Barbara Illowsky; Susan Dean

One simple graph, the stem-and-leaf graph or stemplot, comes from the field of exploratory data analysis. It is a good choice when the data sets are small. To create the plot, divide each observation of data into a stem and a leaf. The stem consists of the leading digit(s), while the leaf consists of a final significant digit. For example, 23 has stem two and leaf three. The number 432 has stem 43 and leaf two. Likewise, the number 5,432 has stem 543 and leaf two. The decimal 9.3 has stem nine and leaf three. Write the stems in a vertical line from smallest to largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their corresponding stem. Make sure the leaves show a space between values, so that the exact data values may be easily determined. The frequency of data values for each stem provides information about the shape of the distribution.

Example 2.1

For Susan Dean's spring precalculus class, scores for the first exam were as follows (smallest to largest):
33, 42, 49, 49, 53, 55, 55, 61, 63, 67, 68, 68, 69, 69, 72, 73, 74, 78, 80, 83, 88, 88, 88, 90, 92, 94, 94, 94, 94, 96, 100

Stem	Leaf
3	3
4	2 9 9
5	3 5 5
6	1 3 7 8 8 9 9
7	2 3 4 8
8	0 3 8 8 8
9	0 2 4 4 4 4 6
10	0

Table 2.1 Stem-and-Leaf Graph

The stemplot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores or approximately 26 percent $(\frac{8}{31})$ were in the 90s or 100, a fairly high number of As.

Try It 2.1

For the Park City basketball team, scores for the last 30 games were as follows (smallest to largest):
32, 32, 33, 34, 38, 40, 42, 42, 43, 44, 46, 47, 47, 48, 48, 48, 49, 50, 50, 51, 52, 52, 52, 53, 54, 56, 57, 57, 60, 61
Construct a stemplot for the data.

The stemplot is a quick way to graph data and gives an exact picture of the data. You want to look for an overall pattern and any outliers. An outlier is an observation of data that does not fit the rest of the data. It is sometimes called an extreme value. When you graph an outlier, it will appear not to fit the pattern of the graph. Some outliers are due to mistakes, for example, writing 50 instead of 500, while others may indicate that something unusual is happening. It takes some background information to explain outliers, so we will cover them in more detail later.

Example 2.2

The data are the distances (in kilometers) from a home to local supermarkets. Create a stemplot using the data.
1.1, 1.5, 2.3, 2.5, 2.7, 3.2, 3.3, 3.3, 3.5, 3.8, 4.0, 4.2, 4.5, 4.5, 4.7, 4.8, 5.5, 5.6, 6.5, 6.7, 12.3

Problem

Do the data seem to have any concentration of values?

The leaves are to the right of the decimal.

Solution

The value 12.3 may be an outlier. Values appear to concentrate at 3 and 4 kilometers.

Stem	Leaf
1	1 5
2	3 5 7
3	2 3 3 5 8
4	0 2 5 5 7 8
5	5 6
6	5 7
7
8
9
10
11
12	3

Table 2.2

Try It 2.2

The data below show the distances (in miles) from the homes of high school students to the school. Create a stemplot using the following data and identify any outliers.

0.5, 0.7, 1.1, 1.2, 1.2, 1.3, 1.3, 1.5, 1.5, 1.7, 1.7, 1.8, 1.9, 2.0, 2.2, 2.5, 2.6, 2.8, 2.8, 2.8, 3.5, 3.8, 4.4, 4.8, 4.9, 5.2, 5.5, 5.7, 5.8, 8.0

Example 2.3

Problem

A side-by-side stem-and-leaf plot allows a comparison of the two data sets in two columns. In a side-by-side stem-and-leaf plot, two sets of leaves share the same stem. The leaves are to the left and the right of the stems. Table 2.3 and Table 2.4 show the ages of presidents at their inauguration and at their death. Construct a side-by-side stem-and-leaf plot using these data.

President	Age	President	Age	President	Age
Washington	57	Lincoln	52	Hoover	54
J. Adams	61	A. Johnson	56	F. Roosevelt	51
Jefferson	57	Grant	46	Truman	60
Madison	57	Hayes	54	Eisenhower	62
Monroe	58	Garfield	49	Kennedy	43
J. Q. Adams	57	Arthur	51	L. Johnson	55
Jackson	61	Cleveland	47	Nixon	56
Van Buren	54	B. Harrison	55	Ford	61
W. H. Harrison	68	Cleveland	55	Carter	52
Tyler	51	McKinley	54	Reagan	69
Polk	49	T. Roosevelt	42	G.H.W. Bush	64
Taylor	64	Taft	51	Clinton	47
Fillmore	50	Wilson	56	G. W. Bush	54
Pierce	48	Harding	55	Obama	47
Buchanan	65	Coolidge	51

Table 2.3 Presidential Ages at Inauguration

President	Age	President	Age	President	Age
Washington	67	Lincoln	56	Hoover	90
J. Adams	90	A. Johnson	66	F. Roosevelt	63
Jefferson	83	Grant	63	Truman	88
Madison	85	Hayes	70	Eisenhower	78
Monroe	73	Garfield	49	Kennedy	46
J. Q. Adams	80	Arthur	56	L. Johnson	64
Jackson	78	Cleveland	71	Nixon	81
Van Buren	79	B. Harrison	67	Ford	93
W. H. Harrison	68	Cleveland	71	Reagan	93
Tyler	71	McKinley	58
Polk	53	T. Roosevelt	60
Taylor	65	Taft	72
Fillmore	74	Wilson	67
Pierce	64	Harding	57
Buchanan	77	Coolidge	60

Table 2.4 Presidential Age at Death

Solution

Ages at Inauguration		Ages at Death
9 9 8 7 7 7 6 3 2	4	6 9
8 7 7 7 7 6 6 6 5 5 5 5 4 4 4 4 4 2 1 1 1 1 1 0	5	3 6 6 7 7 8
9 5 4 4 2 1 1 1 0	6	0 0 3 3 4 4 5 6 7 7 7 8
	7	0 0 1 1 1 4 7 8 8 9
	8	0 1 3 5 8
	9	0 0 3 3

Table 2.5

Notice that the leaf values increase in order, from right to left, for leaves shown to the left of the stem, while the leaf values increase in order from left to right, for leaves shown to the right of the stem.

Try It 2.3

The table shows the number of wins and losses a sports team has had in 42 seasons. Create a side-by-side stem-and-leaf plot of these wins and losses.

Losses	Wins	Year	Losses	Wins	Year
34	48	1968–1969	41	41	1989–1990
34	48	1969–1970	39	43	1990–1991
46	36	1970–1971	44	38	1991–1992
46	36	1971–1972	39	43	1992–1993
36	46	1972–1973	25	57	1993–1994
47	35	1973–1974	40	42	1994–1995
51	31	1974–1975	36	46	1995–1996
53	29	1975–1976	26	56	1996–1997
51	31	1976–1977	32	50	1997–1998
41	41	1977–1978	19	31	1998–1999
36	46	1978–1979	54	28	1999–2000
32	50	1979–1980	57	25	2000–2001
51	31	1980–1981	49	33	2001–2002
40	42	1981–1982	47	35	2002–2003
39	43	1982–1983	54	28	2003–2004
42	40	1983–1984	69	13	2004–2005
48	34	1984–1985	56	26	2005–2006
32	50	1985–1986	52	30	2006–2007
25	57	1986–1987	45	37	2007–2008
32	50	1987–1988	35	47	2008–2009
30	52	1988–1989	29	53	2009–2010

Table 2.6

Another type of graph that is useful for specific data values is a line graph. In the particular line graph shown in Example 2.4, the x-axis (horizontal axis) consists of data values and the y-axis (vertical axis) consists of frequency points. The frequency points are connected using line segments.

Example 2.4

In a survey, 40 mothers were asked how many times per week a teenager must be reminded to do his or her chores. The results are shown in Table 2.7 and in Figure 2.2.

Number of Times Teenager Is Reminded	Frequency
0	2
1	5
2	8
3	14
4	7
5	4

Table 2.7

A line graph showing the number of times a teenager needs to be reminded to do chores on the x-axis and frequency on the y-axis.

Figure 2.2

Try It 2.4

In a survey, 40 people were asked how many times per year they had their car in the shop for repairs. The results are shown in Table 2.8. Construct a line graph.

Number of Times in Shop	Frequency
0	7
1	10
2	14
3	9

Table 2.8

Bar graphs consist of bars that are separated from each other. The bars can be rectangles, or they can be rectangular boxes, used in three-dimensional plots, and they can be vertical or horizontal. The bar graph shown in Example 2.5 has age-groups represented on the x-axis and proportions on the y-axis.

Example 2.5

Problem

By the end of 2011, a social media site had more than 146 million users in the United States. Table 2.9 shows three age-groups, the number of users in each age-group, and the proportion (percentage) of users in each age-group. Construct a bar graph using this data.

Age-Groups	Number of Site Users	Proportion (%) of Site Users
13–25	65,082,280	45%
26–44	53,300,200	36%
45–64	27,885,100	19%

Table 2.9

Solution

This is a bar graph that matches the supplied data. The x-axis shows age groups, and the y-axis shows the percentages of Facebook users.

Figure 2.3

Try It 2.5

The population in Park City is made up of children, working-age adults, and retirees. Table 2.10 shows the three age-groups, the number of people in the town from each age-group, and the proportion (%) of people in each age-group. Construct a bar graph showing the proportions.

Age-Groups	Number of People	Proportion of Population
Children	67,059	19%
Working-age adults	152,198	43%
Retirees	131,662	38%

Table 2.10

Example 2.6

Problem

The columns in Table 2.11 contain the race or ethnicity of students in U.S. public schools for the class of 2011, percentages for the Advanced Placement (AP) examinee population for that class, and percentages for the overall student population. Create a bar graph with the student race or ethnicity (qualitative data) on the x-axis and the AP examinee population percentages on the y-axis.

Race/Ethnicity	AP Examinee Population	Overall Student Population
1 = Asian, Asian American, or Pacific Islander	10.3%	5.7%
2 = Black or African American	9.0%	14.7%
3 = Hispanic or Latino	17.0%	17.6%
4 = American Indian or Alaska Native	0.6%	1.1%
5 = White	57.1%	59.2%
6 = Not reported/other	6.0%	1.7%

Table 2.11

Solution

This is a bar graph that matches the supplied data. The x-axis shows race and ethnicity, and the y-axis shows the percentages of AP examinees.

Figure 2.4

Try It 2.6

Park City is broken down into six voting districts. The table shows the percentage of the total registered voter population that lives in each district as well as the percentage of the entire population that lives in each district. Construct a bar graph that shows the registered voter population by district.

District	Registered Voter Population	Overall City Population
1	15.5%	19.4%
2	12.2%	15.6%
3	9.8%	9.0%
4	17.4%	18.5%
5	22.8%	20.7%
6	22.3%	16.8%

Table 2.12

Example 2.7

Problem

Table 2.13 is a two-way table showing the types of pets owned by men and women.

	Dogs	Cats	Fish	Total
Men	4	2	2	8
Women	4	6	2	12
Total	8	8	4	20

Table 2.13

Given these data, calculate the marginal distributions of pets for the people surveyed.

Solution

$Dogs = 8/20 = 0.4$

$Cats = 8/20 = 0.4$

$Fish = 4/20 = 0.2$

Note—The sum of all the marginal distributions must equal one. In this case, $0.4 + 0.4 + 0.2 = 1;$ therefore, the solution checks.

Example 2.8

Problem

Table 2.14 is a two-way table showing the types of pets owned by men and women.

	Dogs	Cats	Fish	Total
Men	4	2	2	8
Women	4	6	2	12
Total	8	8	4	20

Table 2.14

Given these data, calculate the conditional distributions for the subpopulation of men who own each pet type.

Solution

$Men who own dogs = 4/8 = 0.5$

$Men who own cats = 2/8 = 0.25$

$Men who own fish = 2/8 = 0.25$

Note—The sum of all the conditional distributions must equal one. In this case, $0.5 + 0.25 + 0.25 = 1;$ therefore, the solution checks.

2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs

Problem

Solution

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