-score corresponding to 524 is (524 - 512) / 15.6 ≈ 0.77. Using a z-table or a calculator, we find that the area under the curve between -0.77 and 0.77 is approximately 0.7389. Therefore, the probability that a randomly selected student scored between 500 and 524 on the test is 0.7389, or 73.89%.
C. To find the probability that the mean score for a sample of 16 students is between 500 and 524 on the test, we need to use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means for large sample sizes will be approximately normal, regardless of the shape of the population distribution. The mean of the sample means will be equal to the population mean, and the standard deviation of the sample means will be equal to the population standard deviation divided by the square root of the sample size. Therefore, the mean score for a sample of 16 students will still have a mean of 512, but the standard deviation will be 15.6 / sqrt(16) ≈ 3.9. We can then use the z-score formula to find the z-scores corresponding to 500 and 524, and find the area under the normal distribution curve between those z-scores using a z-table or a calculator. The probability can be found by subtracting the area to the left of 500 from the area to the left of 524. Performing these calculations, we find that the probability that the mean score for a sample of 16 students is between 500 and 524 on the test is approximately 0.3928, or 39.28%.
D. The probabilities from B and C are different because they are measuring different scenarios. In part B, we are finding the probability that a single randomly selected student scored between 500 and 524 on the test. In part C, we are finding the probability that the mean score for a sample of students is between 500 and 524 on the test. The Central Limit Theorem tells us that the distribution of sample means will be less spread out and closer to a normal distribution compared to the distribution of individual scores. Therefore, the probability of finding a mean score within a certain range will be higher than the probability of finding an individual score within the same range.