📚 Year 11 SQA Statistics: Unit Test Mock Paper Analysis | SQA 统计 Year 11 单元测试模拟卷解析
This article provides a comprehensive walkthrough of a unit test mock paper for the Year 11 SQA Statistics course. Each question is presented with a detailed solution and an explanation, covering key topics such as descriptive statistics, correlation, regression, probability, discrete random variables and the normal distribution. Working through these model answers will reinforce your understanding and boost your confidence for the actual exam.
本文为 Year 11 SQA 统计课程单元测试模拟卷提供了全面解析。每道题目都配有详细的解答与解释，涵盖了描述统计、相关性、回归、概率、离散随机变量以及正态分布等重要专题。通过研读这些模拟答案，你将巩固理解并增强应对真实考试的信心。
1. How This Mock Paper Is Structured | 本模拟卷的整体结构
The mock paper below follows the shape of a typical Year 11 SQA Statistics unit test: a short-answer section that rewards clear working, followed by longer context-based questions that combine two or more topics. The paper is worth 40 marks, lasts 60 minutes, and a scientific or graphical calculator is permitted throughout. Marks are awarded for method as well as for the final answer, so the working you write down matters as much as the number you circle.
下面的模拟卷按照 Year 11 SQA 统计单元测试的典型结构编排：先是重视解题过程的简答题部分，随后是结合两个或更多专题的情境长题。全卷满分 40 分，时长 60 分钟，全程允许使用科学计算器或图形计算器。分数既给方法也给答案，因此你写下的推导过程和最终圈出的数字同样重要。
The six topic areas examined are descriptive statistics, correlation, regression, probability, discrete random variables and the normal distribution. The questions are ordered roughly from routine to demanding, so if you run out of time, you should still be able to bank the early marks.
所考查的六大专题是：描述统计、相关性、回归、概率、离散随机变量和正态分布。题目大致由常规到有难度排序，因此如果时间不够，你至少应当稳稳拿下前面的分数。
- Section A (Q1-Q4): descriptive statistics, correlation and regression, 16 marks.
A 部分（第1-4题）：描述统计、相关性与回归，共 16 分。 - Section B (Q5-Q7): probability and conditional probability, 12 marks.
B 部分（第5-7题）：概率与条件概率，共 12 分。 - Section C (Q8-Q9): discrete random variables and the normal distribution, 12 marks。
C 部分（第8-9题）：离散随机变量与正态分布，共 12 分。
2. Q1: Mean, Median and Mode | 第1题：平均数、中位数与众数
Question: The test scores of eight students are: 12, 15, 11, 18, 12, 14, 16, 12. Calculate the mean, median and mode.
题目：八名学生的测试成绩为：12, 15, 11, 18, 12, 14, 16, 12。计算平均数、中位数和众数。
Solution (mean): 11 + 18 + 12 + 14 + 16 + 12 = 110. Divide by the number of values, n = 8: 110 ÷ 8 = 13.75.
解答（平均数）：首先求平均数：将所有值相加 12 + 15 + 11 + 18 + 12 + 14 + 16 + 12 = 110。除以数值个数 n = 8：110 ÷ 8 = 13.75。
Solution (median): the median is the mean of the two middle values, the 4th and the 5th: (12 + 14) ÷ 2 = 13.
解答（中位数）：先把数据从小到大排序：11, 12, 12, 12, 14, 15, 16, 18。因为数据个数为偶数，中位数是两个中间值（第 4 个和第 5 个）的平均数：(12 + 14) ÷ 2 = 13。
Solution (mode): The mode is the most frequent value. The value 12 appears three times, more than any other, so the mode is 12.
解答（众数）：众数出现次数最多的数值。12 出现了三次，比任何其他值都多，所以众数是 12。
Mean = Σx ÷ n = 110 ÷ 8 = 13.75, Median = 13, Mode = 12
3. Q2: Range, Interquartile Range and Standard Deviation | 第2题：极差、四分位距与标准差
Question: Using the same eight scores, find the range, the interquartile range and the population standard deviation.
题目：仍然使用这八个成绩，求极差、四分位距和总体标准差。
lest: 18 − 11 = 7. This is easy to calculate but very sensitive to a single extreme value.
极差是最大值减最小值：18 − 11 = 7。它计算简单，但对单个极端值非常敏感。
ian of the upper half (14, 15, 16, 18), which is 15.5. So IQR = Q₃ − Q₁ = 15.5 − 12 = 3.5.
求四分位距时，下四分位数 Q₁ 是下半部分数据 (11, 12, 12, 12) 的中位数，即 12；上四分位数 Q₃ 是上半部分数据 (14, 15, 16, 18) 的中位数，即 15.5。所以 IQR = Q₃ − Q₁ = 15.5 − 12 = 3.5。
.25, 2.25 and −1.75, and the sum of their squares is 41.5. Dividing by n = 8 gives a variance of 5.1875, so the standard deviation is about 2.28.
总体标准差衡量各数据与平均数之间的典型距离。八个数据与 13.75 的偏差分别是 −1.75、1.25、−2.75、4.25、−1.75、0.25、2.25、−1.75，它们平方的总和为 41.5。除以 n = 8 得到方差 5.1875，因此标准差约为 2.28。
σ = √(Σ(x − μ)² ÷ n) = √41.5 ÷ 8 = √5.1875 ≈ 2.28
tead, giving about 2.43. Read the question wording carefully to see which one is required.
如果同样的数据被当作样本而非总体处理，就要除以 n − 1 = 7，结果约为 2.43。答题时务必看清题目措辞，判断该用哪一种。
4. Q3: Scatterplots and Correlation | 第3题：散点图与相关性
Question: r upward trend and the calculator gives r = 0.93. Describe and interpret this correlation.
题目：一项研究记录六名学生每周学习时数与考试成绩。散点图显示明显的上升趋势，计算器给出 r = 0.93。描述并解释这一相关性。
The correlation coefficient r always lies between −1 and 1. A value of 0.93 is close to +1, so there is a strong positive linear correlation between study hours and test scores: students who study more hours tend to score higher.
相关系数 r 的取值范围总是介于 −1 与 1 之间。0.93 接近于 +1，说明学习时数与考试成绩之间存在强的正线性相关：学习时数越多的学生，成绩往往越高。
Three points earn the interpretation marks. First, state the strength (strong, moderate or weak) and the direction (positive or negative). Second, say that the relationship is linear, since r only detects straight-line patterns. Third, note that correlation does not prove causation, and an unmeasured third variable such as prior attainment or attendance could explain part of the pattern.
解释部分有三个得分点。第一，说明强度（强、中等或弱）与方向（正或负）。第二，说明这一关系是线性的，因为 r 只能识别直线型趋势。第三，指出相关并不等于因果，诸如原有基础或出勤率等未被测量的第三变量也可能解释部分现象。
- |r| above about 0.8: strong correlation. |r| 约大于 0.8：强相关。
- |r| around 0.4 to 0.8: moderate correlation. |r| 约在 0.4 到 0.8：中等相关。
- |r| below about 0.4: weak correlation. |r| 约小于 0.4：弱相关。
5. Q4: The Least-Squares Regression Line | 第4题：最小二乘回归直线
Question: here x is weekly study hours and y is the test score. Interpret the gradient and predict the score of a student who studies 7 hours.
题目：同一组数据由计算器得到回归直线 y = 24.5 + 6.2x，其中 x 为每周学习时数，y 为考试成绩。解释斜率含义，并预测一名学习 7 小时学生的成绩。
of study per week is associated with a rise of 6.2 marks. The intercept 24.5 is the model' but it should only be quoted in context if x = 0 is within the range of the observed data.
斜率 6.2 表示：平均而言，每周每多学习一小时，成绩大约提高 6.2 分。截距 24.5 是模型对零小时学习的预测成绩，但只有当 x = 0 落在实际观测数据的范围内时，才适合结合情境引用它。
score is about 68 marks. Always finish a prediction question with a sentence in context and mention that this is only an estimate.
代入 x = 7 得 y = 24.5 + 6.2 × 7 = 24.5 + 43.4 = 67.9，因此预测成绩约为 68 分。做预测题时，最后一定要用情境语言写一句结论，并说明这只是一个估计值。
y = 24.5 + 6.2x → y = 24.5 + 6.2 × 7 = 67.9
The line is called the least-squares regression line because it is the straight line that makes the sum of squared vertical distances Σ(y − ŷ)² as small as possible. Be careful with extrapolation: predicting far outside the observed range of x is unreliable, and a high r does not guarantee that the prediction is accurate for an individual student.
这条线被称为最小二乘回归直线，因为它是使纵向离差平方和 Σ(y − ŷ)² 达到最小的直线。要注意外推问题：在 x 的观测范围之外做远距离预测是不可靠的，r 值高也不保证对某一个具体学生的预测一定准确。
6. Q5: Basic Probability Rules | 第5题：概率基本法则
Question: A bag contains 5 red, 3 blue and 2 green counters. One counter is drawn at random. Find P(red), P(red or blue) and P(not green).
题目：一个袋中有 5 个红球、3 个蓝球、2 个绿球。随机取出一个球。求 P(红)、P(红或蓝) 和 P(非绿)。
= 0.3 and P(green) = 0.2. These three probabilities add to 1, which is a good first check.
袋中一共有 10 个球，因此 P(红) = 5 ÷ 10 = 0.5，P(蓝) = 0.3，P(绿) = 0.2。这三个概率之和为 1，这是一个很好的初步检验。
= 0.5 + 0.3 = 0.8. Using the complement rule, P(not green) = 1 − P(green) = 1 − 0.2 = 0.8.
红与蓝是互斥事件，因此加法法则简化为 P(红或蓝) = P(红) + P(蓝) = 0.5 + 0.3 = 0.8。使用补集法则，P(非绿) = 1 − P(绿) = 1 − 0.2 = 0.8。
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
If the first counter were replaced, the calculation would instead be 0.5 × 0.3 = 0.15, because the second draw would be independent of the first.
现在假设不放回地抽取两个球。第一个是红球且第二个是蓝球的概率为 5/10 × 3/9 = 15/90 = 1/6 ≈ 0.167。如果放回后抽取，计算则变为 0.5 × 0.3 = 0.15，因为第二次抽取与第一次相互独立。
7. Q6: Conditional Probability and Tree Diagrams | 第6题：条件概率与树形图
Question: In a class, 60% of students complete their homework. Of those who complete it, 80% pass the test; of those who do not, only 30% pass. Find the probability that a randomly chosen student passes, and the probability that a student completed the homework given that they passed.
题目：某班级 60% 的学生完成作业。在完成作业的学生中，80% 通过测试；在未完成作业的学生中，只有 30% 通过。求随机抽取一名学生通过测试的概率，以及在已知其通过测试的条件下该学生完成了作业的概率。
A tree diagram is the safest way to handle this. The first set of branches splits into completed (0.6) and not completed (0.4), and each of these splits again into pass and fail.
用树形图处理这类题最稳妥。第一层分支分为完成（0.6）与未完成（0.4），每一支再分为通过与未通过。
the branches and adding gives P(pass) = 0.6 × 0.8 + 0.4 × 0.3 = 0.48 + 0.12 = 0.60. So 60% of students pass overall.
沿着分支相乘再相加，得到 P(通过) = 0.6 × 0.8 + 0.4 × 0.3 = 0.48 + 0.12 = 0.60。也就是说，全体学生中有 60% 通过测试。
the reverse conditional probability, restrict attention to the passing group: P(completed | the condition changes the denominator from the whole class to just the students who passed.
求反向条件概率时，把注意力限制在通过测试的这一组：P(完成 | 通过) = 0.48 ÷ 0.60 = 0.8。注意条件如何把分母从全班改成仅通过测试的学生。
P(A | B) = P(A ∩ B) ÷ P(B) = 0.48 ÷ 0.60 = 0.8
8. Q7: Discrete Random Variables | 第7题：离散随机变量
Question: 1) = 0.3, P(X = 2) = 0.4 and P(X = 3) = 0.1. Find E(X), Var(X) and the standard deviation.
题目：离散随机变量 X 的概率分布为 P(X = 0) = 0.2，P(X = 1) = 0.3，P(X = 2) = 0.4，P(X = 3) = 0.1。求 E(X)、Var(X) 及标准差。
First check that the probabilities sum to 1: 0.2 + 0.3 + 0
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图