2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
SLO 2
Identify appropriate graphs and summary statistics for variables and relationships between them and correctly interpret information from graphs and summary statistics.
This is where you learn to match a graph to the data you have: stemplots for small numeric sets, line graphs for frequencies of a count, bar graphs for categories. Reading each one back — finding the peak, the spread, the outlier — is the interpreting half of the outcome.
Learning Objectives
By the end of this section, you will be able to:
- build a stem-and-leaf plot from a raw list of numbers and read the shape of the data off it;
- spot an outlier in a stemplot and say what makes it stand apart from the rest of the data;
- construct a side-by-side stemplot to compare two data sets that share the same scale;
- build a line graph from a frequency table and describe the pattern it traces;
- build a bar graph for categorical data and choose between plotting counts and plotting percentages.
In Chapter 1 you learned how to collect data and how to sort it into a frequency table. A table is honest, but it is slow to read — your eye has to walk down a column of numbers and assemble the pattern in your head. A graph does that assembling for you. This section covers three of the simplest and most useful graphs in statistics: the stem-and-leaf plot, the line graph, and the bar graph. Each one takes a pile of numbers and turns it into a picture whose shape you can read in a second.
2.1.1 Stem-and-Leaf Graphs (Stemplots)
A stem-and-leaf graph, or stemplot, is a display in which each observation is split into two parts: a stem, consisting of all the digits except the final significant digit, and a leaf, consisting of that final significant digit. The stems are written in a vertical column from smallest to largest with a vertical line to their right, and each observation's leaf is written in increasing order next to its corresponding stem.
Turn a stemplot on its side and the rows of leaves become bars — you get the shape of the data just like a histogram. The difference is that a stemplot never throws the actual numbers away. Every original value is still sitting there in the picture, ready to be read back.
The splitting rule is mechanical once you see it a few times. The number 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, and then write the leaves in increasing order next to their corresponding stem.
The reason this graph is worth learning first is that it costs you nothing but a pencil, and it answers two questions at once: what shape is this data, and what are the actual values. Most graphs make you trade one for the other.
Definition 2.1.1 — Each score splits into a stem and a leaf, and the leaves stack into their stem's row.
An outlier is an observation of data that does not fit the rest of the data. It is sometimes called an extreme value. When an outlier is graphed, it appears not to fit the pattern of the graph.
Finding a value far from the others does not tell you to delete it. It tells you to go ask why it is there. Sometimes the answer is a typo; sometimes the answer is the most interesting thing in the whole data set.
Some outliers are due to mistakes — for example, writing down 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.
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.
Definition 2.1.2 — An outlier is defined by the empty stretch beside it, not by being large.
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 stem plot for the data.
Solution
Step 1 — Pick the stems. All 30 scores are two-digit numbers between 32 and 61, so the tens digit is the stem and the ones digit is the leaf. The stems run 3, 4, 5, 6.
Step 2 — Deal each score to its stem, keeping the leaves in increasing order:
| Stem | Leaf |
|---|---|
| 3 | 2 2 3 4 8 |
| 4 | 0 2 2 3 4 6 7 7 8 8 8 9 |
| 5 | 0 0 1 2 2 2 3 4 6 7 7 |
| 6 | 0 1 |
Step 3 — Check your work. Count the leaves: \(5 + 12 + 11 + 2 = 30\), which matches the 30 games we started with.
Answer: The stemplot above. The team scored in the 40s and 50s most often, and there is no value that sits far away from the rest — nothing here looks like an outlier.
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.
Try it in rāSHio
Open rāSHio and paste the thirty game scores (32; 32; 33; 34; 38; …) straight into File → Delimited List… — the semicolons parse as-is, so you never retype a value. Once the scores are in a column you can sort them, which is exactly the ordering a stemplot needs before you start dealing leaves into rows.
Figure 2.1.1 — Loading a raw list into rāSHio with File → Delimited List, the first step before building a stemplot by hand.
For Professor Vang's spring pre-calculus class, scores on her 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
Construct a stemplot for these scores, then describe what it tells you about how the class did.
Solution
Step 1 — Pick the stems. The scores are two- and three-digit whole numbers, so the leaf is the ones digit and the stem is everything to its left. A score of 33 has stem 3 and leaf 3; a score of 100 has stem 10 and leaf 0. The stems therefore run from 3 up to 10.
Step 2 — Deal each score to its stem. Walk the list once, writing each score's leaf next to its stem. Because the list is already sorted, the leaves come out in increasing order automatically.
| 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 |
Step 3 — Read the shape. The rows with the most leaves are the 60s, 70s, 80s, and 90s, so that is where most of the class landed. The single leaf on stem 3 sits well below everything else.
Answer: The stemplot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores, or approximately 26%, were in the 90s or 100 — a fairly high number of As.
The following data show the distances (in miles) from the homes of off-campus statistics students to the college. Create a stem plot using the 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
Solution
Step 1 — Pick the stems. The leaf is the tenths digit and the stem is the whole-number part. Distances run from 0.5 to 8.0, so the stems run 0 through 8 — and we write down the empty stems too.
Step 2 — Deal each distance to its stem:
| Stem | Leaf |
|---|---|
| 0 | 5 7 |
| 1 | 1 2 2 3 3 5 5 7 7 8 9 |
| 2 | 0 2 5 6 8 8 8 |
| 3 | 5 8 |
| 4 | 4 8 9 |
| 5 | 2 5 7 8 |
| 6 | |
| 7 | |
| 8 | 0 |
Step 3 — Check the count: \(2 + 11 + 7 + 2 + 3 + 4 + 0 + 0 + 1 = 30\), which matches the 30 students.
Step 4 — Look for values that do not fit. Stems 6 and 7 are empty, so the value 8.0 sits two full stems away from the next-closest distance of 5.8.
Answer: Most students live within about 3 miles of the college, with the heaviest concentration between 1 and 2 miles. The distance 8.0 miles is a likely outlier — it is separated from the rest of the data by two empty stems.
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
Do the data seem to have any concentration of values?
Here the leaves are the digits to the right of the decimal, so 3.3 has stem 3 and leaf 3. The rule has not changed — the leaf is still the final significant digit — the split just happens to land on the decimal point.
Solution
Step 1 — Pick the stems. Each value is a one- or two-digit whole number plus one decimal place, so the leaf is the tenths digit and the stem is the whole-number part. Values run from 1.1 to 12.3, so the stems run 1 through 12. We list every stem in that range, including the empty ones — the gaps are part of what the picture is telling us.
Step 2 — Deal each distance to its stem:
| 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 |
Step 3 — Read the shape. Stems 3 and 4 hold the most leaves, and stems 7 through 11 hold none at all — five empty rows separate 6.7 from 12.3.
Answer: The value 12.3 may be an outlier. Values appear to concentrate at three and four kilometers.
2.1.2 Comparing Two Data Sets: Side-by-Side Stemplots
A side-by-side stem-and-leaf plot displays two data sets that share the same stems in two columns. The two sets of leaves share a single central column of stems, with one data set's leaves written to the left of the stems and the other data set's leaves written to the right.
Think of the stem column as a shared ruler with one data set hanging off each side. Because both sides are measured against the same ruler, you can see at a glance which group sits higher, which is more spread out, and where they overlap.
Reading a side-by-side plot takes one adjustment: the leaves on the left run outward from the stem, so they increase as you move right toward the center. Everything else works exactly as it did before, and the payoff is that you can see both distributions and compare their shapes without flipping between two pictures.
Definition 2.1.3 — Two data sets share one column of stems, with each set's leaves growing outward.
The table shows the number of wins and losses the Atlanta Hawks have had in 42 seasons. Create a side-by-side stem-and-leaf plot of these wins and losses.
| Losses | Wins | Season | Losses | Wins | Season |
|---|---|---|---|---|---|
| 34 | 48 | 1 | 41 | 41 | 22 |
| 34 | 48 | 2 | 39 | 43 | 23 |
| 46 | 36 | 3 | 44 | 38 | 24 |
| 46 | 36 | 4 | 39 | 43 | 25 |
| 36 | 46 | 5 | 25 | 57 | 26 |
| 47 | 35 | 6 | 40 | 42 | 27 |
| 51 | 31 | 7 | 36 | 46 | 28 |
| 53 | 29 | 8 | 26 | 56 | 29 |
| 51 | 31 | 9 | 32 | 50 | 30 |
| 41 | 41 | 10 | 19 | 31 | 31 |
| 36 | 46 | 11 | 54 | 28 | 32 |
| 32 | 50 | 12 | 57 | 25 | 33 |
| 51 | 31 | 13 | 49 | 33 | 34 |
| 40 | 42 | 14 | 47 | 35 | 35 |
| 39 | 43 | 15 | 54 | 28 | 36 |
| 42 | 40 | 16 | 69 | 13 | 37 |
| 48 | 34 | 17 | 56 | 26 | 38 |
| 32 | 50 | 18 | 52 | 30 | 39 |
| 25 | 57 | 19 | 45 | 37 | 40 |
| 32 | 50 | 20 | 35 | 47 | 41 |
| 30 | 52 | 21 | 29 | 53 | 42 |
Solution
Step 1 — Collect all 42 values for each variable. The table stores two seasons per row, so read the left three columns down first (seasons 1–21), then the right three columns (seasons 22–42).
Step 2 — Find the shared stems. Losses run from 19 to 69 and wins run from 13 to 57, so the shared stem column runs 1 through 6 (the tens digit).
Step 3 — Put losses on the left and wins on the right, with the left-hand leaves increasing toward the center:
| Losses | Stem | Wins |
|---|---|---|
| 9 | 1 | 3 |
| 9 6 5 5 | 2 | 5 6 8 8 9 |
| 9 9 9 6 6 6 5 4 4 2 2 2 2 0 | 3 | 0 1 1 1 1 3 4 5 5 6 6 7 8 |
| 9 8 7 7 6 6 5 4 2 1 1 0 0 | 4 | 0 1 1 2 2 3 3 3 6 6 6 7 8 8 |
| 7 6 4 4 3 2 1 1 1 | 5 | 0 0 0 0 2 3 6 7 7 |
| 9 | 6 |
Step 4 — Check the counts. Losses: \(1 + 4 + 14 + 13 + 9 + 1 = 42\). Wins: \(1 + 5 + 13 + 14 + 9 = 42\). Both match the 42 seasons.
Answer: The side-by-side plot above. The two distributions are near mirror images of each other, which makes sense — every season has a fixed number of games, so a season with many losses is automatically a season with few wins. The 69-loss season on the left and the 13-win season on the right are the same season (season 37), and it stands apart from the rest on both sides.
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 2.1.7 and Table 2.1.8 show the ages of presidents at their inauguration and at their death. Construct a side-by-side stem-and-leaf plot using this 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 |
| 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 |
Solution
Step 1 — Find the stems both data sets need to share. Inauguration ages run from 42 to 69, so those need stems 4, 5, and 6. Ages at death run from 46 to 93, so those need stems 4 through 9. The shared stem column has to cover both, so it runs 4 through 9.
Step 2 — Put the inauguration leaves on the left and the death leaves on the right. On the left side the leaves are written outward from the stem, so they increase as you read toward the center column.
| Ages at Inauguration | Stem | 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 2 1 1 1 1 1 0 | 5 | 3 6 6 7 7 8 |
| 9 8 5 4 4 2 1 1 1 0 | 6 | 0 0 3 3 4 4 5 6 7 7 7 8 |
| 7 | 0 1 1 1 2 3 4 7 8 8 9 | |
| 8 | 0 1 3 5 8 | |
| 9 | 0 0 3 3 |
Step 3 — Compare the two sides. The inauguration leaves pile up almost entirely on stem 5 — most presidents took office in their fifties, and none of them took office past 69. The death leaves are spread across stems 4 through 9, sitting highest on stems 6, 7, and 8.
Answer: The side-by-side plot above. Ages at inauguration are tightly clustered in the fifties, while ages at death are both higher and far more spread out — which is what you would expect, since every president is inaugurated inside a narrow window of a political career but can die at almost any age afterward.
2.1.3 Line Graphs
A line graph displays a set of data values by plotting each value as a point and connecting consecutive points with line segments. In the line graphs used here, the \(x\)-axis (horizontal axis) consists of data values and the \(y\)-axis (vertical axis) consists of frequency points.
In a graph like this the categories are whole counts — you cannot be reminded 2.5 times. The segment between two points does not assert that anything happened in between; it is there so your eye can follow the trend from one count to the next.
The plotted points are exactly the rows of a frequency table — the data value tells you how far right to go, and the frequency tells you how far up. Connecting them is what makes the graph readable: your eye follows the line up to a peak and back down, and that rise-and-fall is the shape of the distribution.
Another type of graph that is useful for specific data values is a line graph.
Definition 2.1.4 — Each frequency-table row becomes a point, and the connecting segments trace the distribution's shape.
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 2.1.10. Construct a line graph.
| Number of times in shop | Frequency |
|---|---|
| 0 | 7 |
| 1 | 10 |
| 2 | 14 |
| 3 | 9 |
Solution
Step 1 — Set up the axes. Number of repair visits goes on the horizontal axis, running 0 through 3. Frequency goes on the vertical axis; the largest frequency is 14, so scale it 0 to 16 in steps of 2.
Step 2 — Plot one point per row: \((0, 7)\), \((1, 10)\), \((2, 14)\), \((3, 9)\).
Step 3 — Connect consecutive points with straight line segments, moving left to right.
Step 4 — Check the total. The four frequencies add to \(7 + 10 + 14 + 9 = 40\), which matches the 40 people surveyed.
Answer: A line graph that rises from \((0, 7)\) through \((1, 10)\) to a peak at \((2, 14)\), then drops to \((3, 9)\). Two repair visits per year is the most common answer, and the counts fall off on both sides of it.
Try it in rāSHio
Open rāSHio, paste the raw repair counts into File → Delimited List…, then choose Graph → Frequency Table with Discrete values checked. rāSHio builds the value-and-frequency pairs for you — the same two columns this exercise hands you ready-made — so you can check a frequency table you tallied yourself before you plot it.
Figure 2.1.2 — Building a frequency table in rāSHio with Graph → Frequency Table, the two columns a line graph plots.
In a survey, 40 parents were asked how many times per week a teenager must be reminded to do their chores. The results are shown in Table 2.1.11. Display these results as a line graph and describe the pattern.
| Number of times teenager is reminded | Frequency |
|---|---|
| 0 | 2 |
| 1 | 5 |
| 2 | 8 |
| 3 | 14 |
| 4 | 7 |
| 5 | 4 |
Solution
Step 1 — Set up the axes. Put the number of reminders on the horizontal axis, running 0 through 6, and the frequency on the vertical axis, running 0 through 16 so that the tallest value (14) fits comfortably.
Step 2 — Plot one point per row of the table: \((0, 2)\), \((1, 5)\), \((2, 8)\), \((3, 14)\), \((4, 7)\), \((5, 4)\).
Step 3 — Connect consecutive points with straight segments.
Figure 2.1.3 — Line graph of how many times per week a teenager must be reminded to do chores.
Step 4 — Read the pattern. The line climbs steadily from 2 up to a peak of 14 at three reminders, then falls away to 4 at five reminders.
Answer: The line graph above. Three reminders per week is by far the most common answer, and responses fall off on either side of it — the distribution has a single clear peak.
2.1.4 Bar Graphs
A bar graph displays data using rectangular bars that are separated from one another, with one bar per category. The length of each bar is proportional to the value it represents — a frequency, a count, or a percentage.
Separated bars mean categories that are genuinely distinct. Touching bars mean a number line that has been chopped into intervals. Before you read any bar chart, check the gaps — they tell you whether the horizontal axis is a list or a ruler.
The gaps between the bars are not decoration; they carry meaning. A bar graph is for data whose categories do not flow into one another — age brackets, regions, brands, school names. Because there is nothing "between" the South and the West, the bars stand apart. In Chapter 2.2 you will meet the histogram, which looks similar but pushes its bars together precisely because its categories do run continuously into each other.
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.
Definition 2.1.5 — Bar graph bars are separated on purpose: the gaps mark the categories as distinct.
The population in Park City is made up of children, working-age adults, and retirees. Table 2.1.12 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% |
Solution
Step 1 — Pick the right column. The question asks for proportions, so we plot the third column and ignore the raw counts.
Step 2 — Set up the axes. The three age groups go on the horizontal axis. The vertical axis is the proportion of the population; the largest value is 43%, so scale it 0% to 45% in steps of 5%.
Step 3 — Draw one separated bar per group: Children at 19%, Working-age adults at 43%, Retirees at 38%.
Step 4 — Sanity-check the total. The three proportions add to \(19\% + 43\% + 38\% = 100\%\), so every resident is counted exactly once.
Answer: A bar graph with three separated bars — Children (19%), Working-age adults (43%), and Retirees (38%). Working-age adults are the largest group, but retirees are close behind, and together the two adult groups make up 81% of the town.
The percentage of U.S.-based TikTok users by age is shown in Table 2.1.13. Construct a bar graph using this data.
| Age groups | Proportion (%) of TikTok users |
|---|---|
| 10–19 | 32.5% |
| 20–29 | 29.5% |
| 30–39 | 16.4% |
| 40–49 | 13.9% |
| 50+ | 7.1% |
Solution
Step 1 — Set up the axes. The age groups are the categories, so they go on the horizontal axis in the order the table lists them. The proportions go on the vertical axis; the largest is 32.5%, so scale it from 0% to 35% in steps of 5%.
Step 2 — Draw one bar per age group, with each bar's height equal to that group's percentage, and leave a visible gap between neighboring bars.
Figure 2.1.4 — Bar graph of the proportion of U.S.-based TikTok users in each age group.
Step 3 — Read the result. The bars fall from left to right without exception, and the first two together account for \(32.5\% + 29.5\% = 62\%\) of users.
Answer: The bar graph above. TikTok's U.S. user base skews strongly young — nearly two-thirds of users are under 30, and each successive age group makes up a smaller share than the one before it.
Park City is broken down into six voting districts. Table 2.1.14 shows the percent of the total registered voter population that lives in each district as well as the percent total 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% |
Solution
Step 1 — Pick the right column. The question asks for the registered voter population, so we plot the second column only and leave the overall city population aside.
Step 2 — Set up the axes. The six districts go on the horizontal axis, numbered 1 through 6. The vertical axis is the percentage of registered voters; the largest value is 22.8%, so scale it 0% to 25% in steps of 5%.
Step 3 — Draw one separated bar per district: District 1 at 15.5%, District 2 at 12.2%, District 3 at 9.8%, District 4 at 17.4%, District 5 at 22.8%, District 6 at 22.3%.
Step 4 — Sanity-check the total. The six percentages add to \(15.5 + 12.2 + 9.8 + 17.4 + 22.8 + 22.3 = 100\%\), so every registered voter is counted once.
Answer: A bar graph with six separated bars at the heights listed above. Districts 5 and 6 hold the largest shares of registered voters at 22.8% and 22.3%, while District 3 holds the smallest at 9.8%.
The columns in Table 2.1.15 show the projected data for the year 2030 for the number and percentages of high school graduates by geographic region in the United States. Create a bar graph for this data with the geographic region (qualitative data) on the \(x\)-axis and the percentage of high school graduates (quantitative data) on the \(y\)-axis.
| Region | Number of Graduates | Percentage of Graduates |
|---|---|---|
| Northeast | 517,720 | 16.1% |
| Midwest | 695,170 | 21.6% |
| South | 1,253,540 | 39.0% |
| West | 749,400 | 23.3% |
Solution
Step 1 — Set up the axes. The four regions are qualitative categories, so they go on the horizontal axis. The percentage of graduates is the quantitative variable, so it goes on the vertical axis; the largest value is 39.0%, so scale it 0% to 45% in steps of 5%.
Step 2 — Draw one separated bar per region, with each bar's height equal to that region's percentage.
Figure 2.1.5 — Bar graph of the percentage of projected 2030 high school graduates by U.S. geographic region.
Step 3 — Read the result. The South's bar is roughly the height of the Northeast's and Midwest's put together.
Answer: The bar graph above. The South is projected to produce 39.0% of the country's high school graduates in 2030 — more than either of the next two regions and nearly two and a half times the Northeast's share.
Problem Set 2.1
For each of the following data sets, create a stem plot and identify any outliers.
Problem 1. The miles per gallon rating for 30 cars are shown below (lowest to highest).
19, 19, 19, 20, 21, 21, 25, 25, 25, 26, 26, 28, 29, 31, 31, 32, 32, 33, 34, 35, 36, 37, 37, 38, 38, 38, 38, 41, 43, 43
Solution
Step 1 — Split each value. The ratings are two-digit whole numbers, so the tens digit is the stem and the ones digit is the leaf. The values run from 19 to 43, so the stems are 1, 2, 3, and 4.
Step 2 — Deal each rating to its stem, keeping the leaves in increasing order:
| Stem | Leaf |
|---|---|
| 1 | 9 9 9 |
| 2 | 0 1 1 5 5 5 6 6 8 9 |
| 3 | 1 1 2 2 3 4 5 6 7 7 8 8 8 8 |
| 4 | 1 3 3 |
Step 3 — Check the count: \(3 + 10 + 14 + 3 = 30\), matching the 30 cars.
Step 4 — Look for outliers. Every stem from 1 to 4 carries leaves, so there is no gap anywhere in the display. The bulk of the cars sit on stem 3 (the 30s).
Answer: The stemplot above. Most cars rate in the 30s mpg, and there are no outliers — the values run continuously from 19 to 43 with no isolated gaps.
Problem 2. The height in feet of 25 trees is shown below (lowest to highest).
25, 27, 33, 34, 34, 34, 35, 37, 37, 38, 39, 39, 39, 40, 41, 45, 46, 47, 49, 50, 50, 53, 53, 54, 54
Solution
Step 1 — Split each value. Heights are two-digit whole numbers, so the tens digit is the stem and the ones digit is the leaf. Values run from 25 to 54, so the stems are 2, 3, 4, and 5.
Step 2 — Deal each height to its stem:
| Stem | Leaf |
|---|---|
| 2 | 5 7 |
| 3 | 3 4 4 4 5 7 7 8 9 9 9 |
| 4 | 0 1 5 6 7 9 |
| 5 | 0 0 3 3 4 4 |
Step 3 — Check the count: \(2 + 11 + 6 + 6 = 25\), matching the 25 trees.
Step 4 — Look for outliers. No stem in the range is empty, so nothing is stranded away from the rest of the data.
Answer: The stemplot above. Tree heights concentrate in the 30s, and there are no outliers.
Problem 3. The data are the prices of different laptops at an electronics store. Round each value to the nearest ten.
249, 249, 260, 265, 265, 280, 299, 299, 309, 319, 325, 326, 350, 350, 350, 365, 369, 389, 409, 459, 489, 559, 569, 570, 610
Solution
Step 1 — Round each price to the nearest ten first, as the problem instructs. Rounding half up:
250, 250, 260, 270, 270, 280, 300, 300, 310, 320, 330, 330, 350, 350, 350, 370, 370, 390, 410, 460, 490, 560, 570, 570, 610
Step 2 — Split each rounded value. Because every value now ends in a zero, the useful split is the hundreds digit as the stem and the tens digit as the leaf. So $250 has stem 2 and leaf 5, and $610 has stem 6 and leaf 1.
Step 3 — Deal each rounded price to its stem:
| Stem (hundreds) | Leaf (tens) |
|---|---|
| 2 | 5 5 6 7 7 8 |
| 3 | 0 0 1 2 3 3 5 5 5 7 7 9 |
| 4 | 1 6 9 |
| 5 | 6 7 7 |
| 6 | 1 |
Step 4 — Check the count: \(6 + 12 + 3 + 3 + 1 = 25\), matching the 25 laptops.
Step 5 — Look for outliers. The display is heavily bunched on stems 2 and 3 — 18 of the 25 laptops are priced under $400 — and then thins out sharply. The single laptop at $610 sits by itself on stem 6.
Answer: The stemplot above. Prices concentrate between $250 and $390. The $610 laptop (and to a lesser extent the $560–$570 cluster) sits well above the main body of the data, so $610 is a likely outlier.
Problem 4. The data are daily high temperatures in a town for one month.
61, 61, 62, 64, 66, 67, 67, 67, 68, 69, 70, 70, 70, 71, 71, 72, 74, 74, 74, 75, 75, 75, 76, 76, 77, 78, 78, 79, 79, 95
Solution
Step 1 — Split each value. Temperatures are two-digit whole numbers, so the tens digit is the stem and the ones digit is the leaf. Values run from 61 to 95, so the stems are 6, 7, 8, and 9 — and we write stem 8 down even though it will end up empty, because the gap is the point.
Step 2 — Deal each temperature to its stem:
| Stem | Leaf |
|---|---|
| 6 | 1 1 2 4 6 7 7 7 8 9 |
| 7 | 0 0 0 1 1 2 4 4 4 5 5 5 6 6 7 8 8 9 9 |
| 8 | |
| 9 | 5 |
Step 3 — Check the count: \(10 + 19 + 0 + 1 = 30\), matching the 30 days.
Step 4 — Look for outliers. Stem 8 is completely empty, so the 95-degree day is separated from the next-highest temperature (79) by an entire stem.
Answer: The stemplot above. Daily highs cluster in the 70s, with a secondary group in the 60s. The high of 95 is an outlier — the empty 80s row shows there is nothing bridging it to the rest of the month.
For the next three exercises, use the data to construct a line graph.
Problem 5. In a survey, 40 people were asked how many times they visited a store before making a major purchase. The results are shown in Table 2.1.16.
| Number of times in store | Frequency |
|---|---|
| 1 | 4 |
| 2 | 10 |
| 3 | 16 |
| 4 | 6 |
| 5 | 4 |
Solution
Step 1 — Set up the axes. The number of store visits goes on the horizontal axis, running 1 through 5. Frequency goes on the vertical axis; the largest frequency is 16, so scale it 0 to 18 in steps of 2.
Step 2 — Plot one point per row of the table: \((1, 4)\), \((2, 10)\), \((3, 16)\), \((4, 6)\), \((5, 4)\).
Step 3 — Connect consecutive points with straight line segments, moving left to right.
Step 4 — Check the total. The frequencies add to \(4 + 10 + 16 + 6 + 4 = 40\), matching the 40 people surveyed.
Answer: A line graph rising from \((1, 4)\) to a peak at \((3, 16)\), then dropping to \((4, 6)\) and \((5, 4)\). Three visits before a major purchase is by far the most common answer, and the drop after the peak is much steeper than the climb up to it.
Problem 6. In a survey, several people were asked how many years it has been since they purchased a mattress. The results are shown in Table 2.1.17.
| Years since last purchase | Frequency |
|---|---|
| 0 | 2 |
| 1 | 8 |
| 2 | 13 |
| 3 | 22 |
| 4 | 16 |
| 5 | 9 |
Solution
Step 1 — Set up the axes. Years since the last mattress purchase goes on the horizontal axis, running 0 through 5. Frequency goes on the vertical axis; the largest frequency is 22, so scale it 0 to 24 in steps of 4.
Step 2 — Plot one point per row: \((0, 2)\), \((1, 8)\), \((2, 13)\), \((3, 22)\), \((4, 16)\), \((5, 9)\).
Step 3 — Connect consecutive points with straight line segments.
Step 4 — Note the total. The frequencies add to \(2 + 8 + 13 + 22 + 16 + 9 = 70\), so 70 people answered the survey.
Answer: A line graph climbing steadily from \((0, 2)\) to a peak at \((3, 22)\), then falling away through \((4, 16)\) to \((5, 9)\). Three years is the most common answer, and the shape has a single clear peak with a gentler decline on the right than the rise on the left.
Problem 7. Several children were asked how many TV shows they watch each day. The results of the survey are shown in Table 2.1.18.
| Number of TV Shows | Frequency |
|---|---|
| 0 | 12 |
| 1 | 18 |
| 2 | 36 |
| 3 | 7 |
| 4 | 2 |
Solution
Step 1 — Set up the axes. The number of TV shows watched per day goes on the horizontal axis, running 0 through 4. Frequency goes on the vertical axis; the largest frequency is 36, so scale it 0 to 40 in steps of 5.
Step 2 — Plot one point per row: \((0, 12)\), \((1, 18)\), \((2, 36)\), \((3, 7)\), \((4, 2)\).
Step 3 — Connect consecutive points with straight line segments.
Step 4 — Note the total. The frequencies add to \(12 + 18 + 36 + 7 + 2 = 75\), so 75 children were surveyed.
Answer: A line graph rising from \((0, 12)\) through \((1, 18)\) to a sharp peak at \((2, 36)\), then collapsing to \((3, 7)\) and \((4, 2)\). Two shows per day is by far the most common answer, and the fall-off after the peak is far steeper than the climb — almost half the children watch exactly two shows, and very few watch more than that.
Problem 8. The students in Ms. Ramirez's math class have birthdays in each of the four seasons. She recorded them in Table 2.1.19, which shows the four seasons, the number of students who have birthdays in each season, and the percentage (%) of students in each group. Construct a bar graph showing the number of students.
| Seasons | Number of students | Proportion of population |
|---|---|---|
| Spring | 8 | 24% |
| Summer | 9 | 26% |
| Autumn | 11 | 32% |
| Winter | 6 | 18% |
Solution
Step 1 — Pick the right column. The problem asks for the number of students, so we plot the second column and set the percentages aside.
Step 2 — Set up the axes. The four seasons go on the horizontal axis in the order the table lists them. The vertical axis is the number of students; the largest value is 11, so scale it 0 to 12 in steps of 2.
Step 3 — Draw one separated bar per season: Spring at 8, Summer at 9, Autumn at 11, Winter at 6. Leave a visible gap between the bars, because the seasons are distinct categories rather than intervals on a number line.
Step 4 — Note the total. The counts add to \(8 + 9 + 11 + 6 = 34\), so there are 34 students in the class.
Answer: A bar graph with four separated bars at heights 8, 9, 11, and 6. Autumn has the most birthdays and Winter the fewest, but the four seasons are fairly evenly represented — the tallest bar is under twice the height of the shortest.
Problem 9. Using the data from Ms. Ramirez's math class supplied in Table 2.1.19, construct a bar graph showing the percentages.
Solution
Step 1 — Pick the right column. This time the problem asks for the percentages, so we plot the third column instead of the counts.
Step 2 — Set up the axes. The four seasons stay on the horizontal axis in the same order. The vertical axis is now the percentage of students; the largest value is 32%, so scale it 0% to 35% in steps of 5%.
Step 3 — Draw one separated bar per season: Spring at 24%, Summer at 26%, Autumn at 32%, Winter at 18%.
Step 4 — Sanity-check the total. The percentages add to \(24\% + 26\% + 32\% + 18\% = 100\%\), so every student is counted exactly once.
Answer: A bar graph with four separated bars at heights 24%, 26%, 32%, and 18%. Notice that this graph has exactly the same shape as the one in the previous problem — only the vertical scale changed. Plotting counts and plotting percentages of the same data always produces the same profile; the percentage version just lets you compare this class against a class of a different size.
Problem 10. David County has six high schools. Each school sent students to participate in a county-wide science competition. Table 2.1.20 shows the percentage breakdown of competitors from each school, and the percentage of the entire student population of the county that goes to each school. Construct a bar graph that shows the population percentage of competitors from each school.
| High School | Science competition population | Overall student population |
|---|---|---|
| Alabaster | 28.9% | 8.6% |
| Concordia | 7.6% | 23.2% |
| Genoa | 12.1% | 15.0% |
| Mocksville | 18.5% | 14.3% |
| Tynneson | 24.2% | 10.1% |
| West End | 8.7% | 28.8% |
Solution
Step 1 — Pick the right column. The problem asks for the percentage of competitors from each school, so we plot the "Science competition population" column and leave the overall student population aside.
Step 2 — Set up the axes. The six high schools go on the horizontal axis. The vertical axis is the percentage of competitors; the largest value is 28.9%, so scale it 0% to 30% in steps of 5%.
Step 3 — Draw one separated bar per school: Alabaster at 28.9%, Concordia at 7.6%, Genoa at 12.1%, Mocksville at 18.5%, Tynneson at 24.2%, West End at 8.7%.
Step 4 — Sanity-check the total. The percentages add to \(28.9 + 7.6 + 12.1 + 18.5 + 24.2 + 8.7 = 100.0\%\), so every competitor is counted once.
Answer: A bar graph with six separated bars at the heights listed above. Alabaster and Tynneson send the most competitors (28.9% and 24.2%), while Concordia sends the fewest at 7.6%.
Problem 11. Use the data from the David County science competition supplied in Table 2.1.20. Construct a bar graph that shows the county-wide population percentage of students at each school.
Solution
Step 1 — Pick the right column. This time the problem asks for the county-wide student population at each school, so we plot the "Overall student population" column.
Step 2 — Set up the axes. The same six high schools go on the horizontal axis, in the same order as the previous problem so the two graphs can be compared. The vertical axis is the percentage of the county's students; the largest value is 28.8%, so scale it 0% to 30% in steps of 5%.
Step 3 — Draw one separated bar per school: Alabaster at 8.6%, Concordia at 23.2%, Genoa at 15.0%, Mocksville at 14.3%, Tynneson at 10.1%, West End at 28.8%.
Step 4 — Sanity-check the total. The percentages add to \(8.6 + 23.2 + 15.0 + 14.3 + 10.1 + 28.8 = 100.0\%\), so every student in the county is counted once.
Answer: A bar graph with six separated bars at the heights listed above. Comparing this graph with the one from the previous problem is the interesting part: Alabaster holds only 8.6% of the county's students but supplies 28.9% of the competitors, while West End holds the largest share of students (28.8%) and supplies just 8.7% of the competitors. Alabaster is punching far above its weight in the science competition, and West End far below.
Key Terms
stem-and-leaf graph (stemplot) — a display that splits each observation into a stem (all digits but the last) and a leaf (the final significant digit), listing stems in a vertical column with their leaves in increasing order beside them.
stem — the leading digits of an observation, used as the row label in a stemplot.
leaf — the final significant digit of an observation, written beside its stem in a stemplot.
outlier — an observation that does not fit the pattern of the rest of the data; also called an extreme value.
side-by-side stem-and-leaf plot — a stemplot comparing two data sets that share one central column of stems, with one set's leaves to the left and the other's to the right.
line graph — a graph that plots each data value against its frequency as a point and joins consecutive points with line segments.
bar graph — a graph that represents each category with a separated rectangular bar whose length is proportional to the value it represents.