A sample of size 200 is to be taken at random. Given that the population proportion is 0.60, the probability that the sample proportion will be greater than 0.58 is:随机抽取 200 号样本。假设总体比例为 0.60，则样本比例大于 0.58 的概率为：

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BlackTom AI · Accepted Answer

We need to compare the sample proportion p̂ to the population proportion p = 0.60 with a sample size n = 200. First, recognize that for a large sample, p̂ is approximately normal with mean p and standard deviation sqrt[p(1−p)/n] = sqrt(0.60×0.40/200) = sqrt(0.24/200) = sqrt(0.0012) ≈ 0.03464.

Now standardize the target value 0.58:
Z = (p̂ − p) / sqrt[p(1−p)/n] → P(p̂ > 0.58) ≈ P(Z > (0.58 − 0.60)/0.03464) = P(Z > −0.02/0.03464) ≈ P(Z > −0.577).

Because the standard normal distribution is symmetric, P(Z > −0.577) = P(Z < 0.577). Looking up or recalling a standard normal CDF at 0.577 gives about 0.718. So the probability is approximately 0.718, which rounds to 0.719.

Evaluating the options:
- Option a: 0.281 would be the probability of the complement event (p̂ ≤ 0.58) and is not correct for P(p̂ > 0.58).
- Option b: 0.719 matches the calculated probability, based on the normal approximation.
- Option c: 0.580 is not aligned with the standard normal probability we computed for this threshold.
- Option d: 0.762 is farther from the computed value and does not reflect the calculated z-score of −0.577.

In summary, the normal-approximation computation yields a probability near 0.719, making option b the closest match to the result.

COMM111_S2 Quiz 6: Estimation 测验 6：估计

A sample of size 200 is to be taken at random. Given that the population proportion is 0.60, the probability that the sample proportion will be greater than 0.58 is:随机抽取 200 号样本。假设总体比例为 0.60，则样本比例大于 0.58 的概率为：

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