Year 8 Cambridge Statistics: Formula & Theorems Quick Reference Handbook

📚 Year 8 Cambridge Statistics: Formula & Theorems Quick Reference Handbook | 公式定理速查手册

Welcome to your essential quick reference guide for Year 8 Cambridge Statistics. Here you will find clear statements of all the key formulas, theorems, and concepts you need to memorise and apply. Each section pairs a concise explanation in English with its equivalent in Chinese, helping you build confidence whether you are studying at an international school or preparing for the Cambridge lower secondary checkpoint.

欢迎使用八年级剑桥统计必备速查手册。这里汇集了所有你需要记忆和运用的关键公式、定理和概念。每一节都用中英文配对简明解释，无论你是在国际学校学习还是准备剑桥初中段测评，都能帮你树立信心。

1. Mean (Average) | 平均数

The mean is the sum of all data values divided by the number of data points. It is the most common measure of central tendency and can be affected by extreme values.

平均数是所有数据值的总和除以数据点的个数。它是最常见的集中趋势度量，但会受到极端值的影响。

Mean = (Sum of all values) ÷ (Number of values)

If a dataset contains the numbers 4, 8, 6, 5, 9, the mean is (4+8+6+5+9) / 5 = 32/5 = 6.4.

如果数据集包含数字 4、8、6、5、9，平均数就是 (4+8+6+5+9) / 5 = 32/5 = 6.4。

2. Median | 中位数

The median is the middle value when the data are arranged in order. For an even number of data points, the median is the average of the two middle numbers.

中位数是将数据按顺序排列后的中间值。如果数据点个数为偶数，中位数就是中间两个数的平均数。

Place data in order → For odd n: median = middle value.
For even n: median = (n/2ᵗʰ value + (n/2 +1)ᵗʰ value) ÷ 2

Example: Data 3, 5, 7, 12, 13 (odd count: 5). Median = 7.
Data 4, 6, 8, 10 (even count: 4). Median = (6+8)/2 = 7.

例子：数据 3、5、7、12、13（奇数个：5）。中位数 = 7。
数据 4、6、8、10（偶数个：4）。中位数 = (6+8)/2 = 7。

3. Mode | 众数

The mode is the value that appears most often in a data set. A set of data may have one mode, more than one mode (bimodal or multimodal), or no mode at all if all values appear equally often.

众数是数据集中出现次数最多的值。一组数据可能有一个众数、多个众数（双众数或多众数），或者如果所有值出现次数相同则没有众数。

Mode = value with the highest frequency

In the list 2, 3, 3, 4, 5, 5, 5, 6, the mode is 5 because it occurs three times, more than any other number.

在数据列 2、3、3、4、5、5、5、6 中，众数是 5，因为它出现了三次，比任何其他数字都多。

4. Range | 极差（范围）

The range is a measure of spread. It tells you how far apart the data are, and is found by subtracting the smallest value from the largest.

极差是衡量数据分散程度的度量。它告诉你数据有多分散，通过最大值减去最小值得到。

Range = Largest value − Smallest value

For the test scores 45, 67, 82, 91, the range = 91 − 45 = 46. A larger range indicates greater spread.

对于测验分数 45、67、82、91，极差 = 91 − 45 = 46。极差越大说明数据越分散。

5. Frequency Tables & Mean from Frequency | 频数表及用频数求均值

When data are presented in a frequency table, the mean is calculated by multiplying each value by its frequency, summing these products, and then dividing by the total frequency.

当数据以频数表呈现时，计算平均数要将每个值乘以其频数，把这些乘积相加，然后除以总频数。

Mean from frequency table = (Σ(value × frequency)) ÷ (Σ frequency)

Value x: 2, 5, 7; Frequency f: 3, 2, 4
Sum of f × x = (2×3) + (5×2) + (7×4) = 6 + 10 + 28 = 44
Total frequency = 3+2+4 = 9
Mean = 44 ÷ 9 ≈ 4.89

数值 x：2、5、7；频数 f：3、2、4
f × x 之和 = (2×3) + (5×2) + (7×4) = 6 + 10 + 28 = 44
总频数 = 3+2+4 = 9
平均数 = 44 ÷ 9 ≈ 4.89

6. Grouped Data and Estimated Mean | 分组数据与估算平均数

When data are grouped into class intervals, we use the midpoint of each interval as the representative value. The estimated mean is then calculated similarly to the frequency table mean, but with midpoints.

当数据被分成组距时，我们用每一组的中点值作为代表值。然后类似于频数表均值的方式计算估算平均数，但用的是组中值。

Estimated mean = (Σ(midpoint × frequency)) ÷ (Σ frequency)

Class 0 ≤ x < 10, frequency 4, midpoint 5; class 10 ≤ x < 20, frequency 6, midpoint 15.
Sum of products = (5×4) + (15×6) = 20 + 90 = 110. Total frequency = 10. Estimated mean = 11.

组距 0 ≤ x < 10，频数 4，中点 5；组距 10 ≤ x < 20，频数 6，中点 15。
乘积之和 = (5×4) + (15×6) = 20 + 90 = 110。总频数 = 10。估算平均数 = 11。

7. Median and Mode from Frequency Tables | 从频数表找中位数与众数

To find the median from a frequency table, use the cumulative frequency to locate the middle position. The mode is simply the value with the highest frequency.

从频数表中找中位数，需要利用累积频数来定位中间位置。众数就是频数最高的那个值。

Median position = (Total frequency + 1) ÷ 2

Data: x=1 (f=3), x=2 (f=5), x=3 (f=2). Total f=10. Median position = (10+1)/2 = 5.5, so median is between the 5th and 6th value. Cumulative: 1→3, 2→8, so both 5th and 6th are 2, median=2. Mode=2 because frequency 5 is highest.

数据：x=1 (f=3)、x=2 (f=5)、x=3 (f=2)。总频数=10。中位数位置 = (10+1)/2 = 5.5，因此中位数在第5和第6个值之间。累积：1→3，2→8，所以第5和第6个值都是2，中位数=2。众数=2，因为频数5最高。

8. Bar Charts and Pictograms | 柱状图与象形图

Bar charts represent categorical data with rectangular bars of height proportional to the frequency. The bars must be of equal width and gaps between bars. Pictograms use symbols to represent a certain number of items, and a key must be given.

柱状图用高度与频数成正比的矩形条表示分类数据。条宽必须相等，且条间有间隔。象形图用符号表示一定数量的项目，必须给出图例说明。

Height of bar = frequency of the category

If one symbol represents 4 books, and the category ‘Fiction’ shows 7 symbols, that represents 7×4 = 28 books. Always check the key.

如果一个符号代表4本书，类别“小说”显示7个符号，就代表7×4=28本书。务必检查图例。

9. Pie Charts and Angles | 饼图与角度计算

In a pie chart, each sector angle is proportional to the frequency of the category. The total angle in a circle is 360°, so the angle for a category is calculated by (category frequency ÷ total frequency) × 360°.

在饼图中，每个扇区的角度与该类别的频数成比例。圆的总角度是360°，所以一个类别的角度等于（该类别的频数÷总频数）×360°。

Sector angle = (Frequency of category ÷ Total frequency) × 360°

Total students = 30; 12 prefer football. Angle for football = (12/30) × 360° = 144°. Use a protractor to draw the sector.

学生总数=30；12人喜欢足球。足球扇区角度 = (12/30) × 360° = 144°。用量角器画出该扇区。

10. Probability Scale and Basic Probability | 概率标度与基本概率

Probability is a measure of how likely an event is to happen. It is always a number between 0 and 1 inclusive, where 0 means impossible and 1 means certain. Probability can be written as a fraction, decimal, or percentage.

概率是衡量事件发生可能性的度量，总是介于0和1之间的一个数，包括0和1，0表示不可能，1表示必然发生。概率可以用分数、小数或百分数表示。

Probability of an event = (Number of favourable outcomes) ÷ (Total number of equally likely outcomes)

When rolling a fair six-sided die, the probability of rolling a 4 is 1/6. The probability of rolling an even number is 3/6 = 1/2.

抛掷一个均匀的六面骰子，掷出4点的概率是1/6。掷出偶数的概率是3/6 = 1/2。

11. Experimental Probability and Expected Frequency | 实验概率与期望频数

Experimental probability is based on actual trials or experiments. Expected frequency predicts how many times an event would occur if an experiment is repeated many times.

实验概率基于实际的试验或实验。期望频数预测如果实验重复很多次，事件预计会发生多少次。

Experimental probability = Number of times event occurs ÷ Total number of trials

Expected frequency = Probability of event × Number of trials

If a coin is flipped 150 times, the expected number of heads is 0.5 × 150 = 75. If a spinner is spun 80 times and lands on red 24 times, the experimental probability of red is 24/80 = 0.3.

如果一枚硬币抛掷150次，正面朝上的期望次数是 0.5 × 150 = 75。如果一个转盘旋转80次，红色出现24次，那么红色的实验概率是 24/80 = 0.3。

12. Mutually Exclusive Events and Addition Rule | 互斥事件与加法法则

Two events are mutually exclusive if they cannot happen at the same time. The probability that either one or the other happens is the sum of their individual probabilities.

如果两个事件不能同时发生，它们就是互斥的。任一事件发生的概率是它们各自概率的和。

For mutually exclusive events A and B: P(A or B) = P(A) + P(B)

In a bag with 3 red, 5 blue, and 2 green marbles, if you pick one marble, the probability of red or green is P(red) + P(green) = 3/10 + 2/10 = 5/10 = 1/2. Red and green are mutually exclusive.

袋中有3个红球、5个蓝球和2个绿球，随机抽取一个球，抽到红球或绿球的概率是 P(红)+P(绿)=3/10+2/10=5/10=1/2。红和绿是互斥事件。

The sum of probabilities of all mutually exclusive and exhaustive outcomes is 1. This is used to find missing probabilities.

所有互斥且穷举的结果的概率之和为1。这可用于求缺失的概率。

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Year 8 Cambridge Statistics: Formula & Theorems Quick Reference Handbook | 公式定理速查手册