📚 A Quick Reference Handbook of Formulas and Theorems for Year 8 SQA Statistics | Year 8 SQA 统计：公式定理速查手册
This quick reference handbook collects all the key formulas, definitions and theorems you need for Year 8 SQA Statistics. It is written as a revision companion: read it before a test, keep it beside you while you practise past-paper questions, and use the summary tables at the end to check that nothing has been forgotten.
本速查手册汇总了 Year 8 SQA 统计学所需的全部关键公式、定义与定理。它是一份复习搭档：考前通读一遍，做练习题时放在手边随时查阅，最后再用文末的速查总表检查自己是否遗漏了任何知识点。
Every section follows the same pattern - the definition first, then the formula in a centred box, then a worked example. Where a formula has a common trap, the trap is pointed out explicitly, because in SQA assessments marks are usually lost through small slips rather than through not knowing the method at all.
每一节都遵循同样的结构——先给定义，再给出居中排版的公式，然后配一道例题。凡是公式常见易错的地方，都会明确指出，因为在 SQA 的考评中，失分通常来自小疏忽，而不是完全不会方法。
1. The Data Handling Cycle | 数据处理循环
Statistics in Year 8 is not a list of separate tricks; it is a cycle. You start with a question, collect data that can answer it, organise and display that data, analyse it with the measures in this handbook, and finally communicate a conclusion that refers back to the original question.
Year 8 的统计不是一堆互不相关的技巧，而是一个循环：从一个问题出发，收集能够回答该问题的数据，整理并展示数据，用本手册中的各种量数进行分析，最后给出回到原问题的结论。
Formulate a question that data can answer, for example: How long does a Year 8 pupil spend on homework each night?
提出一个数据能够回答的问题，例如：Year 8 学生每晚做作业花多长时间？
Plan and collect data using a survey, an experiment, a sensor or a published source, and record the sample size n.
用问卷调查、实验、传感器或公开数据来规划并收集数据，并记录样本容量 n。
Process and represent the data using tables, bar charts, pie charts, line graphs, scatter graphs or box plots.
用表格、条形图、饼图、折线图、散点图或箱线图来处理与呈现数据。
Analyse and interpret the data using an average (mean, median or mode) together with a measure of spread (range or interquartile range).
用集中量数（平均数、中位数或众数）配合离散程度的量数（极差或四分位距）来分析并解读数据。
Communicate the conclusion in context, stating the comparison, the evidence and any uncertainty.
结合具体情境表述结论，说明比较结果、支撑证据以及存在的不确定性。
A useful rule throughout the course is that a single number never tells the whole story. A claim such as the average is 12 minutes is incomplete until you also say how spread out the times were and who was measured.
贯穿整个课程的一条实用原则是：单个数字永远讲不完整个故事。像"平均值为 12 分钟"这样的说法并不完整，你还必须说明这些时间的离散程度以及测量对象是谁。
2. Mean | 平均数
The mean (or average) is a measure of central tendency. It is found by adding all the data values together and then dividing by the total number of values. The mean uses every value in the data set, which makes it powerful but also sensitive to extreme values.
平均数（或平均值）是一种集中量数。计算方法是先将所有数据值相加，再除以数据值的总个数。平均数用到了数据集中的每一个数值，因此信息量大，但也容易受极端值影响。
If a data set contains n values, listed as x₁, x₂, …, xₙ, then the mean is given by the formula:
如果数据集包含 n 个值，记为 x₁, x₂, …, xₙ，则平均数由以下公式给出：
Mean = (x₁ + x₂ + … + xₙ) ÷ n
The same idea is written more compactly using sigma notation, where Σx means the sum of all the values of x:
用 Σ 记号可以写得更加紧凑，其中 Σx 表示 x 的所有取值之和：
x̄ = Σx ÷ n
For example, the mean of 4, 7, 9, 10 is (4 + 7 + 9 + 10) ÷ 4 = 30 ÷ 4 = 7.5.
例如，数据集 4、7、9、10 的平均数为 (4 + 7 + 9 + 10) ÷ 4 = 30 ÷ 4 = 7.5。
d again and again. First, the sum of the values equals the mean multiplied by n, so Σx = x̄ the rest, the mean is pulled towards it, so the mean can be larger than most of the data.
有两个性质反复出现在考题中。第一，各数值之和等于平均数乘以 n，即 Σx = x̄ × n，这是找回缺失总和的方法。第二，若某个数值远大于其他数值，平均数会被拉向它，因此平均数可能比大多数数据都大。
of the values are 5, 6, 9 and 11, their sum is 31, so the fifth value must be 40 − 31 = 9.
验算示例：五个数的平均数为 8，则 Σx = 8 × 5 = 40。若其中四个数是 5、6、9、11，其和为 31，那么第五个数必须是 40 − 31 = 9。
3. Median | 中位数
The median is the middle value when the data are arranged in order. It splits the data into two equal halves. Because it depends on position rather than on size, the median is not dragged about by one extreme value.
中位数是一组数据排序后位于中间位置的数值，它将数据分成相等的两部分。由于它取决于位置而非数值大小，因此不会被某个极端值"拉走"。
To find the median: first order the numbers from smallest to largest. If there is an odd number of values, the median is the middle one. If there is an even number of values, the median is the mean of the two middle numbers.
求中位数的步骤：先将数据从小到大排列。如果数据个数为奇数，则中位数为最中间的那个数；如果个数为偶数，则中位数为中间两个数的平均数。
The position of the median can be located directly with this formula:
中位数的位置可以直接用下面的公式确定：
Position of median = (n + 1) ÷ 2
ple: for the data 3, 5, 7, the median is 5. For 2, 4, 6, 8, the median is (4 + 6) ÷ 2 = 5.
示例：数据集 3、5、7 的中位数是 5。数据集 2、4、6、8 的中位数是 (4 + 6) ÷ 2 = 5。
, so the median lies halfway between the fifth and sixth values, giving (6 + 7) ÷ 2 = 6.5.
位置公式验算：对于 2、3、4、5、6、7、8、9、10、11 这十个数据，中位数位置为 (10 + 1) ÷ 2 = 5.5，因此中位数位于第 5 个和第 6 个数值之间，即 (6 + 7) ÷ 2 = 6.5。
Remember that the median must be read from ordered data. Sorting the list is not optional, and forgetting to sort is one of the most common sources of lost marks in this topic.
请记住，中位数必须从排序后的数据中读取。给数据排序不是可省略的步骤，而"忘记排序"正是本主题最常见的失分原因之一。
4. Mode | 众数
The mode is the value that appears most frequently in a data set. A set of data may have one mode, more than one mode, or no mode at all if all values occur equally often. The mode is the only average that can be used with non-numerical (categorical) data, such as favourite colours or types of transport.
众数是数据集中出现次数最多的数值。一组数据可能有一个众数、多个众数，或者如果所有数值出现次数相同，则没有众数。众数是唯一可用于非数值型（分类）数据的集中量数，例如最喜欢的颜色或出行方式。
For example, in the list 2, 3, 3, 5, 7, the mode is 3. In 1, 2, 2, 4, 4, 6, the modes are 2 and 4 (bimodal).
例如，在数据集 2、3、3、5、7 中，众数是 3。在 1、2、2、4、4、6 中，众数为 2 和 4（双众数）。
One mode: the value with the highest frequency, for example 3 in the list above.
单众数：频数最高的数值，例如上例中的 3。
Two or more modes: two or more values tie for the highest frequency, so report all of them.
多众数：两个或更多数值并列最高频数，需要全部写出。
No mode: every value occurs the same number of times, for example 1, 2, 3, 4 each occurring once.
无众数：每个数值出现次数相同，例如 1、2、3、4 各出现一次。
For grouped or tabulated data, the modal class is the class interval with the largest frequency. The modal class is reported as an interval, such as 20 to 29 minutes, not as a single number.
对于分组数据或频数表数据，众数所在组是频数最大的组区间。众数所在组应以区间形式报告，例如 20 至 29 分钟，而不是单个数字。
5. Range | 极差
The range is a measure of spread. It shows how spread out the data values are. It is calculated as the difference between the largest and smallest values, and it is always a single number with the same units as the data.
极差是用来衡量离散程度的量数，反映数据值的分散程度。它的计算方法是最大值与最小值之差，结果是一个与数据单位相同的单一数值。
Range = largest value − smallest value
, 11, 19, 14, the largest value is 19 and the smallest is 11, so the range is 19 − 11 = 8.
示例：对于数据 12、15、11、19、14，最大值为 19，最小值为 11，因此极差为 19 − 11 = 8。
The range is quick to calculate and easy to interpret, but it has a serious weakness: it uses only two values and ignores everything in between. A single unusually high reading can make the range look enormous even when the rest of the data are tightly clustered.
极差计算快捷、易于解释，但有一个明显缺点：它只用到两个数值，忽略了中间的所有数据。哪怕其余数据非常集中，只要出现一个异常高的读数，极差就会显得极大。
Worked comparison: set A is 10, 11, 12, 13, 14 and set B is 10, 12, 12, 12, 14. Both have range 4, but set B is clearly more concentrated in the middle. This is exactly why the interquartile range in the next section exists.
对比示例：数据集 A 为 10、11、12、13、14，数据集 B 为 10、12、12、12、14。两者的极差都是 4，但数据集 B 在中部明显更集中。这正是下一节四分位距存在的理由。
6. Quartiles and the Interquartile Range | 四分位数与四分位距
Quartiles split ordered data into four equal parts. Q1, the lower quartile, has roughly one quarter of the data below it; Q2 is the median; Q3, the upper quartile, has roughly three quarters of the data below it.
四分位数将排序后的数据分成四等份。下四分位数 Q1 之下约有四分之一的数据；Q2 就是中位数；上四分位数 Q3 之下约有四分之三的数据。
The interquartile range (IQR) measures the spread of the middle half of the data, so it ignores the extreme quarters at each end:
四分位距（IQR）衡量的是数据中间一半的离散程度，因此忽略了首尾各四分之一的极端数据：
IQR = Q3 − Q1
The IQR also gives a test for outliers. Any value outside the fences below is usually treated as an outlier:
四分位距还可以用来判断异常值。凡超出下面两个界限的数值，通常视为异常值：
Lower fence = Q1 − 1.5 × IQR Upper fence = Q3 + 1.5 × IQR
15, 18, 20, so Q3 = 15. The median is (9 + 11) ÷ 2 = 10, so Q2 = 10 and IQR = 15 − 6 = 9.
例题：考虑排序后的十个数值 2、5、6、7、9、11、13、15、18、20。下半部分为 2、5、6、7、9，故 Q1 = 6；上半部分为 11、13、15、18、20，故 Q3 = 15。中位数为 (9 + 11) ÷ 2 = 10，即 Q2 = 10，因此 IQR = 15 − 6 = 9。
15 + 13.5 = 28.5. Every value lies inside these fences, so this data set has no outliers.
异常值检验：下界限为 6 − 1.5 × 9 = 6 − 13.5 = −7.5，上界限为 15 + 13.5 = 28.5。所有数值都落在界限之内，因此该数据集没有异常值。
A box plot displays the five-figure summary - minimum, Q1, median, Q3 and maximum - with the box drawn from Q1 to Q3 and the median marked inside it. The longer the box, the more spread out the middle half of the data.
箱线图展示五数概括——最小值、Q1、中位数、Q3、最大值——箱体从 Q1 画到 Q3，中位数标在箱内。箱体越长，说明中间一半数据越分散。
7. Frequency Tables and Grouped Data | 频数表与分组数据
Raw data are often tidied into a frequency table. If the data are discrete and ungrouped, the mean can be calculated exactly by multiplying each value by its frequency and dividing by the total frequency:
原始数据通常会被整理成频数表。如果数据是离散且未分组的，可以用"每个数值乘以它的频数，再除以总频数"来精确计算平均数：
x̄ = Σ(f × x) ÷ Σf where Σf = n
For grouped data you do not know the individual values, so you use the midpoint of each class interval as a representative value. This gives an estimate of the mean rather than an exact value, which is why exam answers should say estimate.
对于分组数据，由于不知道各个原始数值，只能用每个组区间的组中值作为代表值。这样得到的是平均数的估计值而非精确值，因此答题时应写明是估计值。
| Time (minutes) | 时间（分钟） | Midpoint x | 组中值 x | Frequency f | 频数 f | f × x |
|---|---|---|---|
| 0 to 9 | 4.5 | 3 | 13.5 |
| 10 to 19 | 14.5 | 7 | 101.5 |
| 20 to 29 | 24.5 | 6 | 147 |
| 30 to 39 | 34.5 | 4 | 138 |
20 = 20 minutes. The modal class is 10 to 19 minutes because it has the highest frequency.
由表格可得 Σf = 3 + 7 + 6 + 4 = 20，Σ(f × x) = 13.5 + 101.5 + 147 + 138 = 400，因此平均数的估计值为 400 ÷ 20 = 20 分钟。众数所在组为 10 至 19 分钟，因为它的频数最大。
Two cautions. Midpoints must be consistent: for the class 0 to 9 the midpoint is 4.5, not 5, if the data are continuous. And when classes have unequal widths, frequency alone can mislead you, so compare frequency density instead.
有两点需要注意。组中值必须保持一致：若数据是连续的，0 至 9 这一组的组中值是 4.5，而不是 5。另外，当各组区间宽度不等时，只看频数会产生误导，应改为比较频数密度。
8. Comparing Distributions and Identifying Outliers | 分布比较与异常值识别
A statistical comparison is a sentence, not a number. The standard structure is: state the average for each group, say which is higher, then compare the spread, then interpret both in the context of the question.
统计比较是一句话，而不是一个数字。标准结构是：先给出各组的平均数，说明哪一组更高，再比较离散程度，最后结合问题情境解读两者。
Model answer: the median time for Year 8 pupils is 20 minutes compared with 15 minutes for Year 7 pupils, so Year 8 pupils typically spend longer on homework. The interquartile range for Year 8 is 9 minutes compared with 4 minutes for Year 7, which shows that Year 8 homework times are more variable as well as longer.
范答：Year 8 学生的中位数为 20 分钟，而 Year 7 学生为 15 分钟，说明 Year 8 学生做作业时间通常更长。Year 8 的四分位距为 9 分钟，Year 7 为 4 分钟，说明 Year 8 的作业时间不仅更长，波动也更大。
Use the median and IQR when the data are skewed or contain outliers, because they are resistant to extreme values.
当数据偏斜或含有异常值时，应使用中位数和四分位距，因为它们不易受极端值影响。
Use the mean and range when the data are reasonably symmetric and you want to use every value.
当数据大致对称且希望利用每一个数值时，应使用平均数和极差。
Always name the units and the group, and never simply write the mean is bigger without saying bigger than what.
务必写明单位和组别，绝不能只写"平均数更大"而不说明比什么更大。
Outliers deserve a comment rather than automatic deletion. They may be recording errors, or they may be genuine and important. If an outlier is removed, say so and explain why; if it is kept, explain its effect on the mean.
对异常值应当加以说明，而不是自动删除。它可能是记录错误，也可能是真实且重要的数据。如果删除了异常值，要说明并解释原因；如果保留，要说明它对平均数的影响。
Effect to remember: adding a very large value increases the mean and the range but leaves the median
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