Final answer:
The sampling distribution of the sample proportion has a mean of the population proportion (0.02) and a standard deviation calculated using the square root of (pq/n). With a large sample size and np and nq greater than 5, the distribution approximates a normal distribution according to the Central Limit Theorem.
Step-by-step explanation:
The sampling distribution of the sample proportion measures how the proportion of a given sample compares to the overall population proportion. In this case, we are examining the proportion of students who drink more than 3 cups of coffee each day. Given that 2% (0.02) of students is the claimed population proportion, the sample of 123 students with 3 drinking more than 3 cups yields a sample proportion (p') of 3/123 = 0.0244. The standard error (SE) of p' is calculated using the formula SE = sqrt(pq/n), where p is the population proportion, q = 1 - p, and n is the sample size.
Using the provided figures, we have:
- Population proportion (p) = 0.02
- q = 1 - p = 0.98
- Sample size (n) = 123
Therefore, the SE is sqrt(0.02 * 0.98 / 123). The mean of the sampling distribution is the population proportion (0.02), and the standard deviation is the calculated SE.
The Central Limit Theorem implies that for large samples, the sampling distribution of the sample proportion will approximate a normal distribution, particularly when np and nq are both greater than 5. Since the sample size here is fairly large (123) and both np (2.46) and nq (120.54) are greater than 5, it is reasonable to use the normal approximation for the sampling distribution of p'.