📚 Unit Test Mock Paper Analysis: Year 9 SQA Statistics | Year 9 SQA 统计：单元测试模拟卷解析
Welcome to this detailed walkthrough of a mock unit test designed for Year 9 SQA Statistics. This paper brings together the most important concepts you have studied: averages and spread, data representation, probability, and comparative analysis. By working through each question, you will see exactly what examiners are looking for, learn how to structure clear solutions, and pick up valuable tips for avoiding common errors. Use this resource to identify your strengths and pinpoint areas where you still need a little extra practice.
欢迎阅读这份为 Year 9 SQA 统计课程设计的单元测试模拟卷详细解析。本试卷涵盖了你所学过的最核心的概念：平均数与离散程度、数据表示、概率以及数据集的比较。通过逐题剖析，你将清晰地看到考官在寻找什么，学会如何组织清晰规范的解答，并掌握避免常见错误的实用技巧。请用这篇解析来发现你的优势，并找到还需要加强练习的知识点。
1. Mean, Median, Mode and Range | 平均数、中位数、众数和极差
Question 1: A student recorded the number of books read by seven friends over the summer: 12, 15, 18, 22, 25, 18, 30. Calculate the mean, median, mode, and range for this data set.
问题1：一名学生记录了她七位朋友在暑假读过的书的数量：12, 15, 18, 22, 25, 18, 30。请计算这组数据的平均数、中位数、众数和极差。
Mean: First add all the values together. Sum = 12 + 15 + 18 + 22 + 25 + 18 + 30 = 140. Since there are 7 values, divide by 7.
平均数：首先将所有数值相加。总和 = 12 + 15 + 18 + 22 + 25 + 18 + 30 = 140。因为有7个数据，除以7。
Mean = 140 ÷ 7 = 20
Median: Write the numbers in ascending order: 12, 15, 18, 18, 22, 25, 30. The median is the middle value. Here it is the 4th number, which is 18.
中位数：将数字从小到大排列：12, 15, 18, 18, 22, 25, 30。中位数位于正中间，即第4个数，为18。
Mode: The mode is the value that occurs most often. The number 18 appears twice while all others appear once, so the mode is 18.
众数：众数是出现频率最高的数值。18出现了两次，其他数字只出现一次，所以众数为18。
Range: Subtract the smallest value from the largest. Range = 30 – 12 = 18.
极差：用最大值减去最小值。极差 = 30 – 12 = 18。
2. Frequency Tables and Estimated Mean | 频数表与估算平均数
Question 2: The table below shows the lengths of phone calls (in minutes) made by a group of students. Use it to estimate the mean call length and state the median class interval.
问题2：下面的频数表给出了一组学生打电话的时长（单位：分钟）。请用它估算平均通话时长，并指出中位数所在的组距。
| Length (min) | Frequency |
|---|---|
| 0–9 | 4 |
| 10–19 | 6 |
| 20–29 | 10 |
| 30–39 | 5 |
Estimated mean: Find the midpoint of each interval (midpoint = (lower bound + upper bound) ÷ 2). Multiply each midpoint by its frequency to get fx, then sum these values.
估算平均数：求每个区间的组中值（组中值 = (下限+上限)÷2）。将每个组中值乘以其频数得到 fx，然后求和。
- 0–9: midpoint = 4.5, fx = 4.5 × 4 = 18
- 10–19: midpoint = 14.5, fx = 14.5 × 6 = 87
- 20–29: midpoint = 24.5, fx = 24.5 × 10 = 245
- 30–39: midpoint = 34.5, fx = 34.5 × 5 = 172.5
Sum of frequencies, Σf = 4 + 6 + 10 + 5 = 25. Sum of fx, Σfx = 18 + 87 + 245 + 172.5 = 522.5.
频数总和 Σf = 4+6+10+5 = 25。fx 的总和 Σfx = 18+87+245+172.5 = 522.5。
Estimated mean = 522.5 ÷ 25 ≈ 20.9 minutes
Median interval: Find cumulative frequencies: 4, 10 (4+6), 20 (10+10), 25 (20+5). The median position is (25+1) ÷ 2 = 13th value. This lies in the interval where the cumulative frequency first reaches or exceeds 13, which is 20–29.
中位数所在区间：求累积频数：4，10（4+6），20（10+10），25（20+5）。中位数的位置为 (25+1)÷2 = 第13个数据。它在累积频数第一次达到或超过13的区间内，即 20–29 这一组。
3. Bar Charts and Pie Charts | 柱状图与饼图
Question 3: A survey asked 50 students to choose their favourite colour. The frequencies were: Red 10, Blue 15, Green 5, Yellow 20. Interpret the bar chart drawn from this data and calculate the angle needed for the Yellow sector in a pie chart.
问题3：一项调查询问50名学生最喜欢的颜色。频数为：红色10，蓝色15，绿色5，黄色20。请解读由此数据绘制的柱状图，并计算在饼图中黄色扇形的角度。
Bar charts show categorical data with gaps between bars. The height of each bar represents frequency. By reading the bar chart, you can instantly see that Yellow is the most popular colour (frequency 20), while Green is the least popular (5). All bars must be equally wide and clearly labelled.
柱状图用有间隔的直条展示分类数据，每个直条的高度代表频数。通过观察柱状图，你可以立即看出黄色最受欢迎（频数为20），绿色最不受欢迎（频数为5）。所有直条宽度必须相等，并且要有清晰的标签。
Pie chart calculation: The total number of students is 50. The angle for a sector = (frequency ÷ total) × 360°. For Yellow: (20 ÷ 50) × 360° = 0.4 × 360° = 144°. The table below shows the angles for all colours.
饼图计算：学生总数为50。扇形角度 = (频数÷总数) × 360°。黄色扇形的角度：(20÷50)×360° = 0.4×360° = 144°。下表给出了所有颜色扇形的角度。
| Colour | Frequency | Calculation | Angle (°) |
|---|---|---|---|
| Red | 10 | (10 ÷ 50) × 360 | 72° |
| Blue | 15 | (15 ÷ 50) × 360 | 108° |
| Green | 5 | (5 ÷ 50) × 360 | 36° |
| Yellow | 20 | (20 ÷ 50) × 360 | 144° |
4. Stem-and-Leaf Diagrams | 茎叶图
Question 4: An ordered stem-and-leaf diagram shows the test scores of 13 pupils. Key: 4|3 means 43.
问题4：一个有序茎叶图显示了13名学生的测验分数。图例： 4|3 表示 43。
3 | 2 5 4 | 0 3 3 7 5 | 1 4 8 9 6 | 2 5 5
List the raw scores, then find the median, lower quartile, upper quartile, and interquartile range (IQR).
列出原始分数，然后求中位数、下四分位数、上四分位数和四分位距 (IQR)。
Raw scores in order: 32, 35, 40, 43, 43, 47, 51, 54, 58, 59, 62, 65, 65. There are 13 data points.
原始分数从小到大排列：32, 35, 40, 43, 43, 47, 51, 54, 58, 59, 62, 65, 65。共有13个数据点。
Median: position (13+1) ÷ 2 = 7th value → 51.
中位数：位置 (13+1)÷2 = 第7个数值 → 51。
Lower quartile (LQ): median of first half below 51. First 6 numbers: 32,35,40,43,43,47. LQ = (40+43) ÷ 2 = 41.5. (Some methods may take the 3.5th position, giving 41.5; here we use splitting method: often taught as LQ = 43, let’s adopt standard SQA: n=13, LQ at (n+1)/4 = 3.5th, so average of 3rd and 4th: 40 and 43, average 41.5. Accept 43 if taught differently; for clarity use median of lower half: lower half is first 6, median of 6 is average of 3rd and 4th → (40+43)/2 = 41.5. We’ll present 41.5.)
下四分位数 (LQ)：51 以下的前一半数据的中位数。前6个数为：32,35,40,43,43,47。LQ = (40+43)÷2 = 41.5。（如果采用位置法，第 3.5 个数据也是 41.5。）
Upper quartile (UQ): upper half after 51: 54,58,59,62,65,65. UQ = (59+62) ÷ 2 = 60.5.
上四分位数 (UQ)：51 以上的后一半数据：54,58,59,62,65,65。UQ = (59+62)÷2 = 60.5。
Interquartile range (IQR) = UQ – LQ = 60.5 – 41.5 = 19.
四分位距 (IQR) = UQ – LQ = 60.5 – 41.5 = 19。
5. Box Plots | 箱线图
Question 5: Using the data from the stem-and-leaf diagram, construct a box plot and comment on the distribution. The five-number summary is: Minimum = 32, LQ = 41.5, Median = 51, UQ = 60.5, Maximum = 65.
问题5：利用茎叶图的数据，绘制一个箱线图并评论其分布特征。五数概括法结果为：最小值 = 32，LQ = 41.5，中位数 = 51，UQ = 60.5，最大值 = 65。
A box plot is drawn on a scale. The box spans from LQ to UQ, with a line inside at the median. Whiskers extend to the minimum and maximum, provided there are no outliers. Here, IQR = 19. Boundaries for outliers: lower fence = LQ – 1.5 × IQR = 41.5 – 28.5 = 13; upper fence = UQ + 1.5 × IQR = 60.5 + 28.5 = 89. All data lie within these fences, so no outliers. The whiskers go to 32 and 65.
在数轴上绘制箱线图。矩形盒从 LQ 延伸到 UQ，中间用竖线标出中位数。若无异常值，须线延伸至最小值和最大值。此处 IQR = 19。异常值的界限：下限 = LQ – 1.5 × IQR = 41.5 – 28.5 = 13；上限 = UQ + 1.5 × IQR = 60.5 + 28.5 = 89。所有数据均在界限内，因此没有异常值，须线直接画到 32 和 65。
Distribution shape: The median is closer to the lower quartile than to the upper quartile, and the right whisker is longer than the left. This suggests a slight positive skew (the data is stretched towards higher values). The box plot also shows a fairly concentrated middle 50% of scores between 41.5 and 60.5.
分布形状：中位数更靠近下四分位数而非上四分位数，且右侧须线比左侧长。这表明分布略呈正偏态（数据向较高值方向拉伸）。箱线图也显示中间 50% 的分数相对集中，分布在 41.5 至 60.5 之间。
6. Scatter Graphs and Correlation | 散点图与相关性
Question 6: The table shows hours of revision and test scores for six students.
问题6：下表显示了六名学生的复习时数和考试成绩。
| Revision (h) | 2 | 3 | 5 | 7 | 8 | 10 |
|---|---|---|---|---|---|---|
| Score (%) | 50 | 55 | 70 | 78 | 82 | 92 |
Describe the correlation and add a line of best fit. Use your line to estimate the score for a student who revises for 6 hours.
请描述相关性，并添加一条最佳拟合线。利用该线预估一位复习了6小时的学生的分数。
Plotting the points: (2,50), (3,55), (5,70), (7,78), (8,82), (10,92). The points show a clear upward trend – as revision hours increase, scores increase. This indicates a strong positive correlation.
描点：(2,50), (3,55), (5,70), (7,78), (8,82), (10,92)。这些点呈现明显的上升趋势——随着复习时间增加，分数也随之升高。这表明存在强正相关。
A line of best fit should pass through the middle of the points. Choose two well‑separated points on the line, for example (3,55) and (9,85). Calculate the gradient:
最佳拟合线应穿过点群的中部。在线上的两点，选择距离较远的点，例如 (3,55) 和 (9,85)。计算斜率：
Gradient = (85 – 55) ÷ (9 – 3) = 30 ÷ 6 = 5
Equation of the line: y – 55 = 5(x – 3), which simplifies to y = 5x + 40.
直线方程：y – 55 = 5(x – 3)，简化得 y = 5x + 40。
Estimate for 6 hours: substitute x = 6 into the equation → y = 5×6 + 40 = 30 + 40 = 70. The estimated score is 70%. This is an interpolation because 6 hours lies within the range of the data, so the estimate is reliable.
复习6小时的预估：将 x = 6 代入方程 → y = 5×6 + 40 = 30 + 40 = 70。预估分数为 70%。这是一种内插估计，因为6小时位于数据范围之内，因此该估计值较为可靠。
7. Probability Basics | 概率基础
Question 7: A bag contains 5 red, 3 blue, and 2 green counters. A counter is taken at random. Find: (a) P(red), (b) P(not green), (c) If a counter is taken out, the colour noted, and then replaced, do 20 trials – how many red counters would you expect?
问题7：一个袋子里装有5个红色、3个蓝色和2个绿色的筹码。随机取出一个筹码。求：(a) P(红色)，(b) P(非绿色)，(c) 如果重复实验：取出一个筹码记录颜色后放回，进行20次，预期红色筹码会出现多少次？
Total number of counters = 5 + 3 + 2 = 10.
筹码总数 = 5 + 3 + 2 = 10。
(a) P(red) = number of red ÷ total = 5 ÷ 10 = 1/2 or 0.5.
(a) P(红色) = 红色个数÷总数 = 5÷10 = 1/2 或 0.5。
(b) P(not green) = 1 – P(green) = 1 – 2/10 = 8/10 = 4/5 = 0.8. Alternatively, count counters that are not green: 5 red + 3 blue = 8, so 8/10 = 4/5.
(b) P(非绿色) = 1 – P(绿色) = 1 – 2/10 = 8/10 = 4/5 = 0.8。也可以直接计数非绿色筹码：5红+3蓝=8，所以概率为 8/10 = 4/5。
(c) Expected frequency = number of trials × probability = 20 × 1/2 = 10. You would expect about 10 red counters.
(c) 期望频数 = 试验次数 × 概率 = 20 × 1/2 = 10。预期红色筹码大约出现10次。
8. Comparing Data Sets |
Published by TutorHao | Year 9 统计 Revision Series | aleveler.com
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