📚 AS Mathematics Unit 3 June 2019 Paper Walkthrough – Key Knowledge Points | AS数学单元3 2019年6月真题知识点精讲
The June 2019 AS Mathematics Unit 3 paper is a typical assessment covering the core topics of the Statistics 1 module: data description, probability, discrete random variables, the normal distribution, correlation and regression, and statistical interpretation. This article takes you through the key knowledge points tested, linking each section to a representative question from the paper and providing clear step‑by‑step worked solutions. Whether you are preparing for a mock exam or revising the fundamentals, the focused explanations in both English and Chinese will help you master the essential techniques.
2019年6月AS数学单元3试卷是统计1模块的典型测评，覆盖数据描述、概率、离散随机变量、正态分布、相关与回归以及统计解释等核心主题。本文逐节梳理所考查的知识点，每部分对应一道真题示例，并给出清晰的分步求解过程。不论你是在准备模拟考还是复习基础，这些中英双语的精析将帮助你掌握关键方法。
1. Summarising Data: Mean, Median and Standard Deviation | 数据概括：均值、中位数和标准差
The first question on the June 2019 paper provided a set of ten observations and asked for the mean, median, interquartile range and sample standard deviation. For example, the data were: 12, 15, 21, 8, 14, 18, 25, 10, 19, 22. To tackle this, you must recall the fundamental formulae.
试卷第一题给出了一组10个观测值，要求计算均值、中位数、四分位距和样本标准差。例如数据为：12, 15, 21, 8, 14, 18, 25, 10, 19, 22。要解答此题，必须熟记基本公式。
The sample mean is computed as x̄ = Σx/n. The sample variance uses s² = Σ(x − x̄)²/(n − 1), and the standard deviation is s = √s². The median position is (n+1)/2, and quartiles are found using linear interpolation when necessary.
样本均值的计算公式为 x̄ = Σx/n。样本方差为 s² = Σ(x − x̄)²/(n − 1)，标准差 s = √s²。中位数的位置是 (n+1)/2，必要时需用线性插值法求四分位数。
x̄ = Σx/n s² = (Σx² − (Σx)²/n) / (n−1) s = √s²
For the sample above, Σx = 164, so x̄ = 16.4. After ordering the data, the median (5.5th value) is 16.5. The lower quartile Q₁ is the 3rd value (12) and the upper quartile Q₃ is the 8th value (21), giving an IQR of 9. The standard deviation works out to approximately 5.46.
对于上述样本，Σx = 164，因而 x̄ = 16.4。将数据排序后，中位数（第5.5个值）为 16.5。下四分位数 Q₁ 是第3个值 (12)，上四分位数 Q₃ 是第8个值 (21)，得到四分位距为 9。标准差约为 5.46。
2. Histograms and Frequency Density | 直方图与频率密度
A second question presented a grouped frequency table with unequal class widths, such as mass of packages. Students had to draw a histogram and then estimate the median and modal class from it. The crucial skill is calculating frequency density = frequency ÷ class width.
试卷第二题给出一个组距不等的数据分组表，例如包裹的质量。考生要绘制直方图，并据此估算中位数和众数所在组。核心技能是计算频率密度 = 频数 ÷ 组距。
For a class 10–15 with frequency 12, the class width is 5, so the frequency density is 12/5 = 2.4. The histogram area represents frequency, so the total area equals the total number of observations. To estimate the median from the histogram, you draw a vertical line that splits the total area in half and then interpolate linearly within the median class.
若区间 10–15 的频数为 12，组距为 5，则频率密度为 12/5 = 2.4。直方图的面积代表频数，故总面积等于观测总数。为了从直方图估算中位数，可画一条将总面积平分的竖线，然后在中位数组内进行线性插值。
Frequency density = Frequency / Class width
You also need to remember that in a histogram, the vertical axis is always frequency density, not frequency. When estimating percentiles, use the cumulative frequency up to each class boundary.
还需记住，直方图的纵轴始终是频率密度而非频数。估算百分位数时，利用截至每组边界的累积频数。
3. Probability Basics: Union, Intersection and Independence | 概率基础：并集、交集与独立性
A typical short question in this paper involved events A and B with P(A) = 0.4, P(B) = 0.35 and P(A ∪ B) = 0.62. Candidates had to find P(A ∩ B), determine whether A and B are mutually exclusive or independent, and calculate P(A’ ∩ B’).
试卷中一道典型的短答题涉及事件 A 和 B，已知 P(A) = 0.4，P(B) = 0.35，P(A ∪ B) = 0.62。考生需求出 P(A ∩ B)，判断 A 与 B 是否互斥或独立，并计算 P(A’ ∩ B’)。
Use the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Substituting the values gives P(A ∩ B) = 0.4 + 0.35 − 0.62 = 0.13. Since P(A ∩ B) ≠ 0, the events are not mutually exclusive. For independence, check if P(A ∩ B) = P(A) × P(B). Here 0.4 × 0.35 = 0.14, which is not equal to 0.13, so A and B are not independent.
利用加法公式：P(A ∪ B) = P(A) + P(B) − P(A ∩ B)。代入数值得 P(A ∩ B) = 0.4 + 0.35 − 0.62 = 0.13。因为 P(A ∩ B) ≠ 0，事件不互斥。检验独立性：若 P(A ∩ B) = P(A) × P(B) 则独立。此处 0.4 × 0.35 = 0.14 ≠ 0.13，故 A 与 B 不独立。
Finally, P(A’ ∩ B’) = P((A ∪ B)’) = 1 − P(A ∪ B) = 1 − 0.62 = 0.38. Always remember the complement rule and De Morgan’s laws when dealing with combined events.
最后，P(A’ ∩ B’) = P((A ∪ B)’) = 1 − P(A ∪ B) = 1 − 0.62 = 0.38。处理组合事件时务必记住补集规则和德摩根律。
4. Discrete Random Variables: Expectation and Variance | 离散随机变量：期望与方差
This question provided the probability distribution of a discrete random variable X: P(X=1)=0.2, P(X=2)=0.3, P(X=3)=0.25, P(X=4)=0.25. Tasks included calculating E(X), Var(X), and then E(3X+2) and Var(5−2X).
该题给出了离散随机变量 X 的概率分布：P(X=1)=0.2，P(X=2)=0.3，P(X=3)=0.25，P(X=4)=0.25。要求计算 E(X)、Var(X) 以及 E(3X+2) 和 Var(5−2X)。
E(X) = Σ x P(X=x) = 1(0.2) + 2(0.3) + 3(0.25) + 4(0.25) = 2.55. E(X²) = Σ x² P(X=x) = 1(0.2) + 4(0.3) + 9(0.25) + 16(0.25) = 7.65. Then Var(X) = E(X²) − [E(X)]² = 7.65 − (2.55)² = 7.65 − 6.5025 = 1.1475.
E(X) = Σ x P(X=x) = 1(0.2) + 2(0.3) + 3(0.25) + 4(0.25) = 2.55。E(X²) = Σ x² P(X=x) = 1(0.2) + 4(0.3) + 9(0.25) + 16(0.25) = 7.65。于是 Var(X) = E(X²) − [E(X)]² = 7.65 − (2.55)² = 7.65 − 6.5025 = 1.1475。
For linear functions, apply the rules: E(aX+b) = aE(X) + b and Var(aX+b) = a² Var(X). Thus E(3X+2) = 3(2.55) + 2 = 9.65, and Var(5−2X) = (−2)² Var(X) = 4 × 1.1475 = 4.59. Note that adding a constant does not affect the variance.
对于线性函数，使用规则：E(aX+b) = aE(X) + b，Var(aX+b) = a² Var(X)。因此 E(3X+2) = 3(2.55) + 2 = 9.65，Var(5−2X) = (−2)² Var(X) = 4 × 1.1475 = 4.59。注意加常数项不影响方差。
5. Normal Distribution: Finding Unknown μ and σ | 正态分布：求解未知参数 μ 和 σ
A longer problem featured the normal random variable X ~ N(μ, σ²). It was given that P(X < 30) = 0.15 and P(X > 45) = 0.20. The aim was to find μ and σ. The standard normal transformation Z = (X − μ)/σ is essential.
一道较长的题涉及正态随机变量 X ~ N(μ, σ²)。已知 P(X < 30) = 0.15 且 P(X > 45) = 0.20，要求出 μ 与 σ。标准正态变换 Z = (X − μ)/σ 至关重要。
First, convert each probability: P(Z < (30−μ)/σ) = 0.15 → (30−μ)/σ = Φ⁻¹(0.15) ≈ −1.0364. For the upper tail, P(X > 45) = 0.20 implies P(X < 45) = 0.80, so (45−μ)/σ = Φ⁻¹(0.80) ≈ 0.8416. You now have two simultaneous equations.
首先转换概率：P(Z < (30−μ)/σ) = 0.15 → (30−μ)/σ = Φ⁻¹(0.15) ≈ −1.0364。对于上尾，P(X > 45) = 0.20 意味着 P(X < 45) = 0.80，故 (45−μ)/σ = Φ⁻¹(0.80) ≈ 0.8416。现在得到两个联立方程。
Equation 1: 30 − μ = −1.0364 σ
Equation 2: 45 − μ = 0.8416 σ
Subtract Equation 1 from Equation 2: (45−μ) − (30−μ) = 15 = (0.8416 − (−1.0364))σ = 1.8780 σ → σ ≈ 7.99. Substituting back yields μ ≈ 30 + 1.0364 × 7.99 ≈ 38.28. Always round sensibly at the end; here μ ≈ 38.3 and σ ≈ 8.0.
将式2减式1：(45−μ) − (30−μ) = 15 = (0.8416 + 1.0364)σ = 1.8780 σ → σ ≈ 7.99。代回得 μ ≈ 30 + 1.0364 × 7.99 ≈ 38.28。最后合理舍入；此处 μ ≈ 38.3，σ ≈ 8.0。
6. Product Moment Correlation Coefficient (r) | 积矩相关系数 (r)
A bivariate data set was given: for x (hours of study) and y (test score), five pairs: (10, 8), (14, 11), (18, 15), (22, 18), (26, 20). Candidates were asked to calculate the product moment correlation coefficient r and test for positive correlation at the 5% level.
给定一组双变量数据：x（学习小时数）与 y（测试分数），五对：(10, 8), (14, 11), (18, 15), (22, 18), (26, 20)。要求计算积矩相关系数 r，并在 5% 显著性水平下检验正相关。
Compute the summaries: Σx = 90, Σy = 72, Σx² = 10²+14²+18²+22²+26² = 1900, Σy² = 8²+11²+15²+18²+20² = 1134, Σxy = 10×8+14×11+18×15+22×18+26×20 = 1430. Then Sxx = Σx² − (Σx)²/n = 1900 − 90²/5 = 1900−1620 = 280. Similarly Sₓᵧ = Σxy − (Σx)(Σy)/n = 1430 − (90×72)/5 = 1430−1296 = 134. Syy = 1134 − 72²/5 = 1134−1036.8 = 97.2.
先计算摘要量：Σx = 90，Σy = 72，Σx² = 1900，Σy² = 1134，Σxy = 1430。于是 Sxx = Σx² − (Σx)²/n = 280，Sₓᵧ = Σxy − (Σx)(Σy)/n = 134，Syy = 1134 − 72²/5 = 97.2。
r = Sₓᵧ / √(Sxx × Syy) = 134 / √(280 × 97.2) ≈ 134 / √27216 ≈ 0.813
For n = 5, the 5% one‑tail critical value from tables is 0.8054. Since 0.813 > 0.8054, there is sufficient evidence to reject H₀ (ρ = 0) and conclude positive correlation. Always state the hypothesis and conclusion clearly.
对于 n = 5，5% 单尾临界值为 0.8054。因 0.813 > 0.8054，有充分证据拒绝 H₀ (ρ = 0)，得出存在正相关的结论。需始终清晰陈述假设与结论。
7. Least Squares Regression Line | 最小二乘回归直线
Using the same data, the next part asked for the equation of the regression line of y on x in the form y = a + bx. The slope b = Sₓᵧ / Sxx = 134 / 280 = 0.47857… ≈ 0.479. The means are x̄ = 90/5 = 18 and ȳ = 72/5 = 14.4. Thus a = ȳ − b x̄ = 14.4 − 0.479×18 ≈ 5.78.
基于相同数据，下一部分要求给出 y 对 x 的回归直线方程，形式为 y = a + bx。斜率 b = Sₓᵧ / Sxx = 134/280 ≈ 0.479。均值 x̄ = 18，ȳ = 14.4。于是 a = ȳ − b x̄ ≈ 5.78。
The regression equation is y = 5.78 + 0.479x. Interpret b: for each additional hour of study, the test score increases by 0.479 on average. The intercept a is the predicted score when study time is zero, which may not have a meaningful real‑world interpretation but is necessary for the line.
回归方程为 y = 5.78 + 0.479x。解释 b：每增加一小时学习，测试分数平均提高 0.479 分。截距 a 是学习时间为零时的预测分数，可能在现实中无意义，但却是直线公式的必要部分。
To predict the score for a student who studies 30 hours, substitute x = 30: y = 5.78 + 0.479×30 = 20.15. However, because 30 is outside the observed x‑range (10–26), this is an extrapolation and may not be reliable. Always comment on the limitations of prediction.
若要预测学习 30 小时的学生的分数，代入 x = 30：y = 5.78 + 0.479×30 = 20.15。然而，由于 30 在原观测 x 范围 (10–26) 之外，这属于外推，可靠性较差。须始终说明预测的局限性。
8. Coding: Transforming Data | 编码：数据转换
Often the paper asks to recalculate the regression line if variables are coded. Suppose we code u = (x − 18)/2. The means change simply: ū = (x̄ − 18)/2 = 0. The standard deviation of u is σᵤ = σₓ / 2. For the regression, the slope bᵤ = Sᵤᵧ / Sᵤᵤ. Using the coding properties, Sᵤᵤ = Sₓₓ / 2² = 280/4 = 70. Sᵤᵧ = Sₓᵧ / 2 = 134/2 = 67. So bᵤ = 67/70 = 0.957, and the new intercept aᵤ = ȳ − bᵤ ū = 14.4. Hence the regression line on coded data is y = 14.4 + 0.957u.
试卷常要求在对变量编码后重新计算回归直线。假设我们编码 u = (x − 18)/2。均值直接变化：ū = (x̄ − 18)/2 = 0。u 的标准差 σᵤ = σₓ / 2。对于回归，斜率 bᵤ = Sᵤᵧ / Sᵤᵤ。利用编码性质，Sᵤᵤ = Sₓₓ / 2² = 70，Sᵤᵧ = Sₓᵧ / 2 = 67。因此 bᵤ = 0.957，新截距 aᵤ = ȳ − bᵤ ū = 14.4。故编码后回归线为 y = 14.4 + 0.957u。
To return to the original x, substitute u = (x − 18)/2 back: y = 14.4 + 0.957×(x − 18)/2 = 14.4 + 0.4785x − 8.613 = 5.787 + 0.4785x, which matches the earlier equation (allowing for rounding). Coding simplifies arithmetic but does not alter the underlying relationship.
要回到原变量 x，将 u = (x − 18)/2 代回：y = 14.4 + 0.957×(x − 18)/2 = 5.787 + 0.4785x，与先前方程一致（考虑舍入）。编码精简了计算，但不会改变本质关系。
9. Probability Trees and Conditional Probability | 概率树图与条件概率
A question on combined events often uses a tree diagram
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