Unit 12 · Statistics　统计

By the end of this chapter you can:

1. Calculate the mean, median, and mode of a data set and explain when each measure is most appropriate
2. Compute variance $\sigma^{2}$ and standard deviation $\sigma$ from raw data using the population formula
3. Construct and read a frequency distribution table and convert it into a frequency density histogram
4. Identify the effect of outliers on mean vs. median and choose the right central measure for skewed data

Exam weight on past CSCA papers: ~3% (1–2 of 48 MCQs). Quick high-yield wins.

12.1　Descriptive Statistics　描述统计

Statistics begins with a data set — a list of $n$ numerical values $x_1, x_2, \ldots, x_n$. Three numbers summarise where the "centre" is.

Mean　平均数

$$\bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} = \frac{1}{n}\sum_{i=1}^{n} x_i$$

The mean is the balance point of the data. Every value contributes equally, so a single extreme value (outlier 异常值) can pull the mean far from the rest of the group.

Median　中位数

First sort the values in ascending order (从小到大排列), then:

The median is resistant to outliers — one extreme value cannot shift it by more than one rank.

Mode　众数

The value that appears most frequently. A data set may have one mode, more than one mode (bimodal / multimodal), or no mode if all values appear equally often.

Central tendency at a glance

⚠️ Mean vs. median when outliers appear:
If one student scores 5 on a test where everyone else scores 80–90, the mean drops sharply while the median barely moves. CSCA problems often ask which measure "better represents" the data (更能代表数据). Answer: when data is skewed or has clear outliers, the median is a better representative.

🔑 Sort first, always. The exam almost always gives you an unsorted list on purpose. Sort before computing median or identifying mode.

Worked Example 12.1.A

The scores of 7 students are: $72,\ 85,\ 90,\ 60,\ 85,\ 78,\ 5$.

Find the mean, median, and mode. Which measure best represents the class?

Solution.

Step 1 — Sort: $5,\ 60,\ 72,\ 78,\ 85,\ 85,\ 90$.

Step 2 — Mean:
$$\bar{x} = \frac{5 + 60 + 72 + 78 + 85 + 85 + 90}{7} = \frac{475}{7} \approx 67.9$$

Step 3 — Median: $n = 7$ (odd), so position $\dfrac{7+1}{2} = 4$. The 4th sorted value is $\mathbf{78}$.

Step 4 — Mode: $85$ appears twice; all other values appear once. Mode $= \mathbf{85}$.

Which best represents the class?
The score of $5$ is an outlier (e.g. absent or unwell student). The mean $\approx 67.9$ is dragged well below the main cluster of $60$–$90$. The median $= 78$ better represents a typical student's result.

$$\boxed{\bar{x} \approx 67.9,\qquad \text{median} = 78,\qquad \text{mode} = 85}$$

⚠️ Even $n$ trap: When $n = 6$, for instance, students often pick just one middle value. Always average positions 3 and 4.

12.2　Variance and Standard Deviation　方差与标准差

The mean tells us where the centre is; variance (方差) measures how far data spreads around that centre.

Population variance　方差

$$\sigma^{2} = \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^{2}$$

In words: compute each data value's deviation from the mean, square it, then average all the squares.

Standard deviation　标准差

$$\sigma = \sqrt{\sigma^{2}} = \sqrt{\frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^{2}}$$

Standard deviation has the same units as the original data, making it more interpretable than variance.

Key properties

🔑 Shortcut formula — mean of squares minus square of mean:
$$\sigma^{2} = \frac{1}{n}\sum x_i^{2} - \bar{x}^{2} = \overline{x^{2}} - \bar{x}^{2}$$
This avoids computing $(x_i - \bar{x})$ for each value and is faster on the exam.

⚠️ Population vs. sample formula:
The CSCA syllabus uses the population formula with denominator $n$. The sample variance (样本方差) uses $n-1$ (Bessel's correction). Unless the problem explicitly says "sample variance," use denominator $n$.

Worked Example 12.2.A

Five students' scores are: $70,\ 75,\ 80,\ 85,\ 90$. Find $\sigma^{2}$ and $\sigma$.

Solution.

Step 1 — Mean:
$$\bar{x} = \frac{70+75+80+85+90}{5} = \frac{400}{5} = 80$$

Step 2 — Deviations and squared deviations:

Step 3 — Variance:
$$\sigma^{2} = \frac{250}{5} = 50$$

Step 4 — Standard deviation:
$$\sigma = \sqrt{50} = 5\sqrt{2} \approx 7.07$$

$$\boxed{\sigma^{2} = 50,\qquad \sigma = 5\sqrt{2}}$$

Verification using the shortcut:
$$\overline{x^{2}} = \frac{70^{2}+75^{2}+80^{2}+85^{2}+90^{2}}{5} = \frac{32250}{5} = 6450$$
$$\sigma^{2} = 6450 - 80^{2} = 6450 - 6400 = 50 \checkmark$$

12.3　Frequency Distributions　频率分布

When a data set is large, we group values into class intervals (组距) and summarise with a frequency distribution table and histogram.

Frequency distribution table　频率分布表

• Frequency 频数 $f$: raw count of values in the interval.
• Relative frequency 频率 $f/n$: proportion of all data in that interval; all relative frequencies sum to $1$.
• Class width 组距 $h$: length of each interval.

Frequency density histogram　频率/组距直方图

On a standard CSCA histogram the vertical axis is frequency density (频率/组距), not raw frequency:

$$\text{Frequency density} = \frac{\text{Relative frequency}}{\text{Class width}} = \frac{f/n}{h}$$

The area of each bar equals the relative frequency of that class:
$$\text{Bar area} = \text{frequency density} \times h = \frac{f/n}{h} \times h = \frac{f}{n}$$

All bar areas together sum to $1$.

🔑 Reading a histogram: The $y$-axis label "频率/组距" is your signal that bar height $\ne$ relative frequency. You must multiply height by class width to get the relative frequency, then multiply by $n$ to recover the count.

⚠️ Unequal class widths: If intervals have different widths, bars of equal relative frequency will have different heights. Frequency density corrects for this — wider intervals get shorter bars, keeping areas honest.

Worked Example 12.3.A

A sample of $n = 40$ measurements is grouped as follows. All class widths are $h = 10$.

(a) Complete the relative frequency and frequency density columns.

(b) Which bar is tallest in the histogram?

(c) How many values are at least 30?

Solution.

(a)

(b) The $[30, 40)$ bar has the highest frequency density ($0.040$), so it is the tallest.

(c) Values in $[30,40)$ plus $[40,50)$: $16 + 8 = 24$.

$$\boxed{24 \text{ values are } \ge 30.}$$

Try it!　自测练习

Q1. Find the mean, median, and mode of: $3,\ 7,\ 7,\ 9,\ 4$.

Q2. The data set $\{a,\ 4,\ 6,\ 8,\ 12\}$ has mean $\bar{x} = 7$. Find $a$.

Q3. Calculate the variance of $\{2,\ 4,\ 4,\ 4,\ 6\}$.

Q4. A frequency density histogram has a bar over $[20, 30)$ with height $0.025$. The total sample size is $n = 80$. How many values fall in $[20, 30)$?

Q5. Two classes both have mean score 75. Class A has $\sigma^{2} = 4$ and Class B has $\sigma^{2} = 36$. Which class is more consistent, and why?

📌 Chapter summary

Topic	Key formula / concept	Common trap
Mean 平均数	$\bar{x} = \frac{1}{n}\sum x_i$	Pulled by outliers; prefer median for skewed data
Median 中位数	Middle of sorted list; average two middles when $n$ is even	Forgetting to sort first; wrong middle position for even $n$
Mode 众数	Most frequent value	May not exist, or may be non-unique
Variance 方差	$\sigma^{2} = \frac{1}{n}\sum(x_i-\bar{x})^2 = \overline{x^2}-\bar{x}^2$	Using $n-1$ (sample) instead of $n$ (population)
Standard deviation 标准差	$\sigma = \sqrt{\sigma^{2}}$; same units as data	$\sigma^{2}=0$ iff all values equal
Relative frequency 频率	$f/n$; all classes sum to 1	Confusing frequency count with relative frequency
Frequency density 频率/组距	$(f/n)\div h$; bar area = relative frequency	Reading histogram bar height as frequency or probability directly

Connection to earlier units → Variance's shortcut form $\overline{x^2} - \bar{x}^2$ is a disguised difference of squares — the same algebraic identity $(a^2 - b^2)$ pattern that appears throughout Unit 1 and Unit 3. Recognising it lets you skip the deviation table entirely on the exam.