Final answer:
a. The proportion of students who would get higher than 80 on the exam is approximately 0.1056 (or 10.56%). b. The mean of the sampling distribution is 75 and the standard deviation is approximately 0.82. c. The proportion of samples of size 30 that would have an average score higher than 80 is approximately 0.00007 (or 0.007%).
Step-by-step explanation:
a. To find the proportion of students who would get higher than 80 on the exam, we need to calculate the z-score for a score of 80 and then find the area under the normal curve to the right of that z-score. The z-score formula is: z = (x - mean) / standard deviation. Plugging in the values, we get: z = (80 - 75) / 4 = 1.25. Using a standard normal distribution table or calculator, we can find that the proportion of students who would get higher than 80 is approximately 0.1056 (or 10.56%).
b. The mean of the sampling distribution is equal to the population mean, which is 75. The standard deviation of the sampling distribution is equal to the population standard deviation divided by the square root of the sample size. So, the standard deviation of the sampling distribution is 4.5 / sqrt(30) ≈ 0.82.
c. To find the proportion of samples of size 30 that would have an average score higher than 80, we can use the z-score formula again. The z-score is calculated using the sample mean, so z = (80 - 75) / (4.5 / sqrt(30)) ≈ 3.82. Using a standard normal distribution table or calculator, we can find that the proportion of samples with an average score higher than 80 is approximately 0.00007 (or 0.007%).