📚 IGCSE OCR Statistics: Unit Test Mock Paper Analysis | IGCSE OCR 统计：单元测试模拟卷解析
Welcome to our in-depth analysis of an IGCSE OCR Statistics unit test mock paper. This session dissects a typical paper, question by question, offering clear, step-by-step solutions. By working through these examples, you will strengthen your grasp of core statistical methods and boost your confidence for the actual examination.
欢迎阅读我们针对 IGCSE OCR 统计单元测试模拟卷的深度解析。本文按题目逐一剖析一份典型试卷，提供清晰、逐步的解答。通过这些示例，你将加强对核心统计方法的掌握，并提升应对真实考试的信心。
1. Sampling Methods – Identifying Design Types | 抽样方法——识别设计类型
The first question of the mock paper presents three scenarios: (a) Names are drawn from a hat, (b) every 5th student on an alphabetical register is chosen, (c) students are selected proportionally from each year group. We must name the sampling method used in each case.
模拟试卷的第一题给出了三种情境：(a) 从帽子里抽取名字，(b) 按字母顺序名单每第5名学生被选中，(c) 按各年级人数比例抽取学生。我们需要说出每种所用的抽样方法。
Scenario (a) illustrates simple random sampling, where every member of the population has an equal chance of selection. Scenario (b) is systematic sampling, with a fixed interval applied. Scenario (c) exemplifies stratified sampling, ensuring each subgroup is represented according to its size. Understanding these distinctions is fundamental to collecting unbiased data.
情境 (a) 说明了简单随机抽样，总体中每个成员被选中的机会均等。情境 (b) 是系统抽样，应用了固定的间隔。情境 (c) 体现了分层抽样，确保每个子组按其规模被代表。理解这些区别是收集无偏数据的基础。
2. Stem-and-Leaf Diagram – Median and Quartiles | 茎叶图——中位数和四分位数
We are given an ordered stem-and-leaf plot with ten data points (key: 1|2 = 12): 1 | 2 3 5 ; 2 | 0 2 4 6 7 ; 3 | 1 3. The task is to find the median, lower quartile (Q1), upper quartile (Q3) and the interquartile range (IQR).
我们提供一个有序的茎叶图，共有十个数据点（图例：1|2 = 12）：1 | 2 3 5 ; 2 | 0 2 4 6 7 ; 3 | 1 3。需要求中位数、下四分位数 (Q1)、上四分位数 (Q3) 和四分位距 (IQR)。
The raw data in ascending order are: 12, 13, 15, 20, 22, 24, 26, 27, 31, 33. With n = 10, the median position is (10+1)/2 = 5.5, so the median is (22+24)/2 = 23. The position of Q1 = (10+1)/4 = 2.75, giving Q1 = 13 + 0.75×(15−13) = 13.5. The position of Q3 = 3×(10+1)/4 = 8.25, giving Q3 = 27 + 0.25×(31−27) = 28. The IQR = 28 − 13.5 = 14.5. This method follows the standard OCR approach for discrete data.
升序排列的原始数据为：12、13、15、20、22、24、26、27、31、33。n=10，中位数位置为 (10+1)/2 = 5.5，因此中位数为 (22+24)/2 = 23。Q1 的位置 = (10+1)/4 = 2.75，得出 Q1 = 13 + 0.75×(15−13) = 13.5。Q3 的位置 = 3×(10+1)/4 = 8.25，得出 Q3 = 27 + 0.25×(31−27) = 28。IQR = 28 − 13.5 = 14.5。该方法遵循 OCR 对离散数据的标准处理方式。
3. Standard Deviation from Raw Data | 原始数据的标准差
A question asks for the standard deviation of the set: 5, 7, 8, 10, 12, using the population formula. Step-by-step calculation demonstrates how variability is measured.
一道题目要求使用总体公式计算数据集 5、7、8、10、12 的标准差。逐步计算展示了如何衡量离散程度。
μ = (5 + 7 + 8 + 10 + 12) / 5 = 42/5 = 8.4
First, compute the mean μ = 8.4. Then find the squared deviations: (5−8.4)² = 11.56, (7−8.4)² = 1.96, (8−8.4)² = 0.16, (10−8.4)² = 2.56, (12−8.4)² = 12.96. Their sum is 29.2. Finally, σ = √(29.2 / 5) = √5.84 ≈ 2.416. Always present the final answer to an appropriate degree of accuracy, usually three significant figures.
首先计算均值 μ = 8.4。然后求各偏差的平方：(5−8.4)² = 11.56、(7−8.4)² = 1.96、(8−8.4)² = 0.16、(10−8.4)² = 2.56、(12−8.4)² = 12.96。总和为 29.2。最后 σ = √(29.2 / 5) = √5.84 ≈ 2.416。最终答案通常应给出适当精度，如三位有效数字。
σ = √( Σ(x − μ)² / N )
The formula applied here treats the data as the whole population; for a sample we would divide by (n−1). The OCR specification expects you to know which denominator to use depending on the context.
这里使用的公式将数据视为整个总体；对于样本，我们会除以 (n−1)。OCR 考纲要求根据语境选择正确的分母。
4. Probability Tree Diagrams – Without Replacement | 概率树图——不放回抽取
A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. Calculate the probability that the second ball is red. A tree diagram helps map all outcomes.
一个袋子里有 5 个红球和 3 个蓝球。不放回地抽取两个球。计算第二次抽到红球的概率。树图有助于展现所有可能的结果。
First draw: P(Red) = 5/8, P(Blue) = 3/8. If the first is red, second-draw probabilities become P(Red) = 4/7, P(Blue) = 3/7. If the first is blue, second-draw probabilities are P(Red) = 5/7, P(Blue) = 2/7. The required probability is P(Red second) = (5/8)×(4/7) + (3/8)×(5/7) = 20/56 + 15/56 = 35/56 = 5/8. Interestingly, the probability equals that of drawing a red on the first pull, a property that holds for any such ‘without replacement’ scenario with two draws when the composition is symmetric.
第一次抽取：P(红) = 5/8，P(蓝) = 3/8。若第一次抽到红球，第二次抽取的概率变为 P(红) = 4/7、P(蓝) = 3/7。若第一次抽到蓝球，第二次概率为 P(红) = 5/7、P(蓝) = 2/7。所求概率 P(第二次红) = (5/8)×(4/7) + (3/8)×(5/7) = 20/56 + 15/56 = 35/56 = 5/8。有趣的是，该概率等于第一次抽到红球的概率，这一性质在两次抽取结构对称且不放回时成立。
5. Binomial Distribution – Exact Probability Calculation | 二项分布——精确概率计算
The mock paper includes a binomial variable X ~ B(12, 0.25). Determine P(X = 4). This requires using the binomial probability formula and the combination coefficient.
模拟卷中包含一个二项变量 X ~ B(12, 0.25)。求 P(X = 4)。这需要运用二项概率公式和组合系数。
P(X = 4) = ¹²C₄ × (0.25)⁴ × (0.75)⁸
Compute ¹²C₄ = 495. (0.25)⁴ = 0.00390625 and (0.75)⁸ ≈ 0.100112915. Multiplying gives 495 × 0.00390625 × 0.100112915 ≈ 0.1936. It is essential to show the substitution into the formula before using a calculator, and to express the answer as a decimal correct to at least three significant figures. This process is typical of OCR binomial questions.
计算 ¹²C₄ = 495。(0.25)⁴ = 0.00390625，而 (0.75)⁸ ≈ 0.100112915。相乘得 495 × 0.00390625 × 0.100112915 ≈ 0.1936。在借助计算器之前，应展示代入公式的过程，并以至少三位有效数字的小数形式给出答案。这符合 OCR 二项分布题目的典型要求。
6. Normal Distribution – Probability Using Z-table | 正态分布——利用 Z 表求概率
A continuous random variable X is normally distributed with mean 50 and standard deviation 4: X ~ N(50, 4²). Find P(X > 55). Standardisation converts the problem to a Z-score, and then a standard normal table provides the probability.
一个连续随机变量 X 服从均值为 50、标准差为 4 的正态分布：X ~ N(50, 4²)。求 P(X > 55)。通过标准化将问题转化为 Z 分数，然后查标准正态表得到概率。
Z = (55 − 50) / 4 = 1.25
P(X > 55) = P(Z > 1.25). From the normal table, Φ(1.25) = 0.8944, which is P(Z < 1.25). Therefore, P(Z > 1.25) = 1 − 0.8944 =
Published by TutorHao | IGCSE 统计 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