Back-To-Back Stemplot vs Stemplot

A back-to-back stemplot runs one column of stems down the middle and hangs one group's leaves off each side. Both sides must use the same stem definition and the same place value, or the two halves are not on one scale. No values are discarded, so each group's median can still be counted straight off the display.

With 7 as the stem holding the tens digit, class A's row begins 2, 4, 5 counting outward from the spine, so those leaves are the values 72, 74 and 75, and on the page they print as 5, 4, 2 before you reach the stem. Class B's leaves of 1, 3 and 8 sit to the right and read in the ordinary direction as 71, 73 and 78. On both sides the smallest leaf is the one nearest the spine.

The error worth naming is a counting one. "Class A has 9 leaves on the stem of 7 and class B has only 7, so the 70s were more common in class A." A leaf count is a raw count, and raw counts only compare groups fairly when the groups are the same size. With 24 students in class A and 16 in class B, the shares are $9 / 24 = 0.375$ against $7 / 16 = 0.4375$, so the 70s are a larger part of class B despite its shorter row. Whenever the two sides have different totals, convert rows to proportions of their own group before comparing them.

The layout holds exactly two groups. A third needs a different display, and the plot stops being readable as $n$ grows, since a row of thirty leaves is a wall of digits rather than a shape. That is where a histogram or a side-by-side boxplot takes over.

The Fall 2026 course lists back-to-back stem-and-leaf plots in topic 1.9, among the displays used to compare the distributions of one quantitative variable across groups.

A stemplot splits each value into a stem, the leading place values, and a leaf, the final digit. Stems run down the left in increasing order and the leaves sit beside their stem, also in increasing order. Each row behaves like a sideways histogram bar whose bin width is set by the stem unit, except that the plot keeps every original value, so the data can be read straight back out.

Eleven test scores, 52, 55, 58, 61, 63, 63, 67, 68, 72, 74 and 81, give stems by tens: 5 with leaves 2, 5, 8; 6 with leaves 1, 3, 3, 7, 8; 7 with leaves 2, 4; and 8 with leaf 1. With $n = 11$ the median is the 6th value. Stem 5 holds three leaves, so the 6th value is the third leaf on stem 6, which is 63.

That counting works only on a sorted plot. Write stem 6's leaves in the order the scores arrived, 8, 3, 7, 1, 3, and the 6th value from the top is still the third leaf on stem 6, which is now 7, so counting returns a median of 67 instead of 63. Sorting the leaves within each row is what makes the count valid; it is not a tidiness step.

What a stemplot conceals is magnitude. Stem 6 with leaf 3 means 63 only because the key says so, and the identical picture reads as 6.3 or 630 under a different key, so a stemplot with no stated key cannot be read at all. The shape is hostage to the stem unit as well. Split each stem here into low leaves 0 to 4 and high leaves 5 to 9 and those eleven scores fill seven rows instead of four, one of them empty, exposing a gap between 74 and 81 that the four-row version buried.

Stemplots are one of the three displays in topic 1.5, Graphical Representations for One Quantitative Variable.