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Chapter Review 3

综合练习题 - 第3章 Representations of Data

Chapter 3 综合练习题

以下是6道综合练习题,涵盖茎叶图、直方图、箱线图、异常值、偏度、数据比较等所有核心知识点。

1

题1:茎叶图与四分位数

The stem and leaf diagram below shows the heights (in cm) of 25 students.

Stem Leaf Key: \( 15 \mid 2 = 152 \, \text{cm} \)
15 2 3 5 6 8 (6)
16 0 1 1 3 4 7 8 (7)
17 0 2 2 5 6 9 (6)
18 1 3 4 8 9 (5)
19 0 2 5 (3)

a) Draw the stem and leaf diagram.

b) Find \( Q_1, Q_2, Q_3 \).

c) Calculate the interquartile range (IQR).

解答过程

b) 四分位数计算:总数据量25

  • \( Q_1 \) 位置:\( 25/4 = 6.25 \),第7项为\( 161 \, \text{cm} \)
  • \( Q_2 \) 位置:\( 25/2 = 12.5 \),第13项为\( 172 \, \text{cm} \)
  • \( Q_3 \) 位置:\( 3×25/4 = 18.75 \),第19项为\( 184 \, \text{cm} \)

c) 四分位距计算

\( \text{IQR} = Q_3 - Q_1 = 184 - 161 = 23 \, \text{cm} \)

答案:b) Q₁=161cm,Q₂=172cm,Q₃=184cm;c) IQR=23cm
2

题2:直方图与频率密度

A histogram shows the time (in minutes) spent on homework by 60 students. The table below summarises the data:

Time (min) \( 10 \leq t < 20 \) \( 20 \leq t < 30 \) \( 30 \leq t < 40 \) \( 40 \leq t < 60 \)
Frequency 10 20 15 15

a) Calculate the frequency density for each class.

b) Draw the histogram and frequency polygon.

解答过程

a) 频率密度计算:频率密度 = 频率 ÷ 组宽

组别 频率 组宽 频率密度
10-20 10 10 1.0
20-30 20 10 2.0
30-40 15 10 1.5
40-60 15 20 0.75

b) 直方图绘制要点

  • 横轴:时间(分钟)
  • 纵轴:频率密度
  • 条形面积与频率成正比
  • 频率多边形:连接各条形顶端中点
答案:a) 频率密度分别为1.0, 2.0, 1.5, 0.75
3

题3:异常值判断

The following data show the number of books read by 15 students in a month:

8, 12, 15, 18, 20, 22, 25, 28, 30, 32, 35, 40, 45, 50, 80

a) Find \( Q_1, Q_2, Q_3 \) and IQR.

b) Determine if there are any outliers using \( k = 1.5 \).

解答过程

a) 四分位数计算:数据已排序

  • \( Q_1 \) 位置:\( 15/4 = 3.75 \),第4项为\( 18 \)
  • \( Q_2 \) 位置:\( 15/2 = 7.5 \),第8项为\( 28 \)
  • \( Q_3 \) 位置:\( 3×15/4 = 11.25 \),第12项为\( 40 \)
  • \( \text{IQR} = 40 - 18 = 22 \)

b) 异常值判断:使用公式 \( Q_3 + k \times \text{IQR} \) 和 \( Q_1 - k \times \text{IQR} \)

  • 异常值下限:\( 18 - 1.5×22 = -15 \)(无异常值)
  • 异常值上限:\( 40 + 1.5×22 = 73 \)
  • 数据中\( 80 > 73 \),故80是异常值
答案:a) Q₁=18,Q₂=28,Q₃=40,IQR=22;b) 80是异常值
4

题4:箱线图与数据比较

Two box plots below show the test scores of students in Class A and Class B.

Class A Class B
\( Q_1 \) 55 60
\( Q_2 \) 65 70
\( Q_3 \) 75 80
Min 40 50
Max 85 90

a) Draw box plots for both classes.

b) Compare the test scores of Class A and Class B in terms of location and spread.

解答过程

b) 数据对比分析

位置度量(Location)

  • Class B的中位数(70)高于Class A(65)
  • Class B的整体成绩中心更高

离散度量(Spread)

  • Class A的IQR = 75 - 55 = 20
  • Class B的IQR = 80 - 60 = 20
  • 两班的四分位距相同,成绩分散程度相近

整体分布

  • Class B的成绩范围(50-90)高于Class A(40-85)
  • Class B的最低分和最高分都更高
答案:b) Class B成绩中心更高,但两班分散程度相近
5

题5:偏度分析

For a data set, the mean is 50, the median is 45, and the standard deviation is 10.

a) Calculate the skewness using the formula \( \frac{3(\text{mean} - \text{median})}{\text{standard deviation}} \).

b) Describe the skewness and justify using the relationship between mean, median, and the skewness formula.

解答过程

a) 偏度计算

\[ \text{偏度} = \frac{3(\text{mean} - \text{median})}{\text{standard deviation}} = \frac{3(50 - 45)}{10} = \frac{3 \times 5}{10} = 1.5 \]

b) 偏度描述与解释

  • 数据呈正偏态(右偏)
  • 因为均值(50)> 中位数(45),且偏度公式结果为正
  • 说明数据多集中在较低值,右侧有长尾
  • 偏度值1.5表示偏度较强
答案:a) 偏度=1.5;b) 数据呈正偏态,均值>中位数
6

题6:背靠背茎叶图与数据对比

The back to back stem and leaf diagram below shows the number of goals scored by two football teams in a season.

Team A Stem Team B
8 6 5 0 7 9
9 7 5 3 1 8 6 4 2
8 6 4 2 2 9 7 5 3
5 3 1 3 8 6 4
2 4 7 5

Key: \( 1 \mid 5 = 15 \) goals

a) Find the median, \( Q_1, Q_3 \) for each team.

b) Compare the goal-scoring performance of Team A and Team B.

解答过程

a) 各队统计量计算

Team A数据:5,6,8,13,15,17,19,22,24,26,28,31,33,35,42(共15个)

  • 中位数:第8项 = 22
  • \( Q_1 \):第4项 = 13
  • \( Q_3 \):第12项 = 28

Team B数据:7,9,12,14,16,18,23,25,27,29,34,36,38,45,47(共15个)

  • 中位数:第8项 = 23
  • \( Q_1 \):第4项 = 14
  • \( Q_3 \):第12项 = 29

b) 进球表现对比

  • 位置度量:Team B的中位数(23)略高于Team A(22)
  • 离散度量:Team A的IQR = 28-13 = 15,Team B的IQR = 29-14 = 15
  • 整体表现:两队的进球分布相似,Team B略优
答案:a) Team A: Q₁=13, Q₂=22, Q₃=28;Team B: Q₁=14, Q₂=23, Q₃=29;b) Team B表现略优