统计

To organise statistical 统计 data 数据, use:

• a tally table 计数表 — make a mark for each value, then count.
• a two-way table 双向表 — sorts data by two features at once (for example, boys/girls against walk/bus).

When you read a diagram, only draw conclusions the data really supports.

Three averages describe a typical value of the data, and the range shows how spread out it is. Each is used for a different purpose:

• mean 平均数 $= \dfrac{\text{sum of all values}}{\text{how many values}}$.
• median 中位数 $=$ the middle value when the data is put in order; for an even number of values, average the two middle values.
• mode 众数 $=$ the value that appears most often.
• range 极差 $=$ largest value $-$ smallest value (it shows how spread out the data is).

Worked example. Find the mean, median, mode and range of $4, 7, 7, 2, 5$.

Order the data: $2, 4, 5, 7, 7$.

Worked example: an even number of values. Find the median of $12, 15, 9, 20, 14, 11$.

In order: $9, 11, 12, 14, 15, 20$. There are two middle values, so the median is halfway between them: $\dfrac{12 + 14}{2} = 13$.

Worked example: a missing value. The mean of five numbers is $8$. Four of them are $4$, $7$, $9$ and $10$. Find the fifth.

The five numbers add to $5 \times 8 = 40$, and the four known ones add to $30$, so the fifth is $40 - 30 = 10$.

Which average to use? The mode suits categories (the most popular colour); the median is not pulled by one extreme value (a few very high salaries); the mean uses every value.

Common mistakes. Finding the median without putting the data in order first, and giving the position $\tfrac{n+1}{2}$ instead of the value at that position.

When data is listed with its frequency 频数 (how many times each value occurs), the mean is

Worked example. Values $1, 2, 3$ occur with frequencies $4, 5, 1$. Find the mean.

Grouped data (Extended)

For grouped data 分组数据, you cannot find the exact mean, so estimate it using the midpoint of each group as the value. The modal class 众数组 is simply the group with the highest frequency.

Worked example: estimate the mean. The times, $t$ minutes, that $30$ students took to travel to school are grouped below.

The modal class is $10 < t \leqslant 20$. The answer is only an estimate, because we do not know the exact times inside each group.

Common mistakes. Multiplying by the end of each class (or the class width) instead of the midpoint, and dividing by the number of classes ($4$) instead of the total frequency ($30$).

When data is in order, the quartiles 四分位数 cut it into four equal parts. The lower quartile (LQ) is one quarter of the way up; the upper quartile (UQ) is three quarters of the way up. The interquartile range 四分位距 measures the spread of the middle half:

The interquartile range is useful because, unlike the range, it ignores extreme values.

Worked example: quartiles from a list. Find the median, quartiles and interquartile range of

There are $n = 11$ values, already in order. The median is the $\tfrac{n+1}{2} = 6$th value, $10$. The lower quartile is the $\tfrac{n+1}{4} = 3$rd value, $6$, and the upper quartile is the $\tfrac{3(n+1)}{4} = 9$th value, $15$. So the interquartile range is $15 - 6 = 9$.

Worked example: compare two distributions. Two cinemas record their audience numbers each night for a month.

Make one comparison of the average and one of the spread, and say what each means in context: "On average Cinema B has larger audiences, because its mean is higher. Cinema A's audience numbers vary more, because its range is larger."

Common mistake. Saying only "B's mean is bigger" without saying what that means for the audiences, or comparing two averages and no spread.

Worked example (pie chart). Out of $120$ people, $30$ chose tea. Find the angle of the tea slice.

Worked example (pie chart, backwards). A pie chart shows how $450$ people travel to work. The bus slice has angle $72^{\circ}$. How many people travel by bus?

A scatter diagram 散点图 plots pairs of values as points to show whether two things are linked. The link is the correlation 相关性:

• positive correlation 正相关 — as one goes up, the other goes up.
• negative correlation 负相关 — as one goes up, the other goes down.
• zero correlation 零相关 — no clear straight-line trend; a curved pattern may still exist.

If there is correlation, draw a line of best fit 最佳拟合线: one straight ruled line through the middle of the points, with about the same number of points on each side. Use it to predict values.

Predict only inside the range of the data (interpolation). Reading the line beyond the plotted points (extrapolation) is unreliable, and correlation does not prove that one thing causes the other.

The cumulative frequency 累积频数 is a running total of the frequencies. Plot it against the upper end of each class and join the points with a smooth curve.

For the travel times above, the running totals are:

Plot $(10, 5)$, $(20, 17)$, $(30, 25)$ and $(50, 30)$, starting from $(0, 0)$. The median is read at $\tfrac{30}{2} = 15$ on the vertical axis, which falls in the class $10 < t \leqslant 20$.

Common mistake. Plotting at the midpoint of each class. Cumulative frequency is plotted at the upper end of the class.

From the curve you can read the median (at half the total), the quartiles (at one quarter and three quarters), and any percentile 百分位数 (for example, the $90$th percentile is at $90\%$ of the total).

A histogram 直方图 looks like a bar chart, but the bars can have different widths and the area of each bar (not its height) shows the frequency. The vertical axis is the frequency density 频数密度:

Worked example. A class has class width 组距 $10$ and frequency $25$. Find the frequency density.

Worked example: frequency from a bar. A histogram bar covers $20 < h \leqslant 35$ and has frequency density $1.6$. How many values are in this class?

Worked example: heights of bars. Classes $0 < x \leqslant 10$, $10 < x \leqslant 15$ and $15 < x \leqslant 30$ have frequencies $15$, $20$ and $18$. The bar heights are $15 \div 10 = 1.5$, $20 \div 5 = 4$ and $18 \div 15 = 1.2$. The narrow middle class has the tallest bar, because its values are packed most closely together.

Common mistake. Drawing the frequency as the bar height when the classes have different widths.

• Know the three averages: mean (add up ÷ how many), median (the middle ordered value, or the mean of the two middle values), mode (the most common). The range is highest − lowest.
• From a frequency table the mean is $\frac{\sum fx}{\sum f}$ — divide by the total frequency, not the number of rows.
• On a scatter diagram, describe the correlation as positive, negative or none, and draw the line of best fit through the mean point.
• For a histogram with unequal class widths, the height is the frequency density (frequency ÷ class width), not the frequency.