📚 Year 8 AQA Statistics: Unit Test Mock Paper Analysis | 八年级 AQA 统计：单元测试模拟卷解析
This mock paper analysis guides Year 8 students through typical AQA Statistics unit test questions, with step-by-step solutions and key revision points in both English and Chinese. Understanding these worked examples will strengthen your data handling skills and prepare you for exam success.
本模拟卷解析带领八年级学生演练典型的 AQA 统计单元测试题，提供逐步求解过程和关键复习要点，并配以中英双语讲解。通过这些范例解析，你可以巩固数据处理能力，为考试做好充分准备。
1. Types of Data | 数据类型
The question asks to classify each item as qualitative or quantitative, and if quantitative, to further label it as discrete or continuous: shoe size, type of car, temperature (°C), number of siblings.
题目要求将各项分类为定性数据或定量数据，如果是定量数据，还需标明是离散还是连续：鞋码、汽车类型、温度（°C）、兄弟姐妹数量。
Shoe size is numerical and can be ordered, so it is quantitative. Although shoe sizes can have half sizes, they take fixed values along a scale, making them discrete data.
鞋码是数字且可以排序，因此是定量数据。尽管有半码，但它们沿刻度取固定值，所以是离散数据。
Type of car is a category without a numerical meaning, so it is qualitative data.
汽车类型是类别，没有数值意义，因此是定性数据。
Temperature measured in degrees Celsius can take any value within a range and can include decimals, so it is quantitative continuous data.
以摄氏度计量的温度在范围内可取任意值且可以含小数，因此是定量连续数据。
Number of siblings is a count of people, which can only be a whole number, so it is quantitative discrete data.
兄弟姐妹数量是人数计数，只能是整数，所以是定量离散数据。
2. Bar Chart Interpretation | 条形图解读
A bar chart shows the favourite fruits of 40 students: apples (10), bananas (12), oranges (8), grapes (6), others (4). The question asks: which fruit is the mode and what fraction of the total prefer bananas?
某条形图显示了 40 名学生最喜欢的水果：苹果 (10)、香蕉 (12)、橙子 (8)、葡萄 (6)、其他 (4)。问题要求：找出众数水果，并计算喜欢香蕉的学生占总数的几分之几。
The mode is the category with the highest frequency. Bananas have the highest bar with a frequency of 12, so the mode is bananas.
众数是频数最高的类别。香蕉的条形最高，频数为 12，所以众数是香蕉。
To find the fraction for bananas, write frequency over total: 12/40. This simplifies by dividing numerator and denominator by 4 to give 3/10.
计算香蕉所占比例，用频数除以总数：12/40。分子分母同除以 4 约简，得到 3/10。
Always check the scale on the vertical axis when reading frequencies from a bar chart, as each small square may represent more than one unit.
从条形图读取频数时，务必检查纵轴刻度，因为每个小格可能代表多个单位。
3. Pie Chart Angles | 饼图角度计算
A survey of 60 people on their preferred sport yields: football 20, tennis 15, basketball 10, swimming 15. A pie chart is to be drawn. Calculate the angle for the football sector.
一项对 60 人的偏好运动调查结果为：足球 20 人，网球 15 人，篮球 10 人，游泳 15 人。需要绘制饼图。请计算足球扇形的圆心角。
The total number of people is 60. The angle for any sector is found using the formula:
总人数为 60。任一扇形的角度计算公式为：
Angle = (Frequency / Total) × 360°
For football: (20 / 60) × 360°. First simplify 20/60 to 1/3. Then 1/3 × 360° = 120°.
足球扇形：(20 / 60) × 360°。先将 20/60 化简为 1/3，然后 1/3 × 360° = 120°。
Similarly, tennis and swimming each give (15/60) × 360° = 90°, and basketball gives (10/60) × 360° = 60°. Always check that all angles sum to 360°.
同理，网球和游泳分别是 (15/60) × 360° = 90°，篮球是 (10/60) × 360° = 60°。务必验证所有角度之和等于 360°。
4. Mode from a Data Set | 数据集的众数
Find the mode from this list of test scores: 7, 9, 6, 7, 8, 7, 10, 6, 7, 9.
从以下测验分数列表中找出众数：7, 9, 6, 7, 8, 7, 10, 6, 7, 9。
The mode is the value that appears most often. Organising the data can help: 6 appears twice, 7 appears four times, 8 once, 9 twice, 10 once. The highest frequency is 4, so the mode is 7.
众数是出现次数最多的数值。整理数据有助于观察：6 出现两次，7 出现四次，8 一次，9 两次，10 一次。最高频数为 4，因此众数是 7。
If two values share the highest frequency, the data set is bimodal. In this case, only 7 is the highest, so a single mode.
如果有两个值同时达到最高频数，则数据集为双众数。这里只有 7 最高，所以是单众数。
5. Median Calculation | 中位数计算
Find the median of the following ordered data: 3, 5, 7, 9, 11. Then find the median of: 4, 6, 8, 10, 12, 14.
求以下有序数据的中位数：3, 5, 7, 9, 11。再求另一组数据的中位数：4, 6, 8, 10, 12, 14。
When the number of data values n is odd, the median is the middle value. Here n = 5, the third value is 7, so median = 7.
当数据个数 n 为奇数时，中位数是正中间的值。此处 n = 5，第 3 个数是 7，所以中位数为 7。
For an even number of values, there are two middle numbers. With 6 values, the middle pair are the 3rd and 4th: 8 and 10. The median is their mean: (8 + 10) / 2 = 9.
当数据个数为偶数时，有两个中间数。6 个数据中，中间对是第 3 和第 4 个：8 和 10。中位数是它们的平均值：(8 + 10) / 2 = 9。
Always sort the data in order first. The median is not affected by extreme values, making it useful for comparing distributions.
一定要先将数据排序。中位数不受极端值影响，因此在比较分布时很有用。
6. Mean from a Frequency Table | 从频数表求平均数
A frequency table shows the number of pets owned by 25 families:
一张频数表显示了 25 个家庭拥有的宠物数量：
| Number of pets (x) | Frequency (f) |
|---|---|
| 0 | 4 |
| 1 | 8 |
| 2 | 9 |
| 3 | 4 |
Calculate the mean number of pets per family.
计算每个家庭平均拥有的宠物数量。
Add an fx column multiplying each x by its frequency: 0×4=0, 1×8=8, 2×9=18, 3×4=12. Sum of fx = 0+8+18+12 = 38. Total frequency Σf = 25.
增加一列 fx，将每个 x 乘以对应频数：0×4=0, 1×8=8, 2×9=18, 3×4=12。fx 总和 = 0+8+18+12 = 38。总频数 Σf = 25。
Mean = Σ(fx) / Σf = 38 / 25 = 1.52 pets
The mean is a suitable average when data are fairly symmetrical without extreme outliers.
平均数在数据分布对称且无极端值时是合适的平均值。
7. Range and Spread | 极差与离散度
Two cricketers’ scores over five innings: Player A: 30, 45, 28, 52, 35. Player B: 10, 80, 12, 75, 18. Compare their ranges and decide who is more consistent.
两位板球运动员五局得分：选手 A: 30, 45, 28, 52, 35。选手 B: 10, 80, 12, 75, 18。比较他们的极差，并判断谁的发挥更稳定。
Range = highest value – lowest value. For Player A: max 52, min 28, range = 52 – 28 = 24.
极差 = 最大值 – 最小值。选手 A：最大 52，最小 28，极差 = 52 – 28 = 24。
For Player B: max 80, min 10, range = 80 – 10 = 70.
选手 B：最大 80，最小 10，极差 = 80 – 10 = 70。
A smaller range indicates more consistent scores. Player A has a much smaller range (24) than Player B (70), so Player A is more consistent, even though both may have similar means.
极差越小表示得分越稳定。选手 A 的极差 (24) 比选手 B (70) 小得多，因此选手 A 更加稳定，即使两人的平均分可能相近。
8. Probability Scale | 概率尺度
Describe the probability of each event using a word from the probability scale: impossible, unlikely, even chance, likely, certain. a) The sun will rise tomorrow. b) Rolling a 7 on a fair six-sided dice. c) Picking a red card from a standard deck of 52 playing cards.
用概率尺度的词语描述每个事件的概率：不可能、不太可能、等可能性、很可能、一定。a) 明天太阳升起。b) 掷一个公平六面骰子得到 7 点。c) 从一副标准的 52 张扑克牌中抽到一张红色牌。
The sun rising tomorrow is a certainty based on all past evidence, so the probability is ‘certain’. In numerical terms, it is 1.
根据所有过往证据，明天太阳升起是必然事件，所以概率是“一定”，数值为 1。
Rolling a 7 on a standard dice that only shows 1 to 6 cannot happen, so it is ‘impossible’. The probability is 0.
标准骰子只有 1 到 6 点，掷出 7 点不可能发生，因此是“不可能”，概率为 0。
A deck has 26 red cards and 26 black cards. The chance of picking a red card is exactly 26/52 = 1/2, described as an ‘even chance’.
一副牌有 26 张红牌和 26 张黑牌。抽到红牌的概率恰好是 26/52 = 1/2，描述为“等可能性”。
9. Probability of an Event | 单一事件的概率
A bag contains 5 blue marbles, 3 red marbles and 2 green marbles. One marble is chosen at random. Find the probability that it is (a) blue, (b) red, (c) not green.
一个袋子里有 5 粒蓝色弹珠、3 粒红色弹珠和 2 粒绿色弹珠。随机抽取一粒。求抽到以下颜色的概率：(a) 蓝色，(b) 红色，(c) 不是绿色。
The total number of marbles is 5 + 3 + 2 = 10. Probability is calculated as:
总弹珠数为 5 + 3 + 2 = 10。概率计算公式为：
P(event) = Number of favourable outcomes / Total number of outcomes
(a) P(blue) = 5 / 10 = 1/2 or 0.5. (b) P(red) = 3 / 10. (c) ‘Not green’ means the marble is either blue or red. Favourable outcomes = 5 + 3 = 8, so P(not green) = 8/10 = 4/5.
(a) P(蓝色) = 5/10 = 1/2 或 0.5。(b) P(红色) = 3/10。(c) “不是绿色”表示弹珠是蓝色或红色，有利结果数 = 5+3=8，因此 P(非绿) = 8/10 = 4/5。
Always check that probabilities of all mutually exclusive outcomes sum to 1: 5/10 + 3/10 + 2/10 = 1.
务必检查所有互斥结果的概率之和为 1：5/10 + 3/10 + 2/10 = 1。
10. Mutually Exclusive Events | 互斥事件
A spinner has four equal sections labelled A, B, C, D. What is the probability of landing on A or B? Explain why events ‘land on A’ and ‘land on B’ are mutually exclusive.
一个转盘被均分为四个区域，分别标有 A, B, C, D。求转到 A 或 B 的概率。解释为什么“转到 A”和“转到 B”是互斥事件。
Events are mutually exclusive if they cannot happen at the same time. The spinner cannot land on both A and B in a single spin, so the events are mutually exclusive.
如果两个事件不可能同时发生，则它们互斥。转盘单次转动不可能同时停在 A 和 B，因此这些事件互斥。
For mutually exclusive events, the probability of either occurring is the sum of their individual probabilities. P(A) = 1/4, P(B) = 1/4. So P(A or B) = 1/4 + 1/4 = 2/4 = 1/2.
对于互斥事件，其中任一发生的概率是各自概率之和。P(A) = 1/4, P(B) = 1/4。所以 P(A 或 B) = 1/4 + 1/4 = 2/4 = 1/2。
This addition rule only works for mutually exclusive events. If events can overlap, you must subtract the overlap probability to avoid double counting.
这一加法规则仅适用于互斥事件。如果事件可能重叠，则必须减去重叠部分的概率，避免重复计算。
Published by TutorHao | Statistics Revision Series | aleveler.com
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