Year 9 CCEA Statistics Unit Test Mock Paper Analysis | Year 9 CCEA 统计单元测试模拟卷解析

📚 Year 9 CCEA Statistics Unit Test Mock Paper Analysis | Year 9 CCEA 统计单元测试模拟卷解析

Mock examinations are a crucial revision tool for Year 9 students preparing for the CCEA Statistics unit test. This article provides a detailed walkthrough of a typical mock paper, breaking down each question type, the correct answers, and the reasoning behind them. By reviewing these worked solutions, you will strengthen your understanding of data collection, presentation, measures of central tendency, probability, and statistical interpretation.

模拟考试是Year 9学生备考CCEA统计单元测试的重要复习工具。本文详细解析一份典型的模拟试卷，逐题分析题目类型、正确答案及其背后的推理过程。通过复习这些解题过程，你将加深对数据收集、数据展示、集中趋势测量、概率和统计解释的理解。

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1. Question 1: Types of Data and Collection Methods | 问题1：数据类型与收集方法

Question: A PE teacher records the time taken for each student in Year 9 to run 100m during a lesson. (a) State whether this data is primary or secondary. (b) Give one advantage of collecting data this way.

Answer and explanation (a): The data is primary because the teacher is collecting it first-hand directly from the original source – the students themselves. No existing data set is being used.

答案与解析（a）：这些数据是初级数据，因为体育老师直接从原始来源（学生本人）处一手收集。并未使用任何已有的数据集。

(b) Advantage: Primary data can be designed specifically to answer the research question. In this case, the teacher can control the exact timing method and ensure all students are tested under the same conditions, which makes the data more reliable and fit for purpose.

（b）优点：初级数据可以专门设计以回答研究问题。在此例中，老师可以控制精确的计时方法，确保所有学生在相同条件下测试，从而使数据更可靠且合乎用途。

A common error is confusing primary with secondary data. Remember: if you collect it yourself for the first time, it is primary; if you use data someone else has already collected, it is secondary.

常见错误是将初级数据与次级数据混淆。记住：如果是首次自己收集的数据，就是初级数据；如果使用别人已经收集的数据，就是次级数据。

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2. Question 2: Sampling Methods | 问题2：抽样方法

Question: A school with 600 students wants to survey a sample of 30 about a new after-school club. The office gives you an alphabetical list of all students. Suggest a suitable sampling method and explain how you would use it to obtain a representative sample.

Answer: A systematic sampling method would be appropriate here. Take the alphabetical list and select every 20th student (since 600 ÷ 30 = 20). Start by choosing a random number between 1 and 20, say 7, then select the 7th student, 27th, 47th, and so on.

答案：此处适合使用系统抽样法。拿到按字母顺序排列的名单，每隔20名学生选择一人（因为 600 ÷ 30 = 20）。先从1到20之间随机选一个数字，例如选7，然后选择第7、第27、第47……以此类推。

This method is quick and easy to implement, and because there is no pattern in the alphabetical list related to club preferences, it is likely to give a representative spread across the school population.

这种方法快捷且容易实施，并且由于名单的字母顺序与社团偏好无关，因此很可能在学校总体中获得具有代表性的分布。

Another valid method is simple random sampling, using a random number generator on the numbered list, but the examiner often looks for a clear, practical description.

另一种有效方法是简单随机抽样，使用随机数生成器从编号名单中抽取，但考官通常期望看到清晰且可行的描述。

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3. Question 3: Frequency Tables and the Mean | 问题3：频率表与平均数

Question: The table below shows the scores of 25 students in a spelling test. Calculate the mean score.

Score (x)	Frequency (f)
1	3
2	7
3	8
4	5
5	2

Solution: First, add a column to find the total sum of all scores: multiply each score by its frequency. So, (1×3) + (2×7) + (3×8) + (4×5) + (5×2) = 3 + 14 + 24 + 20 + 10 = 71. The total number of students (total frequency) is 3+7+8+5+2 = 25. Therefore, the mean = total sum ÷ number of students = 71 ÷ 25 = 2.84.

解题过程：首先添加一列求所有分数的总和：各分数乘以其频数。即 (1×3) + (2×7) + (3×8) + (4×5) + (5×2) = 3 + 14 + 24 + 20 + 10 = 71。学生总数（总频数）为 3+7+8+5+2=25。因此，平均数 = 总和 ÷ 学生数 = 71 ÷ 25 = 2.84。

Many students forget to multiply before adding, which would give the wrong answer of 15÷5=3. Always use the formula Mean = Σ(fx) / Σf to structure your working.

许多学生忘记先乘再加，从而得到错误答案 15÷5=3。始终使用公式 平均数 = Σ(fx) / Σf 来搭建计算过程。

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4. Question 4: Bar Charts and Pie Charts | 问题4：柱状图与饼图

Question: The bar chart shows the number of books read by Year 9 students in one month: 0 books (5 students), 1 book (12 students), 2 books (15 students), 3 books (8 students). (a) How many students were surveyed? (b) What fraction of students read at least 2 books? (c) If a pie chart is drawn, what angle represents the ‘2 books’ category?

Answers: (a) Total students = 5 + 12 + 15 + 8 = 40. (b) Students reading at least 2 books = 15 + 8 = 23. The fraction is 23/40. (c) The ‘2 books’ sector has 15 students out of 40, so its angle = (15/40) × 360° = 135°.

答案：(a) 调查学生总数 = 5 + 12 + 15 + 8 = 40。(b) 阅读至少 2 本书的学生数 = 15 + 8 = 23，占比为 23/40。(c) “2本书”扇区对应 15/40 的学生，其角度 = (15/40) × 360° = 135°。

When reading a bar chart, always check the vertical axis scale carefully. The fraction ‘at least 2’ includes 2 and 3, which is a key distinction examiners test. For pie chart angles, the ratio of the part to the whole multiplied by 360° gives the correct answer.

阅读柱状图时，务必仔细检查纵轴刻度。“至少2本”包括 2 本和 3 本，这是考官常考的关键区别。对于饼图角度，部分占总体的比例乘以 360° 便是正确答案。

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5. Question 5: Median, Mode and Range | 问题5：中位数、众数和极差

Question: Find the mode, median and range of the dataset: 12, 15, 14, 12, 18, 13, 12, 17.

Solution: First, arrange the values in ascending order: 12, 12, 12, 13, 14, 15, 17, 18. The mode is the most frequent value, which is 12. The median is the middle value(s). With 8 numbers, the median is the average of the 4th and 5th values: (13 + 14) ÷ 2 = 13.5. The range is the difference between the largest and smallest values: 18 − 12 = 6.

解答：首先将数值从小到大排列：12, 12, 12, 13, 14, 15, 17, 18。众数是出现次数最多的值，即 12。中位数是中间位置的值。8 个数据时，中位数是第 4 和第 5 个值的平均数：(13 + 14) ÷ 2 = 13.5。极差是最大值与最小值之差：18 − 12 = 6。

A typical mistake is to forget to order the data before finding the median, which would lead to an incorrect middle number. Also, the mode is 12 (unimodal) – some students might mistakenly say there is no mode because there are repeated values, but a dataset can certainly have a mode.

典型错误是在找中位数之前忘记排序，这会导致错误地认定中间的数字。此外，众数为 12（单峰），有些学生可能误以为有重复数字就没有众数，但数据集完全可以拥有众数。

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6. Question 6: Probability Scale and Simple Events | 问题6：概率尺度与简单事件

Question: A bag contains 3 red, 2 blue and 5 green counters. One counter is taken at random. Find the probability that it is (a) red, (b) not blue, (c) red or blue. Give your answers as fractions in simplest form.

Answers: Total number of counters = 3 + 2 + 5 = 10. (a) P(red) = 3/10. (b) P(not blue) = 1 − P(blue) = 1 − 2/10 = 8/10 = 4/5. Alternatively, ‘not blue’ includes red and green, so (3+5)/10 = 8/10 = 4/5. (c) P(red or blue) = (3+2)/10 = 5/10 = 1/2.

答案：总棋子数 = 3 + 2 + 5 = 10。(a) P（红）= 3/10。(b) P（非蓝）= 1 − P（蓝）= 1 − 2/10 = 8/10 = 4/5。另一种思路，“非蓝”包括红和绿，所以 (3+5)/10 = 8/10 = 4/5。(c) P（红或蓝）= (3+2)/10 = 5/10 = 1/2。

Make sure to simplify fractions completely – a common mark deduction is leaving ‘8/10’ instead of ‘4/5’. Remember that probabilities always range from 0 to 1, and the sum of probabilities for all possible mutually exclusive outcomes is 1.

务必将分数化简到最简 —— 常见的扣分项是把 ‘8/10’ 留作答案而非 ‘4/5’。请记住概率总是介于 0 到 1 之间，且所有可能互斥结果的概率之和为 1。

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7. Question 7: Two-Way Tables and Venn Diagrams | 问题7：双向表与维恩图

Question: A survey of 75 students asked whether they like sport and whether they are a boy or girl. 30 boys like sport, 10 boys do not like sport, 20 girls like sport, and the rest are girls who do not like sport. (a) Complete the two-way table. (b) One student is chosen at random. Find the probability that the student is a girl who likes sport. (c) Find the probability that the student is a girl or likes sport.

Table and answers:

	Likes sport	Does not like sport	Total
Boys	30	10	40
Girls	20	15	35
Total	50	25	75

Total girls = 75 − 40 = 35, so girls who do not like sport = 35 − 20 = 15. (b) P(girl and likes sport) = 20/75 = 4/15. (c) P(girl or likes sport) = P(girl) + P(likes sport) − P(girl and likes sport) = 35/75 + 50/75 − 20/75 = 65/75 = 13/15.

总女生数 = 75 − 40 = 35，所以不喜欢体育的女生 = 35 − 20 = 15。(b) P（女生且喜欢体育）= 20/75 = 4/15。(c) P（女生或喜欢体育）= P（女生）+ P（喜欢体育）− P（女生且喜欢体育）= 35/75 + 50/75 − 20/75 = 65/75 = 13/15。

Using a two-way table makes ‘and’ probabilities (intersection) straightforward. For ‘or’ probabilities, remember to avoid double-counting by subtracting the intersection. You could also verify with a Venn diagram: two overlapping circles, one for girls (35) and one for liking sport (50), with intersection 20.

使用双向表让“且”的概率（交集）一目了然。对于“或”的概率，切记通过减去交集来避免重复计算。你也可以用维恩图来验证：两个相交的圆，一个代表女生（35），另一个代表喜欢体育（50），交集为 20。

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8. Question 8: Data Analysis, Conclusions and Reliability | 问题8：数据分析、结论与可靠性

Question: A student conducts a survey to find the average time Year 9 pupils spend on social media per day. She asks 20 volunteers from her friendship group and calculates a mean of 3.2 hours. She concludes that all Year 9 students in the school spend about 3.2 hours on social media daily. Comment on the reliability of her conclusion and suggest two improvements.

Comment: The conclusion is not reliable. The sample size of 20 is very small compared to the whole year group, so it may not capture the variety of experiences. Moreover, the sample is biased because it only includes volunteers from her friendship group – these students are not representative of the wider Year 9 population. Their social media habits are likely similar, which skews the data.

评价：该结论不可靠。与整个年级相比，20 的样本量太小，可能无法反映行为的多样性。此外，样本存在偏差，因为它仅包含来自她朋友圈的志愿者——这些学生不能代表更广的 Year 9 人群。她们的社交媒体使用习惯很可能相似，从而使数据产生偏斜。

Improvements: (1) Use a larger, more representative sample, ideally using a probability sampling method such as simple random or systematic sampling from the whole year group list. (2) Ensure the survey is anonymous and conducted in a way that encourages honest answers, perhaps by asking a more diverse group of students at different times, not just volunteers.

改进建议：（1）使用更大、更具代表性的样本，最好从全年级名单中采用概率抽样法，如简单随机抽样或系统抽样。（2）确保调查匿名并以鼓励诚实回答的方式进行，或许可以在不同时间向更多样的学生群体（而非仅仅是志愿者）发放调查。

This question tests your ability to evaluate statistical investigations. Always discuss sample size, sampling bias, and whether the sample reflects the target population. Just saying ‘the sample is too small’ is not enough – explain why it matters.

这道题考查你评估统计调查的能力。务必讨论样本量、抽样偏差以及样本是否反映目标总体。只说“样本太小”是不够的，还要解释为什么它重要。

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9. Common Pitfalls and Revision Tips | 常见错误与复习建议

Pitfalls: Students often lose marks by not simplifying fractions in probability, forgetting to order data before finding the median, mixing up mean and mode, or selecting an unrepresentative sampling method. In pie chart questions, using the frequency directly instead of multiplying by 360° is a frequent mistake. Also, watch out for units – always include them where needed.

常见陷阱：学生常因概率分数未化简、未排序就直接找中位数、混淆平均数与众数、或选择无代表性的抽样方法而丢分。在饼图问题中，直接使用频数而不是乘以 360° 是常见错误。此外，注意单位 – 总是在需要的地方带上单位。

Revision tips: Practise completing two-way tables from word problems. Draw simple Venn diagrams to visualise intersections. Create your own mock questions on sampling and critique them. Use the formula mean = Σ(fx) / Σf as a checklist in frequency table problems. Finally, time yourself on past paper questions to build confidence under exam conditions.

复习建议：练习根据文字类题目填完双向表。绘制简单的维恩图来直观表示交集。自行出一些关于抽样的模拟题并对其进行评析。

Published by TutorHao | Year 9 统计 Revision Series | aleveler.com

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