📚 Year 9 SQA Statistics: Unit Test Mock Exam Analysis | Year 9 SQA 统计：单元测试模拟卷解析
This article provides a detailed walkthrough of a typical Year 9 SQA Statistics unit test mock paper. We will cover key topics such as measures of central tendency, graphical representation, probability, and data comparison, offering step-by-step solutions and exam tips. Working through these worked examples will strengthen your understanding and prepare you for the unit assessment.
本文详细解析了一份典型的 Year 9 SQA 统计单元测试模拟卷。我们将涵盖中心趋势测量、图形表示、概率和数据比较等关键主题，提供逐步解答和考试技巧。通过这些示例的练习，你可以加深理解，为单元评估做好准备。
1. Frequency Table – Mean and Median | 频数表——平均数与中位数
The question: A survey recorded the number of pets owned by 30 households. The results are shown in the frequency table below. Calculate the mean number of pets and find the median.
题目：一项调查记录了30户家庭拥有的宠物数量。结果如下面的频数表所示。计算宠物数量的平均数并求中位数。
| Pets (x) | Frequency (f) |
|---|---|
| 0 | 4 |
| 1 | 10 |
| 2 | 8 |
| 3 | 5 |
| 4 | 3 |
To find the mean, multiply each number of pets by its frequency: (0 x 4) + (1 x 10) + (2 x 8) + (3 x 5) + (4 x 3) = 0 + 10 + 16 + 15 + 12 = 53. Then divide by the total frequency 30. Mean = 53 / 30 ≈ 1.77 pets.
求平均数时，先将每个宠物数量与其频数相乘：(0 × 4) + (1 × 10) + (2 × 8) + (3 × 5) + (4 × 3) = 0 + 10 + 16 + 15 + 12 = 53。然后除以总频数30。平均数 = 53 ÷ 30 ≈ 1.77 只宠物。
For the median, list all 30 values in order or use cumulative frequency. The position of the median is (30+1)/2 = 15.5th value, so we take the average of the 15th and 16th values. The first 4 are 0, next 10 are 1 (positions 5 to 14), so the 15th and 16th are both 2. Median = 2 pets.
求中位数时，可以将全部30个数据按顺序列出，或用累积频数。中位数的位置为 (30+1) ÷ 2 = 15.5，即取第15和第16个值的平均数。前4个为0，接下来10个为1（位置5到14），因此第15和第16个均为2。中位数 = 2只宠物。
2. Bar Chart – Reading and Percentage | 条形图——读取与百分比
The question: The bar chart shows the number of ice creams sold in five flavours during a school fair. (Flavours: Vanilla, Strawberry, Chocolate, Mint, Toffee. Sales: 25, 30, 45, 15, 35). Which flavour sold the most? What percentage of the total sales was Strawberry?
题目：条形图显示了学校游园会五种口味冰淇淋的销量（口味：香草、草莓、巧克力、薄荷、太妃糖。销量：25, 30, 45, 15, 35）。哪种口味销量最高？草莓占总销量的百分之几？
Chocolate had the highest bar, with 45 sales. To find the total, add all sales: 25 + 30 + 45 + 15 + 35 = 150. Strawberry sold 30, so the percentage is (30 / 150) x 100 = 20%. Always check the vertical axis scale and read bar heights accurately.
巧克力口味的条形最高，销量为45。总销量为 25 + 30 + 45 + 15 + 35 = 150。草莓销量为30，因此百分比为 (30 ÷ 150) × 100 = 20%。解答时务必检查纵轴刻度并准确读取条形高度。
3. Pie Chart – Angles and Proportion | 饼图——角度与比例
The question: A pie chart represents how 36 pupils travel to school. The sector for ‘Bus’ has an angle of 150°. How many pupils travel by bus? What fraction is this, in simplest form?
题目：一个饼图表示36名学生上学的方式。’公交车’扇区角度为150°。有多少名学生乘公交车？用最简分数表示这个比例是多少？
A full circle is 360°, so the bus sector fraction is 150/360 = 5/12 after dividing by 30. Number of pupils = (150/360) x 36 = (5/12) x 36 = 15. Alternatively, 36 pupils for 360° means 1° represents 36/360 = 0.1 pupils, so 150° gives 15 pupils. Always simplify fractions for full marks.
整个圆为360°，因此公交车扇区占比为 150/360，约简后等于 5/12。学生人数 = (150/360) × 36 = (5/12) × 36 = 15。也可这样计算：36名学生对应360°，即1°代表 36/360 = 0.1 名学生，所以150°为15名。为得满分，务必约简分数。
4. Scatter Graph – Correlation and Estimation | 散点图——相关性与估算
The question: A scatter graph plots hours of revision (x) against test score (y). The points generally rise from bottom-left to top-right. Describe the correlation. Use a line of best fit to estimate the test score for a student who revised 5.5 hours.
题目：散点图以复习小时数（x轴）对测验分数（y轴）作图。点子大致从左下向右上倾斜。描述相关性。利用最佳拟合线，估算复习5.5小时的学生测验分数。
The correlation is positive: as revision hours increase, test scores tend to increase. To estimate, draw a straight line through the middle of the points. At x = 5.5, read the y-value from your line. If the line passes near points (4, 55) and (7, 85), the gradient is (85-55)/(7-4) = 10, equation y – 55 = 10(x – 4). For x = 5.5, y = 55 + 10(5.5 – 4) = 55 + 15 = 70. So an estimate of around 70 marks is reasonable.
相关性为正相关：复习小时数增加，测验分数也随之上升。估算时，需画一条穿过点群中心的直线。当 x = 5.5 时，从线上读取 y 值。若直线经过点 (4, 55) 和 (7, 85)，斜率为 (85-55)/(7-4) = 10，方程为 y – 55 = 10(x – 4)。代入 x = 5.5，得 y = 55 + 10 × (5.5 – 4) = 55 + 15 = 70。因此合理估算约为70分。
5. Simple Probability – Equally Likely Outcomes | 简单概率——等可能结果
The question: A bag contains 6 red, 4 blue and 2 green counters. One counter is taken at random. What is the probability it is red? What is the probability it is not blue?
题目：一个袋子里有6枚红色、4枚蓝色和2枚绿色筹码。随机取出一枚。取出红色的概率是多少？取出不是蓝色的概率是多少？
Total counters = 6 + 4 + 2 = 12. P(red) = number of red / total = 6/12 = 1/2 or 0.5. P(not blue) means red or green, total 6+2 = 8, so 8/12 = 2/3. Alternatively, P(not blue) = 1 – P(blue) = 1 – 4/12 = 1 – 1/3 = 2/3. Always give probabilities as fractions in simplest form unless asked otherwise.
总筹码数 = 6 + 4 + 2 = 12。P(红色) = 红色个数 / 总数 = 6/12 = 1/2 或 0.5。P(不是蓝色) 即红色或绿色，合计 6+2 = 8，故为 8/12 = 2/3。也可用 P(不是蓝色) = 1 – P(蓝色) = 1 – 4/12 = 1 – 1/3 = 2/3。除非另有要求，概率最好以最简分数表示。
6. Line Graph – Trend and Range | 折线图——趋势与范围
The question: A line graph shows the daily maximum temperature in °C over a week: 12, 14, 15, 13, 11, 10, 9. What is the range of temperatures? Describe the trend from Monday to Sunday.
题目：折线图显示了一周内每日最高气温（°C）：12, 14, 15, 13, 11, 10, 9。气温的范围是多少？描述周一至周日的气温变化趋势。
Range = maximum – minimum = 15 – 9 = 6°C. For the trend, the temperature rose from Monday to Wednesday (12 to 15), then steadily decreased from Wednesday to Sunday (15 to 9). Overall, there was a slight warming then a cooling trend. ‘Trend’ means the general direction over time, not every tiny fluctuation.
范围 = 最大值 – 最小值 = 15 – 9 = 6°C。趋势方面，气温从周一到周三上升（12到15），随后从周三到周日持续下降（15到9）。总体而言，先短暂升温后降温。’趋势’指的是随时间变化的大致走向，不是每个微小波动。
7. Comparing Data Sets – Mean and Consistency | 比较数据集——平均数与一致性
The question: Two athletes, A and B, record their 100 m sprint times (seconds) over five trials. A: 12.1, 11.8, 12.0, 11.9, 12.2. B: 11.5, 12.5, 11.0, 12.8, 12.2. Calculate the mean and range for each. Who is more consistent? Justify your answer.
题目：两位运动员A和B记录了他们五次百米短跑的时间（秒）。A: 12.1, 11.8, 12.0, 11.9, 12.2。B: 11.5, 12.5, 11.0, 12.8, 12.2。分别计算两人的平均数和范围。谁更稳定？说明理由。
Athlete A: sum = 12.1+11.8+12.0+11.9+12.2 = 60.0, mean = 60.0 / 5 = 12.0 s. Range = 12.2 – 11.8 = 0.4 s. Athlete B: sum = 11.5+12.5+11.0+12.8+12.2 = 60.0, mean = 12.0 s. Range = 12.8 – 11.0 = 1.8 s. Both have the same mean time, but A’s range is much smaller. This shows A’s times are less spread out, so A is more consistent.
运动员A：总和 = 12.1+11.8+12.0+11.9+12.2 = 60.0，平均数 = 60.0 ÷ 5 = 12.0秒。范围 = 12.2 – 11.8 = 0.4秒。运动员B：总和 = 11.5+12.5+11.0+12.8+12.2 = 60.0，平均数 = 12.0秒。范围 = 12.8 – 11.0 = 1.8秒。两人平均成绩相同，但A的范围小得多，说明A的成绩离散程度低，因此A更稳定。
8. Experimental vs Theoretical Probability | 实验概率与理论概率
The question: Shannon suspects a dice is biased. She rolls it 300 times and obtains a ‘6’ on 68 occasions. Calculate the relative frequency of rolling a six. Compare this with the theoretical probability and comment on whether the dice seems fair.
题目：Shannon怀疑一个骰子有偏差。她掷了300次，得到68次’6’。计算掷出6的相对频率。将此与理论概率比较，并判断该骰子是否看似公平。
Relative frequency = number of sixes / total rolls = 68/300 = 17/75 ≈ 0.2267. The theoretical probability for a fair dice is 1/6 ≈ 0.1667. The experimental result is higher. For 300 rolls, we expect about 50 sixes (300 x 1/6). She got 68, which is noticeably more. While there is always variation, a difference of 18 might indicate bias, especially with a large number of trials. In an exam, you would state that the dice may not be fair because the relative frequency is quite far from the expected value.
相对频率 = 出现6的次数 / 总投掷次数 = 68/300 = 17/75 ≈ 0.2267。公平骰子的理论概率为 1/6 ≈ 0.1667。实验结果更高。掷300次，我们期望约50次6 (300 × 1/6)。实际得到68次，明显更多。虽然总会有波动，但在大数次试验下18的差距可能表明偏差。考试中应表述为：由于相对频率与期望值相差较远，该骰子可能不公平。
9. Sampling Methods – Avoiding Bias | 抽样方法——避免偏差
The question: A student wants to find out the favourite lunch option of Year 9 pupils. She plans to ask only her friends in the canteen. Explain why this sample might be biased and suggest a better method.
题目：一位学生想了解九年级学生最喜欢的午餐选项。她打算只询问她在食堂的朋友。说明此样本为何可能存在偏差，并提出更好的方法。
This is a convenience sample and is not representative because it only includes her friends, who might share similar tastes. It also excludes those who bring packed lunches or eat elsewhere. To obtain an unbiased sample, she could use a simple random sample: give every Year 9 pupil a number and use a random number generator to select, say, 50 pupils. Alternatively, a stratified sample by tutor group would ensure all groups are fairly represented.
这是一个方便样本，不具代表性，因为只包含她的朋友，他们可能有相似的口味。同时排除了自带午餐或在别处用餐的学生。为获得无偏样本，她可采用简单随机抽样：给每位九年级学生编号，用随机数生成器抽取，比如50名学生。或者按辅导组分层抽样，以确保各群体均有公平代表。
10. Stem-and-Leaf Diagram – Median and Mode | 茎叶图——中位数与众数
The question: The stem-and-leaf diagram below shows the marks of 21 pupils in a test. (Stem: tens, Leaf: units. Data: 3|2, 4|1 5 7, 5|0 2 2 6 8 9, 6|1 3 3 4 7, 7|0 2 5, 8|1 4). Find the mode and the median mark.
题目：下面的茎叶图显示了21名学生的测验分数（茎为十位数，叶为个位数。数据：3|2, 4|1 5 7, 5|0 2 2 6 8 9, 6|1 3 3 4 7, 7|0 2 5, 8|1 4）。找出众数和中位数。
First, list all values in order: 32, 41, 45, 47, 50, 52, 52, 56, 58, 59, 61, 63, 63, 64, 67, 70, 72, 75, 81, 84. Mode is the most frequent value: 52 appears twice, 63 appears twice – so bimodal, 52 and 63. Median position = (21+1)/2 = 11th value. Counting to the 11th: 1st 32, 2nd 41, 3rd 45, 4th 47, 5th 50, 6th 52, 7th 52, 8th 56, 9th 58, 10th 59, 11th 61. So the median is 61 marks.
首先按顺序列出所有数值：32, 41, 45, 47, 50, 52, 52, 56, 58, 59, 61, 63, 63, 64, 67, 70, 72, 75, 81, 84。众数为出现最频繁的值：52出现两次，63出现两次，因此是双众数 52 和 63。中位数位置 = (21+1) ÷ 2 = 第11个值。数到第11个：第1个32，第2个41，第3个45，第4个47，第5个50，第6个52，第7个52，第8个56，第9个58，第10个59，第11个61。因此中位数为61分。
Published by TutorHao | Statistics Revision Series | aleveler.com
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