📚 Year 8 OCR Statistics: Unit Test Mock Paper Walkthrough | 八年级 OCR 统计：单元测试模拟卷解析
This article provides a detailed walkthrough for a mock unit test designed for Year 8 students following the OCR Statistics curriculum. Each question is carefully explained to reinforce key concepts such as data types, averages, charts, probability, and critical evaluation of statistical claims. By working through these solutions, you will strengthen your understanding and exam technique.
本文为遵循 OCR 统计课程的八年级学生精心设计了一份单元测试模拟卷，并逐题进行详细解析。内容涵盖数据类型、平均数、图表、概率以及统计结论的批判性评估等重要知识点。通过研读这些解题步骤，你将巩固理解并提升应试技巧。
1. Identifying Data Types | 识别数据类型
The question asks you to classify whether ‘favourite sport’, ‘number of siblings’, and ‘height in cm’ are categorical, discrete, or continuous data.
题目要求你将“最喜欢的运动”、“兄弟姐妹的数量”和“身高（厘米）”分类为分类数据、离散数据或连续数据。
‘Favourite sport’ is a categorical variable because it describes a quality or name, not a numerical measurement. This type of data can be sorted into groups but cannot be measured on a number scale.
“最喜欢的运动”是分类变量，因为它描述的是性质或名称，而非数值测量。这种数据可以分组归类，但无法在数值标尺上进行测量。
‘Number of siblings’ is discrete numerical data. You count siblings in whole numbers — you cannot have 2.5 siblings in a genuine count. Discrete data usually arises from counting.
“兄弟姐妹的数量”是离散数值数据。兄弟姐妹是用整数计算的——在真实计数中不可能有 2.5 个兄弟姐妹。离散数据通常来自计数过程。
‘Height in centimetres’ is continuous data because it can take any value within a range. A person’s height could be 152.3 cm or 164.7 cm; it is measured, not counted.
“身高（厘米）”是连续数据，因为它可以在一个区间内取任意值。一个人的身高可能是 152.3 厘米或 164.7 厘米；它是测量得到的，而非计数得到的。
So the correct classifications are: categorical, discrete, continuous. Always remember: counted data is discrete, measured data is continuous.
因此正确的分类是：分类、离散、连续。请始终牢记：计数所得为离散数据，测量所得为连续数据。
2. Calculating the Mean from a Frequency Table | 根据频率表计算平均数
A frequency table shows the number of books read by 25 students: 0 books (frequency 3), 1 book (8), 2 books (7), 3 books (4), 4 books (3). Work out the mean number of books read.
一张频率表显示了 25 名学生阅读的书籍数量：0 本（频数 3）、1 本（8）、2 本（7）、3 本（4）、4 本（3）。请计算阅读书籍的平均数。
First, we need the total number of books. Multiply each number of books by its frequency and add the results: (0×3) + (1×8) + (2×7) + (3×4) + (4×3).
首先，我们需要书籍的总数。将每类书籍数乘以其频数，然后相加：(0×3) + (1×8) + (2×7) + (3×4) + (4×3)。
Sum = 0 + 8 + 14 + 12 + 12 = 46 books
The total frequency (number of students) is 3+8+7+4+3 = 25.
总频数（学生人数）为 3+8+7+4+3 = 25。
Mean = Total number of books ÷ Total frequency = 46 ÷ 25.
平均数 = 总书籍数 ÷ 总频数 = 46 ÷ 25。
Mean = 1.84 books
So the mean number of books read is 1.84. Even though nobody read exactly 1.84 books, the mean gives a useful central value for the data.
因此平均阅读量为 1.84 本。虽然没有人恰好读了 1.84 本书，但平均数给出了这组数据中的一个有用的中心值。
3. Finding Median, Mode and Range | 求中位数、众数和极差
The test gives the following set of test scores out of 10: 7, 4, 8, 6, 4, 9, 5, 4. Find the mode, median, and range.
试题给出了一组十分制测验分数：7, 4, 8, 6, 4, 9, 5, 4。请求出众数、中位数和极差。
Mode is the value that appears most often. In this list, 4 appears three times, more than any other number, so the mode is 4.
众数是出现次数最多的值。在这组数据中，4 出现了三次，多于其他任何数字，因此众数为 4。
To find the median, write the numbers in order from smallest to largest: 4, 4, 4, 5, 6, 7, 8, 9. There are 8 values (an even number), so the median is the mean of the 4th and 5th values.
要计算中位数，先将数据从小到大排列：4, 4, 4, 5, 6, 7, 8, 9。共有 8 个值（偶数个），因此中位数是第 4 和第 5 个值的平均数。
The 4th value is 5 and the 5th is 6. Their mean is (5 + 6) ÷ 2 = 5.5. So the median is 5.5.
第 4 个值是 5，第 5 个是 6。它们的平均数为 (5 + 6) ÷ 2 = 5.5。所以中位数为 5.5。
Range = highest value − lowest value = 9 − 4 = 5. The range measures how spread out the data are, and here it is 5 points.
极差 = 最大值 − 最小值 = 9 − 4 = 5。极差衡量数据的离散程度，此处为 5 分。
4. Interpreting Bar Charts and Dual Bar Charts | 解读条形图和双条形图
A dual bar chart compares the number of boys and girls who chose different school lunch options: pasta, salad, pizza, and wrap. The question asks you to identify which option had the biggest difference between boys and girls, and which was most popular overall.
一张双条形图比较了选择不同学校午餐选项（意面、沙拉、披萨和卷饼）的男生和女生人数。题目要求你找出男生与女生选择差异最大的选项，以及总体上最受欢迎的选项。
To find the biggest difference, visually compare the heights of the two bars for each option. For pizza, the boys’ bar is much taller than the girls’ bar — the gap is the largest. Counting the scale, it might be a difference of 8 students.
要找出最大差异，可以直接观察每个选项的两根条形柱高度。对于披萨，男生柱远高于女生柱——差距最大。按刻度数算，可能相差 8 名学生。
Overall popularity is found by adding the boys and girls figures for each option. Pasta: boys 10 + girls 12 = 22; salad: 6 + 14 = 20; pizza: 18 + 10 = 28; wrap: 9 + 11 = 20. Pizza has the highest total, so it is the most popular choice overall.
总体受欢迎程度通过将每个选项的男女生人数相加得出。意面：10 + 12 = 22；沙拉：6 + 14 = 20；披萨：18 + 10 = 28；卷饼：9 + 11 = 20。披萨的总数最高，因此是整体上最受欢迎的选择。
Dual bar charts are excellent for comparing two related data sets side by side. Always check the scale and read the bar heights accurately.
双条形图非常适合于并排比较两组相关数据。请务必检查刻度并准确读取条形柱的高度。
5. Drawing and Analysing Line Graphs | 绘制和分析折线图
A table gives the temperature recorded every two hours from 08:00 to 18:00: 8°C, 12°C, 15°C, 17°C, 14°C, 10°C. The task is to plot these values on a line graph and describe the trend.
一张表格给出了从 08:00 到 18:00 每两小时记录的温度：8°C, 12°C, 15°C, 17°C, 14°C, 10°C。任务是在折线图上标出这些值，并描述其变化趋势。
First, draw axes with ‘Time’ on the horizontal axis and ‘Temperature (°C)’ on the vertical axis. Plot each point precisely, then join the points with straight lines. Label your axes and give the graph a title.
首先，画出坐标轴，横轴为“时间”，纵轴为“温度 (°C)”。精确地标出每个数据点，然后用直线连接各点。给坐标轴加注标签，并为图表拟定标题。
The trend shows temperature increasing from morning until early afternoon (08:00 to 14:00), reaching a peak of 17°C at 14:00, and then decreasing towards the evening. This is a typical pattern driven by the sun’s intensity throughout the day.
趋势显示，从上午到下午较早时段（08:00 至 14:00），气温逐渐上升，在 14:00 达到峰值 17°C，然后向傍晚下降。这是受一天中太阳强度变化影响的典型模式。
When describing a line graph, use phrases like ‘increases’, ‘decreases’, ‘peaks at’, and ‘remains constant’ to be precise.
描述折线图时，要准确使用“上升”、“下降”、“在……达到峰值”、“保持平稳”等表述。
6. Pie Charts: Calculating Angles | 饼图：角度计算
A survey of favourite fruits among 40 students gives: apples 12, bananas 10, oranges 8, grapes 6, others 4. Calculate the angle for each sector in a pie chart.
一项关于 40 名学生最喜欢水果的调查结果如下：苹果 12 人，香蕉 10 人，橙子 8 人，葡萄 6 人，其他 4 人。请计算饼图中每个扇区的角度。
The whole pie chart represents 360° and the total frequency is 40. To find the angle for one item, use: angle = (frequency ÷ total) × 360°.
整个饼图代表 360°，总频数为 40。要计算某一项的扇区角度，可使用公式：角度 = (频数 ÷ 总数) × 360°。
Apples: (12 ÷ 40) × 360 = 0.3 × 360 = 108°. Bananas: (10 ÷ 40) × 360 = 90°. Oranges: (8 ÷ 40) × 360 = 72°. Grapes: (6 ÷ 40) × 360 = 54°. Others: (4 ÷ 40) × 360 = 36°.
苹果：(12 ÷ 40) × 360 = 0.3 × 360 = 108°。香蕉：(10 ÷ 40) × 360 = 90°。橙子：(8 ÷ 40) × 360 = 72°。葡萄：(6 ÷ 40) × 360 = 54°。其他：(4 ÷ 40) × 360 = 36°。
Check: 108 + 90 + 72 + 54 + 36 = 360°, which confirms the calculations are correct. When drawing the pie chart, use a protractor to measure each angle accurately.
检验：108 + 90 + 72 + 54 + 36 = 360°，这确认了计算无误。绘制饼图时，要用量角器精确测量每个角度。
7. Scatter Graphs and Correlation | 散点图与相关性
A scatter graph plots the number of hours spent revising against test score. The points suggest that as revision hours increase, test scores tend to increase. Describe the correlation and draw a line of best fit.
一张散点图描绘了复习小时数与测验分数之间的关系。各点显示，随着复习时间增加，测验分数也倾向于提高。请描述相关性并画出最佳拟合线。
The pattern of points goes from bottom-left to top-right, indicating a positive correlation. This means the two variables move in the same direction.
点的分布模式从左下方向右上方延伸，表明存在正相关。这意味着两个变量同向变动。
A line of best fit is drawn through the middle of the points, balancing the number of points above and below the line. It should be straight and not necessarily pass through the origin.
最佳拟合线应穿过点群的中心，使线上方和线下方的点数大致均衡。这条线应为直线，且不一定经过原点。
Using the line, you can estimate values — for example, predict the test score for 5 hours of revision. Stronger correlation means the points are closer to the line, making predictions more reliable.
利用这条线可以估计数值——例如，预测复习 5 小时对应的测验分数。相关性越强，点就越紧密地聚集在直线周围，预测也就越可靠。
No correlation occurs when the points are randomly scattered, showing no relationship between the variables.
若点呈随机散落状，表明变量之间没有关联，即为无相关。
8. Basic Probability Scale | 基础概率尺度
A question asks: Place the following events on a probability scale marked from 0 to 1: rolling a 7 on a fair six-sided dice, picking a red card from a standard deck, the sun rising tomorrow, flipping a coin and getting tails.
有一道题目要求：将下列事件标注在 0 到 1 的概率尺度上：掷一个公平六面骰子掷出 7；从一副标准扑克牌中抽到红牌；太阳明天升起；抛一枚硬币得到反面。
An event that is impossible has probability 0. Rolling a 7 with a six-sided dice is impossible because the highest number is 6. So this must be placed at 0.
不可能事件的概率为 0。用六面骰子掷出 7 是不可能的，因为最大的点数是 6。因此应放在 0 的位置。
Flipping a fair coin and getting tails has exactly one favourable outcome out of two equally likely outcomes, so its probability is 1/2 or 0.5. Mark this exactly halfway along the scale.
抛掷一枚公平硬币得到反面的情况，在两种等可能结果中有一种有利结果，因此其概率为 1/2 或 0.5。在标尺正中位置标出。
Picking a red card from a standard 52-card deck: there are 26 red cards, so P(red) = 26/52 = 1/2 = 0.5 as well. This also sits at 0.5 on the scale.
从一副标准 52 张扑克牌中抽到红牌：共有 26 张红牌，因此 P(红牌) = 26/52 = 1/2 = 0.5。在标尺上同样位于 0.5。
The sun rising tomorrow is practically certain, with probability 1 (or extremely close to 1). Place this at the 1 end of the scale.
太阳明天升起几乎是必然事件，概率为 1（或极其接近 1）。将其标注在标尺的 1 端。
9. Experimental Probability and Expected Outcomes | 实验概率与期望结果
A spinner with four equal sections coloured red, blue, green, and yellow is spun 80 times. The results are recorded: red 18, blue 25, green 17, yellow 20. Calculate the experimental probability of landing on blue, and compare it with the theoretical probability.
一个被平分为红、蓝、绿、黄四色的转盘被转动了 80 次。记录结果：红色 18 次，蓝色 25 次，绿色 17 次，黄色 20 次。请计算落在蓝色的实验概率，并与理论概率进行比较。
Experimental probability = number of times blue occurred ÷ total number of trials = 25 ÷ 80 = 0.3125 (or 31.25%).
实验概率 = 蓝色出现的次数 ÷ 总试验次数 = 25 ÷ 80 = 0.3125（或 31.25%）。
The theoretical probability for a fair four-section spinner is 1/4 = 0.25 (or 25%). The experimental probability is slightly higher, which is common in a small number of trials. If you spun the spinner 1000 times, you would expect the experimental probability to get closer to 0.25.
对于一个公平的四色转盘，理论概率为 1/4 = 0.25（或 25%）。实验概率略高，这在小量试验中很常见。如果转动转盘 1000 次，实验概率预计会更接近 0.25。
To find the expected number of times the spinner should land on green in 200 spins, multiply the theoretical probability by the number of spins: 0.25 × 200 = 50 times.
要计算转盘在 200 次转动中预期落在绿色的次数，用理论概率乘以总转动次数：0.25 × 200 = 50 次。
10. Evaluating Statistical Claims and Misleading Graphs | 评估统计结论和误导性图表
A bar chart titled ‘Our Sales Are Booming!’ shows monthly sales figures but the vertical axis starts at 80 instead of 0, making the increase appear much steeper. Explain why the graph is misleading and what a fair graph should show.
一张标题为“我们的销售额正在激增！”的条形图展示了月度销售数据，但纵轴起点为 80 而非 0，使得增长幅度看起来陡峭得多。请解释该图表为何具有误导性，以及一份公正的图表应如何呈现。
By starting the axis at 80, the differences between bars are exaggerated. A small increase from 82 to 86 looks huge because the axis magnifies the top portion of the bars. This can mislead the viewer into thinking sales have grown dramatically.
由于纵轴从 80 起步，柱状条之间的差异被夸大了。从 82 到 86 的微小增长看起来幅度巨大，因为纵轴放大了柱状条的顶部区域。这可能会误导人们，让人以为销售额出现了大幅增长。
A fair bar chart should always start the vertical scale at 0 so that the bar lengths are proportional to the actual values. This allows the reader to make an accurate visual comparison.
公正的条形图纵轴的标度始终应从 0 开始，从而使条形高度与实际数值成比例。这样阅读者才能做出准确的视觉比较。
Another common trick is using inconsistent intervals on the scale or using 3D effects that distort the sizes. Always check the axes, the scale, and the source of the data before believing a statistical claim.
其他常见的误导手法还包括使用不一致的刻度间隔，或采用扭曲尺寸的 3D 效果。在相信任何统计结论之前，一定要查验坐标轴、刻度以及数据的来源。
Published by TutorHao | Statistics Revision Series | aleveler.com
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