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Graphs for Quantitative Data · Tutorial 133 of 996

Splitting Stems for Dense Stemplots

See how splitting stems can make a crowded stemplot easier to read while preserving every value.

Beginner 9 min read

What You'll Learn

  • Identify when a stemplot is crowded enough to benefit from splitting.
  • Apply the 0–4 and 5–9 leaf-range convention to repeated stems.
  • Construct split and unsplit stemplots from the same exam-score data.
  • Check that splitting preserves the original observations and their order.
  • Recognize when empty rows or selective splitting could confuse a display.

Why Split a Stem?

A stemplot keeps the individual data values visible, but that advantage can become harder to appreciate when many leaves crowd into the same row. A row with a long run of digits is difficult to scan, and it may be hard to distinguish repeated leaves or see where the values fall within that stem. Splitting a stem gives its leaves more room without grouping the observations into intervals or discarding their exact values.

The decision is about readability, not about changing the data. Each observation must still appear exactly once, and the stem-and-leaf key must still make it possible to recover the original value. The main choice is how to divide the possible leaves into smaller ranges and how consistently to apply that division.

Definition: A split stemplot repeats each stem in separate rows, assigning each row a different range of leaves. Under the common two-row convention, one row holds leaves 0–4 and the other holds leaves 5–9. The stem retains the same meaning in both rows; the row indicates which leaf range is allowed there.

For example, if the key says \(7|3=73\), then a leaf 3 beside stem 7 represents a score of 73. In a split display using the 0–4 and 5–9 ranges, that leaf belongs in the lower row for stem 7. A leaf 8 beside stem 7 represents 78 and belongs in the upper row. The split changes the placement of the leaves, not the values they encode.

Compare an Unsplit and Split Stemplot

Suppose a hypothetical instructor recorded these 26 exam scores, in points:

52, 54, 57, 58, 61, 62, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 72, 73, 74, 75, 76, 77, 78, 79, 79.

Use the tens digit as the stem and the units digit as the leaf. In an unsplit display, each score from 60 through 69 goes on stem 6, and each score from 70 through 79 goes on stem 7.

StemLeaves
52 4 7 8
61 2 2 3 4 5 6 7 8 9
70 1 2 2 3 4 5 6 7 8 9 9

The row for stem 7 contains 12 leaves, so it is relatively crowded. The row for stem 6 contains 10. The leaves are correctly ordered, but a reader scanning either row has to follow a long string of digits. Splitting the stems creates shorter rows:

Stem and leaf rangeLeaves
5, leaves 0–42 4
5, leaves 5–97 8
6, leaves 0–41 2 2 3 4
6, leaves 5–95 6 7 8 9
7, leaves 0–40 1 2 2 3 4
7, leaves 5–95 6 7 8 9 9

The repeated stem labels are intentional. The first row for stem 6 represents scores from 60 to 64; the second represents scores from 65 to 69. For example, the two leaves 2 on the lower row encode two scores of 62, while the leaf 5 on the upper row encodes 65. Repeated observations remain repeated leaves.

Worked Example: Splitting a Crowded Exam-Score Display

Use the score data above to construct a two-row split stemplot. Then check that the split display represents all 26 scores.

Step 1: Identify the stems and leaves. For a score of 64, the tens digit gives stem 6 and the units digit gives leaf 4. Repeating this for the data gives stems 5, 6, and 7. The leaves for each stem, in order, are \(2,4,7,8\); \(1,2,2,3,4,5,6,7,8,9\); and \(0,1,2,2,3,4,5,6,7,8,9,9\).

Step 2: Divide each stem's leaves by range. Leaves 0, 1, 2, 3, and 4 go in the lower row. Leaves 5, 6, 7, 8, and 9 go in the upper row. Thus, stem 6 divides into \(1,2,2,3,4\) and \(5,6,7,8,9\). Stem 7 divides into \(0,1,2,2,3,4\) and \(5,6,7,8,9,9\). Apply the same rule to stem 5.

Step 3: Count and check. The split rows contain 2, 2, 5, 5, 6, and 6 leaves. Their total is \(2+2+5+5+6+6=26\), matching the 26 observations. Each score is placed in exactly one row: for instance, 64 goes in stem 6's lower row, and 65 goes in stem 6's upper row. Therefore, the split display changes the layout but neither loses nor adds a score.

A Reliable Construction Method

A consistent method helps prevent misplaced leaves. First choose the stem-and-leaf place values and state a key with units. Then decide on the leaf ranges before distributing any observations. With the two-row convention, assign leaves 0–4 to the lower row and 5–9 to the upper row. Sort leaves within each row from smallest to largest, retaining duplicates. Finally, check the number of leaves against the number of observations.

1
Choose the split rule.
Use the 0–4 and 5–9 ranges for a two-row split, and make the rule clear in the key or row labels.
2
Place every observation.
Find its stem and leaf, then put that leaf in the row whose range contains it. Do not round or omit repeated values.
3
Order and audit the rows.
Arrange leaves from least to greatest within each row, then count them. The total must equal the number of observations.

The rule can be justified by noting that the two ranges partition all possible unit digits: every digit from 0 through 9 belongs to one range and only one. In symbols,

$$ \{0,1,2,3,4\}\ \cup\ \{5,6,7,8,9\}=\{0,1,2,3,4,5,6,7,8,9\}, \qquad \{0,1,2,3,4\}\ \cap\ \{5,6,7,8,9\}=\varnothing. $$

Consequently, each original leaf has exactly one permitted row. If every observation is assigned by its stem and leaf, the split display preserves the data. This argument also shows why it matters to state the split rule: without it, a repeated stem label alone does not say which row should receive a particular leaf.

Worked Example: Splitting Scores With Sparse Rows

A second hypothetical class has exam scores of 81, 82, 88, 89, 90, 94, and 96 points. Construct a two-row split display, including a row that has no leaves if the rule calls for it.

Step 1: Separate stems and leaves. The scores in the 80s have stem 8 and leaves 1, 2, 8, and 9. The scores in the 90s have stem 9 and leaves 0, 4, and 6.

Step 2: Apply the same ranges. For stem 8, leaves 1 and 2 belong in the 0–4 row; leaves 8 and 9 belong in the 5–9 row. For stem 9, leaves 0 and 4 belong in the 0–4 row, and leaf 6 belongs in the 5–9 row.

Stem and leaf rangeLeaves
8, leaves 0–41 2
8, leaves 5–98 9
9, leaves 0–40 4
9, leaves 5–96

Step 3: Verify the values. The key \(9|0=90\) shows why a zero leaf must be retained: it represents a score of 90, not a missing observation. The four rows contain \(2+2+2+1=7\) leaves, matching the seven scores. No row is empty here, but an empty row can occur when none of a stem's observations has a leaf in one range. Keeping both rows for each stem can make the rule easier to follow; the empty row should be left visibly blank rather than filled with a value that was not observed.

How Much Should a Stemplot Be Split?

Two rows per stem are a useful standard, but they are not an automatic improvement for every data set. If most stems have only one or two leaves, splitting every stem may create many short or empty rows and make the display taller without making it clearer. If one or two stems are crowded, a display may split only those stems, provided its key or labels clearly identify the leaf ranges and distinguish split from unsplit rows. For a beginner constructing a display, applying one rule consistently is often easier to read and check.

The goal is not to make every row the same length. It is to provide enough room to see the individual leaves while keeping the structure understandable. A reader should be able to tell, without guessing, what values each row can contain. If a display uses a less uniform choice, label the rows explicitly rather than relying on position or an unstated convention.

Worked Example: Choosing a Split for a Dense Stem

A hypothetical set of scores has observations 41, 44, 45, 48, 50, 52, 53, 54, 56, 57, 58, 59, 63, and 67. Stem 5 has eight leaves, while stems 4 and 6 have four and two leaves. Decide how to show the values using the two-row rule.

Step 1: List the leaves by stem. Stem 4 has leaves 1, 4, 5, and 8. Stem 5 has leaves 0, 2, 3, 4, 6, 7, 8, and 9. Stem 6 has leaves 3 and 7.

Step 2: Split each represented stem consistently. Stem 4's lower row gets 1 and 4; its upper row gets 5 and 8. Stem 5's lower row gets 0, 2, 3, and 4; its upper row gets 6, 7, 8, and 9. Stem 6's lower row gets 3 and its upper row gets 7.

Stem and leaf rangeLeaves
4, leaves 0–41 4
4, leaves 5–95 8
5, leaves 0–40 2 3 4
5, leaves 5–96 7 8 9
6, leaves 0–43
6, leaves 5–97

Step 3: Check the count. The rows contain \(2+2+4+4+1+1=14\) leaves, equal to the 14 scores. Although only stem 5 was especially crowded, using the same two-row rule for stems 4 and 6 makes the layout predictable. If only stem 5 were split, the display would need to make clear that stems 4 and 6 are unsplit while stem 5 has two ranges.

Common Mistakes and Why the Choice Matters

Splitting is useful because it can make individual observations easier to locate in a dense row. However, a neat-looking display is not necessarily a correct one. The key, row labels, ordering, and leaf count all need to agree with the actual data.

  • Changing the stem's meaning. A repeated stem does not represent a new place value. Both rows for stem 6 still represent scores in the 60s; the leaf range distinguishes the rows.
  • Putting a boundary leaf in the wrong row. Under the 0–4 and 5–9 rule, leaf 4 belongs in the lower row and leaf 5 in the upper row. State the convention clearly.
  • Removing duplicate leaves. Two observations with the same value require two leaves. Repeated digits are not a formatting error.
  • Dropping zeros or other values that look unusual. With the key \(9|0=90\), a zero leaf represents a real score and must be shown.
  • Assuming a split display changes the observations. It only changes the layout. Each value must still be recoverable from its stem and leaf.
  • Splitting without making the ranges clear. A reader may not know which repeated-stem row contains a particular leaf. Include a key, labels, or an explicit statement of the row ranges.

A common mistaken argument is, “The score 65 is close to 64, so either row is acceptable.” That is incorrect under the stated split rule: the row is determined by the leaf digit, not by which other score seems closest. Since 5 is in the 5–9 range, 65 belongs in stem 6's upper row. Applying the rule mechanically protects the values and makes the display interpretable.

Key takeaway: Split a crowded stem by assigning every possible leaf to a clearly stated range, placing each observation in exactly one row, and keeping the leaves ordered. Then verify that the total number of leaves matches the number of data values.

Check Your Understanding

Try these questions using the two-row convention of leaves 0–4 and 5–9.

  1. Where should the leaf 4 go for stem 8, and what value does it represent if the key is \(8|4=84\)?
  2. A stem has leaves 0, 2, 5, 5, 8, and 9. Write the leaves in its two split rows and state how many observations the rows represent.
  3. Why should two identical exam scores appear as two leaves rather than one?
  4. A stem has observations with leaves 1 and 3 only. What should its 5–9 row contain under a consistent two-row display?
  5. Explain how counting the leaves can help detect an error in a split stemplot.