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3.4 Stem and Leaf Diagrams

茎叶图练习题 - 掌握茎叶图的绘制和解读

Exercise 3D

以下是4道综合练习题，涵盖茎叶图的绘制、统计量计算和比较分析。

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Question 1

Thirty college students were asked how many movies they had in their collection. The results are as follows:

12, 25, 34, 17, 12, 18, 29, 34, 45, 6, 15, 9, 25, 3, 29, 22, 20, 32, 15, 15, 19, 12, 26, 27, 27, 32, 35, 42, 26, 25

Draw a stem and leaf diagram to represent these data.

a) Find the median.

b) Find the lower quartile.

c) Find the upper quartile.

解答过程：

升序排序数据：3, 6, 9, 12, 12, 12, 15, 15, 15, 17, 18, 19, 20, 22, 25, 25, 25, 26, 26, 27, 27, 29, 29, 32, 32, 34, 34, 35, 42, 45

Stem Leaf Key: \( 1 \mid 2 = 12 \)
0 3 6 9
1 2 2 2 5 5 5 7 8 9
2 0 2 5 5 5 6 6 7 7 9 9
3 2 2 4 4 5
4 2 5

统计量计算：

  • a) 中位数：位置\( 30/2 = 15 \)，取第15、16项平均值，即\( (25 + 25)/2 = 25 \)
  • b) 下四分位数（\( Q_1 \)）：位置\( 30/4 = 7.5 \)，取第8项，即15
  • c) 上四分位数（\( Q_3 \)）：位置\( 3×30/4 = 22.5 \)，取第23项，即29
答案：a) 中位数=25；b) Q₁=15；c) Q₃=29
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Question 2

The following stem and leaf diagram shows some information about the marks gained by a group of students in a statistics test.

Stem Leaf Key: \( 2 \mid 3 = 23 \) marks
0 8 9
1 2 5 5 9
2 3 6 6 6 7
3 4 4 5 7 7 7 7 7 9
4 5 8 8 9

a) Write down the highest mark.

b) Write down the lowest mark.

c) How many students scored 26 marks?

d) What is the modal mark?

e) Find the median.

f) Find the lower quartile.

g) Find the upper quartile.

解答过程：

  • a) 最高分为\( 4 \mid 9 = 49 \)
  • b) 最低分为\( 0 \mid 8 = 8 \)
  • c) 26分出现次数：茎2的叶中6出现3次，故有3名学生
  • d) 众数：37（出现次数最多）
  • e) 总人数：\( 2 + 4 + 5 + 9 + 4 = 24 \)，中位数位置\( 24/2 = 12 \)，取第12、13项平均值，即\( (27 + 34)/2 = 30.5 \)
  • f) 下四分位数（\( Q_1 \)）：位置\( 24/4 = 6 \)，取第6项，即15
  • g) 上四分位数（\( Q_3 \)）：位置\( 3×24/4 = 18 \)，取第18项，即37
答案：a) 49；b) 8；c) 3名；d) 37；e) 30.5；f) 15；g) 37
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Question 3

The stem and leaf diagram below shows the median age, in years, of a selection of African elephants in Tanzania.

Stem Leaf Key: \( 1 \mid 8 = 18 \) years
1 5 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9
2 0 0 0 0 0 0 0 1 1 1 1 1 3 3 3 4 5 7
3 4 4
4 1

Find:

a) the median

b) the interquartile range and any outliers.

解答过程：

总数据量：茎1有30个，茎2有18个，茎3有2个，茎4有1个，共\( 30 + 18 + 2 + 1 = 51 \)个。

a) 中位数：

  • 中位数位置\( 51/2 = 25.5 \)，取第25、26项（均为18），故中位数为18

b) 四分位距与异常值：

  • 下四分位数（\( Q_1 \)）：位置\( 51/4 = 12.75 \)，取第13项，即17
  • 上四分位数（\( Q_3 \)）：位置\( 3×51/4 = 38.25 \)，取第39项，即21
  • 四分位距（IQR）：\( 21 - 17 = 4 \)
  • 异常值下限：\( 17 - 1.5×4 = 11 \)
  • 异常值上限：\( 21 + 1.5×4 = 27 \)
  • 数据中34、34、41均大于27，故34、34、41是异常值
答案：a) 中位数=18；b) IQR=4，异常值：34、34、41
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Question 4

A teacher recorded the heights (in cm) of 20 students in her class. The data are shown in the stem and leaf diagram below.

Stem Leaf Key: \( 15 \mid 0 = 150 \) cm
14 5 8 9
15 0 2 3 4 5 6 7 8
16 1 2 3 4 5 6
17 0 2 3

a) Find the median height.

b) Find the range.

c) Calculate the interquartile range.

d) Identify any outliers using the \( 1.5 \times \text{IQR} \) rule.

解答过程：

总数据量：\( 3 + 8 + 6 + 3 = 20 \)个。

a) 中位数：

  • 中位数位置\( 20/2 = 10 \)，取第10、11项平均值，即\( (155 + 156)/2 = 155.5 \) cm

b) 极差：

  • 最高值：173 cm，最低值：145 cm
  • 极差：\( 173 - 145 = 28 \) cm

c) 四分位距：

  • 下四分位数（\( Q_1 \)）：位置\( 20/4 = 5 \)，取第5、6项平均值，即\( (152 + 153)/2 = 152.5 \) cm
  • 上四分位数（\( Q_3 \)）：位置\( 3×20/4 = 15 \)，取第15、16项平均值，即\( (163 + 164)/2 = 163.5 \) cm
  • 四分位距（IQR）：\( 163.5 - 152.5 = 11 \) cm

d) 异常值识别：

  • 异常值下限：\( 152.5 - 1.5×11 = 136 \) cm
  • 异常值上限：\( 163.5 + 1.5×11 = 180 \) cm
  • 所有数据都在此范围内，故无异常值
答案：a) 155.5 cm；b) 28 cm；c) 11 cm；d) 无异常值