📚 The Presidency: Statistical Models for Election Polling | 总统制：选举民调的统计模型
In A-Level Mathematics, statistical ideas are frequently tested through real-world contexts such as presidential approval ratings, election polls and voting behaviour. This article uses the theme of ‘the presidency’ to revise key Edexcel Statistics topics: sampling, the binomial distribution, normal approximation, confidence intervals, hypothesis testing, correlation and regression. Each section pairs an English explanation with a Chinese explanation so that you can strengthen both subject knowledge and academic language.
在 A-Level 数学中，统计思想经常通过总统支持率、选举民调和投票行为等现实情境进行考查。本文以 “总统制” 为主题，复习 Edexcel 统计学部分的核心考点：抽样、二项分布、正态近似、置信区间、假设检验、相关性与回归。每一节均采用英文与中文配对讲解，帮助你同时巩固学科知识与学术语言。
1. Sampling and Bias in Presidential Polling | 总统民调中的抽样与偏差
A national opinion poll for a president usually cannot ask every voter, so a sample must be selected. Edexcel exam questions often ask you to identify simple random sampling, stratified sampling, quota sampling or cluster sampling. A simple random sample gives every voter in the population an equal chance of being chosen, but it can be expensive and time-consuming across a whole country.
针对总统的全国民意调查通常无法询问每一位选民，因此必须抽取样本。Edexcel 考题经常要求你识别简单随机抽样、分层抽样、配额抽样或整群抽样。简单随机样本使总体中每位选民被选中的机会相同，但在全国范围内实施可能成本高且耗时。
Stratified sampling is more useful in presidential polling because the population can be divided into groups by age, region, income or education. A fixed proportion is then sampled from each group, which reduces bias and ensures that key demographic groups are represented in the correct proportions.
分层抽样在总统民调中更为实用，因为总体可以按年龄、地区、收入或教育程度分组。然后从每组中按固定比例抽样，这样可以减少偏差，并确保关键人口群体按正确比例被代表。
A common source of bias is a sampling frame that excludes voters who use only mobile phones or who rarely respond to surveys. This selection bias can make a poll overestimate or underestimate support for a president.
一个常见的偏差来源是抽样框排除了只使用手机或很少回应调查的选民。这种选择偏差可能使民调高估或低估对某位总统的支持率。
2. Binomial Model for Voter Support | 选民支持率的二项模型
When a sample of n voters is taken and each voter either supports the president or does not, the number of supporters X can be modelled by a binomial distribution. We write X ~ B(n, p), where p is the probability that a randomly chosen voter supports the president.
当抽取一个包含 n 位选民的样本，且每位选民要么支持总统要么不支持时，支持者人数 X 可以用二项分布来建模。我们记作 X ~ B(n, p)，其中 p 是随机抽取的一位选民支持总统的概率。
The probability of obtaining exactly k supporters is given by the binomial formula:
恰好得到 k 位支持者的概率由二项分布公式给出：
P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ
In this formula, C(n, k) is the binomial coefficient, often read as ‘n choose k’. The binomial model requires a fixed number of independent trials, only two possible outcomes, and a constant probability p for each trial.
在此公式中，C(n, k) 是二项系数，通常读作 “n 选 k”。二项模型要求试验次数固定、各次试验相互独立、每次试验只有两种可能结果，且每次试验的概率 p 保持不变。
For example, if n = 20 and p = 0.45, the probability that exactly 9 voters support the president is P(X = 9) = C(20, 9) × 0.45⁹ × 0.55¹¹ ≈ 0.177. You may calculate this using the nCr button on a calculator.
例如，若 n = 20 且 p = 0.45，则恰好有 9 位选民支持总统的概率为 P(X = 9) = C(20, 9) × 0.45⁹ × 0.55¹¹ ≈ 0.177。你可以使用计算器上的 nCr 按钮进行计算。
3. Normal Approximation to the Binomial | 二项分布的正态近似
When the sample size n is large, the binomial distribution becomes difficult to calculate directly. Edexcel requires you to use a normal approximation when both np and n(1 − p) are greater than 5. In this case, X is approximately normal with mean np and variance np(1 − p).
当样本量 n 很大时，直接用二项分布计算会变得困难。Edexcel 要求当 np 和 n(1 − p) 都大于 5 时使用正态近似。此时 X 近似服从均值为 np、方差为 np(1 − p) 的正态分布。
X ~ N(np, np(1 − p)) approximately
Because a continuous distribution is used to approximate a discrete one, a continuity correction is required. For example, P(X ≤ 414) is approximated by P(X < 414.5), and P(X ≥ 415) is approximated by P(X > 414.5).
由于用连续分布近似离散分布，需要进行连续性校正。例如，P(X ≤ 414) 近似为 P(X < 414.5)，而 P(X ≥ 415) 近似为 P(X > 414.5)。
Suppose a poll samples n = 1000 voters and the true support rate is p = 0.52. Then the number of supporters X has mean 1000 × 0.52 = 520 and standard deviation √(1000 × 0.52 × 0.48) = √249.6 ≈ 15.80. This normal model can estimate the probability of observing a particular poll result.
假设一项民调抽取 n = 1000 位选民，真实支持率 p = 0.52。则支持者人数 X 的均值为 1000 × 0.52 = 520，标准差为 √(1000 × 0.52 × 0.48) = √249.6 ≈ 15.80。该正态模型可以估计观察到某一民调结果的概率。
4. Confidence Interval for a Proportion | 比例的置信区间
A sample proportion p̂ = X / n is an estimate of the true population proportion p. Because sample results vary, Edexcel often asks for a confidence interval for p. For a large sample, an approximate 95% confidence interval is given by:
样本比例 p̂ = X / n 是总体真实比例 p 的估计值。由于样本结果存在变异，Edexcel 经常要求计算 p 的置信区间。对于大样本，近似的 95% 置信区间由下式给出：
p̂ ± z × √( p̂(1 − p̂) / n )
For a 95% confidence level, the z value is 1.96. If a presidential poll of n = 1200 voters gives p̂ = 0.48, then the standard error is √(0.48 × 0.52 / 1200) ≈ 0.0144. The 95% confidence interval is 0.48 ± 1.96 × 0.0144, which gives approximately (0.452, 0.508).
对于 95% 置信水平，z 值为 1.96。如果一项总统民调 n = 1200 位选民得到 p̂ = 0.48，则标准误为 √(0.48 × 0.52 / 1200) ≈ 0.0144。95% 置信区间为 0.48 ± 1.96 × 0.0144，约为 (0.452, 0.508)。
The correct interpretation is that if many samples of 1200 voters were taken, about 95% of the intervals calculated would contain the true proportion p. It does not mean there is a 95% chance that the true p lies in this single interval.
正确的解释是：如果重复抽取许多个 1200 位选民的样本，大约 95% 计算出的区间会包含真实比例 p。这并不意味着真实 p 有 95% 的概率落在这一个具体区间内。
5. Hypothesis Test for a Claimed Approval Rating | 对声称支持率的假设检验
A political commentator claims that the president’s approval rating is 50%. A polling organisation wants to test this claim. A sample of n = 900 voters is taken, and X = 414 support the president. We test H₀: p = 0.5 against H₁: p ≠ 0.5 at the 5% significance level.
一位政治评论员声称总统的支持率为 50%。一家民调机构想检验这一说法。抽取 n = 900 位选民，其中 X = 414 位支持总统。我们在 5% 显著性水平下检验 H₀: p = 0.5 对 H₁: p ≠ 0.5。
The sample proportion is p̂ = 414 / 900 = 0.46. Under H₀, X ~ B(900, 0.5), which is approximately normal with mean 900 × 0.5 = 450 and standard deviation √(900 × 0.5 × 0.5) = 15.
样本比例为 p̂ = 414 / 900 = 0.46。在 H₀ 下，X ~ B(900, 0.5)，近似服从均值为 900 × 0.5 = 450、标准差为 √(900 × 0.5 × 0.5) = 15 的正态分布。
Using a continuity correction, P(X ≤ 414) is approximated by P(X < 414.5). The z score is (414.5 − 450) / 15 = −2.37. For a two-tailed test, the p-value is 2 × P(Z < −2.37) ≈ 0.0178. Since 0.0178 < 0.05, we reject H₀ at the 5% level and conclude that there is sufficient evidence that the approval rating is not 50%.
使用连续性校正，P(X ≤ 414) 近似为 P(X < 414.5)。z 分数为 (414.5 − 450) / 15 = −2.37。对于双尾检验，p 值为 2 × P(Z < −2.37) ≈ 0.0178。由于 0.0178 < 0.05，我们在 5% 水平下拒绝 H₀，得出结论：有足够证据表明支持率不是 50%。
6. Correlation Between Approval Rating and Unemployment | 支持率与失业率的相关性
Presidential approval ratings often move in response to economic indicators such as unemployment. Edexcel requires you to calculate and interpret the product moment correlation coefficient r. A value of r close to 1 indicates strong positive correlation, while r close to −1 indicates strong negative correlation.
总统支持率往往随失业率等经济指标的变化而波动。Edexcel 要求你计算并解释积矩相关系数 r。r 接近 1 表示强正相关，r 接近 −1 表示强负相关。
Given paired data (xᵢ, yᵢ), where x is unemployment rate and y is approval rating, the formula for r is:
给定成对数据 (xᵢ, yᵢ)，其中 x 为失业率，y 为支持率，r 的公式为：
r = Sxy / √( Sxx × Syy )
Here Sxx = Σ(xᵢ − x̄)², Syy = Σ(yᵢ − ȳ)² and Sxy = Σ(xᵢ − x̄)(yᵢ − ȳ). If r = −0.78, there is a moderately strong negative correlation: as unemployment rises, presidential approval tends to fall.
这里 Sxx = Σ(xᵢ − x̄)²，Syy = Σ(yᵢ − ȳ)²，Sxy = Σ(xᵢ − x̄)(yᵢ − ȳ)。如果 r = −0.78，则存在中等偏强的负相关：随着失业率上升，总统支持率往往下降。
You must be careful not to claim causation from correlation. A high or low r only shows a linear association; it does not prove that unemployment changes cause approval changes.
你必须注意不能从相关性推断因果关系。r 值高或低只能说明存在线性关联，并不能证明失业率的变化导致了支持率的变化。
7. Regression Line for Approval Ratings Over Time | 支持率随时间变化的回归线
When a linear relationship is appropriate, Edexcel expects you to find the regression line of y on x. The least squares regression line has equation y = a + bx, where:
当线性关系适用时，Edexcel 要求你求出 y 对 x 的回归线。最小二乘回归线的方程为 y = a + bx，其中：
b = Sxy / Sxx
a = ȳ − b x̄
For example, suppose x is the number of months since inauguration and y is the approval rating. If the equation is y = 54 − 0.35x, the intercept 54 means the estimated approval rating at month 0 is 54%, and the slope −0.35 means the approval rating falls by 0.35 percentage points on average each month.
例如，假设 x 为自就职以来的月数，y 为支持率。如果方程为 y = 54 − 0.35x，则截距 54 表示在第 0 个月的估计支持率为 54%，斜率 −0.35 表示平均每月支持率下降 0.35 个百分点。
This regression line can be used to estimate y for a given x, but extrapolation beyond the data range is unreliable. You should only use the line for interpolation within the observed range of x values.
该回归线可用于由给定的 x 估计 y，但在数据范围之外进行外推是不可靠的。你只应在已观测到的 x 值范围内使用该直线进行内插。
8. Conditional Probability in Election Scenarios | 选举情形中的条件概率
Presidential election surveys often record voters by region and candidate preference. Conditional probability is used to find the chance that a voter supports the president given that they live in an urban area. The formula is:
总统选举调查常按地区和候选人偏好记录选民信息。条件概率用于求在已知某选民居住在城市地区的条件下，其支持总统的概率。公式为：
P(A | B) = P(A ∩ B) / P(B)
Suppose a survey of 800 voters finds that 360 support the president and 200 of those supporters are urban. If 320 voters are urban, then the probability that a voter supports the president given that they are urban is P(Support | Urban) = P(Support ∩ Urban) / P(Urban) = 200 / 320 = 0.625.
假设一项对 800 位选民的调查发现 360 人支持总统，其中 200 位支持者来自城市地区。若共有 320 位城市选民，则在已知某选民来自城市地区的条件下其支持总统的概率为 P(Support | Urban) = P(Support ∩ Urban) / P(Urban) = 200 / 320 = 0.625。
Tree diagrams are very helpful for conditional probability questions. Label the first set of branches with marginal probabilities and the second set with conditional probabilities, then multiply along branches to find joint probabilities.
树状图对条件概率问题非常有帮助。第一层分支标出边际概率，第二层分支标出条件概率，然后沿分支相乘即可得到联合概率。
9. Expected Value and Variance of Electoral Outcomes | 选举结果的期望值与方差
In a simplified model, the number of swing states won by a president can be treated as a binomial random variable. If there are n = 10 swing states and the probability of winning each is p = 0.55, then the
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导