Math 6

Stem and leaf plots

A list of numbers shows nothing at a glance. How can you draw it so its shape shows, without losing a single value?

You need

Dot plotsGrades 4 to 6 Place valueGrades 1 to 5

Where you are 12 in Data displays

Statistics7

Data displays12

▴ 4 above Reading a graph: titles, labels and units Scatter plots2

Stem and leaf plots

Choosing a display for a question Misleading graphs ▾ Show all 12 (3 more)

When you reach for it

Celyn writes down the best clearance of each of fifteen jumpers at a high jump practice. As a list it says little: where do most jumps sit, and is one far from the rest? A dot plot would answer that.

A stem and leaf plot does the same job with the digits themselves. Each value is split into a stem and a leaf, the stems go down one column, and the leaves sit in rows beside them. The rows make a picture of the data’s shape, and every value can still be read back exactly. It suits a small set of numbers, a few dozen at most. Two groups measured in the same way can share one column of stems, with leaves running out to each side: a back-to-back stem and leaf plot.

The procedure

Splitting each value

By Place value, every digit of a number has a place. The leaf is the value’s last digit, and the stem is all the digits in front of it.

Value	Stem	Leaf
113	11	3
97	9	7
7	0	7
6.4, leaf in tenths	6	4

A one-digit number has stem 0. Values with the same stem lie in the same stretch of ten: 110 to 119 all have stem 11. So one row of the plot holds values that are close together, and every row covers an equal stretch, ten possible leaves from 0 to 9, like the marks on a number line.

Building the plot

Celyn’s fifteen heights, in centimeters, are 118, 104, 125, 131, 112, 97, 120, 116, 108, 127, 113, 122, 135, 119, 110. The smallest is 97 and the largest 135, so the stems run from 9 to 13. They are written once each, down a column from smallest to largest, with a line to their right. Then each value’s leaf goes in its stem’s row: 118 puts an 8 in row 11. Last, the leaves of each row are put in order from smallest to largest, evenly spaced.

Figure 1Turned a quarter turn, the rows are bars. Unlike bars, they are made of the values themselves.

Each of these choices has a reason.

Every stem in the range gets a row, even an empty one. The rows stand for equal stretches of ten. Leaving out an empty row would close up a gap in the data and hide it.
Leaves are spaced evenly. Then the length of a row counts its values, so a longer row really means more values.
Leaves go in order. The whole plot is then sorted: reading row by row, left to right, gives the data from smallest to largest.
A key says what one entry means, here “11 \mid 3 means 113 cm”. The same digits could stand for 11.3 or 1{,}130; only the key settles it, and it carries the unit.

Reading it back

Each entry is read stem first, then leaf: 11 \mid 3 is 113 cm. Turn the page a quarter turn and the rows become bars, as in a histogram: the jumps pile up in the 110s, with fewer above and below. Unlike a histogram, nothing is lost. The bar for the 110s is made of the six actual values 110, 112, 113, 116, 118, 119.

Because the plot is sorted, the smallest value is the first leaf of the top row, 97, and the largest is the last leaf of the bottom row, 135. The median is found by counting. Fifteen values have one in the middle, the eighth. Rows 9 and 10 hold three values, so row 11 holds the fourth to the ninth, and the eighth is 118: the median is 118 cm. When the count is even, the median is halfway between the two middle values, as in Finding the median.

Two groups back to back

To compare two groups measured in the same unit, give them one shared column of stems. One group’s leaves go to the right as usual. The other group’s leaves go to the left, starting next to the stem and running outward, so on the left the smallest leaf is nearest the stem and the leaves grow from right to left. Every leaf is read with the stem it sits beside: on the left, 5\ \ 2 \mid 7 stands for 72 and 75.

Sharing the stems puts equal values at equal heights, so the two groups can be compared row by row. The key says how to read both sides.

Worked examples

Example 1

Building a plot with an empty row

Chioma counts the grapes on sixteen bunches from one vine: 72, 85, 47, 68, 91, 76, 63, 88, 79, 74, 66, 83, 71, 95, 78, 69. Make a stem and leaf plot and say what its shape shows.

Solution

Every value has two digits, so the tens digit is the stem and the ones digit the leaf. The smallest value is 47 and the largest 95, so the stems are 4 to 9. Stem 5 gets a row even though no bunch had fifty-something grapes.

Going through the list in order fills row 7 with 2, 6, 9, 4, 1, 8, from the values in the seventies. Sorting that row gives 1, 2, 4, 6, 8, 9, and the other rows are sorted the same way. The rows hold 1 + 0 + 4 + 6 + 3 + 2 = 16 leaves, one per bunch, so none was missed or written twice.

Figure 2The empty row for stem 5 shows the gap between 47 and the rest.

The key is 7 \mid 4 means 74 grapes. Turned sideways, the plot has its longest row at the 70s, with rows getting shorter on both sides. The bunch of 47 stands apart, cut off from the rest by the empty row: it held far fewer grapes than any other bunch.

Example 2

Reading a plot with a decimal key

Dalton records the lengths of his pea pods on the plot below. How many pods are there, how long are the shortest and the longest, what is the median length, and how many pods are at least 8 cm long?

Figure 3Dalton’s pea pods. Here the leaf is the tenths digit.

Solution

The key, 7 \mid 3 means 7.3 cm, says the stem is the whole centimeters and the leaf the tenths. Counting leaves row by row gives 1 + 4 + 5 + 4 + 2 = 16 pods.

The plot is sorted, so the shortest pod is the first leaf of the top row, 5.8 cm, and the longest is the last leaf of the bottom row, 9.5 cm.

Sixteen is even, so the median is halfway between the eighth and ninth values. Rows 5 and 6 hold 1 + 4 = 5 values, so row 7 holds the sixth to the tenth: 7.0, 7.3, 7.3, 7.6, 7.9. The eighth is 7.3 and the ninth is 7.6, so the median is

\frac{7.3 + 7.6}{2} = 7.45 \text{ cm}.

Pods of at least 8 cm are the leaves in rows 8 and 9: 4 + 2 = 6 pods.

Example 3

Building a back-to-back plot

Danae’s club and Demetra’s club each send ten jumpers to a high jump meeting. Their heights, in centimeters:

Danae’s club: 112, 125, 108, 131, 119, 104, 127, 115, 122, 110
Demetra’s club: 128, 136, 119, 141, 124, 133, 130, 138, 126, 145

Draw a back-to-back stem and leaf plot and say what it shows.

Solution

Both groups are in centimeters, so they can share stems. The smallest value of either is 104 and the largest 145, so the shared stems are 10 to 14, and each leaf is the ones digit.

Demetra’s club goes on the right, as in a single plot. Danae’s club goes on the left, from the stem outward. Its values with stem 11 are 110, 112, 115, 119, so row 11 on the left reads 0, 2, 5, 9 moving away from the stem, which is printed 9\ \ 5\ \ 2\ \ 0.

Figure 4On both sides the leaves grow in the direction of the arrow, away from the shared stems.

Each side has ten leaves, one per jumper. Danae’s leaves crowd rows 10 to 12 and Demetra’s crowd rows 12 to 14, so Demetra’s club jumped higher in general. The groups still overlap: Danae’s club’s best jump, 131, beats Demetra’s club’s lowest, 119.

Example 4

Reading a back-to-back plot

Chioma counts grapes on eleven bunches from a sunny slope and eleven from a shady slope. Find the smallest value, the largest value and the median for each slope, and say in plain words what the plot shows.

Figure 5Grapes per bunch on two slopes of one vineyard.

Solution

The shady slope, on the right, reads like any plot: its first row, 4 \mid 1\ \ 8, is 41 and 48. On the left, each leaf goes with the stem beside it and the smallest is nearest the stem, so the row printed 8\ \ 7\ \ 5\ \ 2 beside stem 7 is 72, 75, 77, 78.

Sunny slope, in order: 64, 69, 72, 75, 77, 78, 81, 83, 86, 92, 97. The smallest is 64 and the largest 97. Eleven values have one in the middle, the sixth, so the median is 78 grapes.

Shady slope, in order: 41, 48, 52, 55, 56, 59, 61, 63, 67, 70, 84. The smallest is 41, the largest 84, and the median 59 grapes.

In plain words, bunches on the sunny slope hold more grapes in general: their median is higher, and on the plot the left-hand leaves sit about two rows further down, where the values are larger, than the right-hand ones. Saying exactly how two groups differ is the work of Comparing two distributions.

Example 5

Recovering a smudged leaf

Rain smudged one leaf on Demetra’s plot of twelve jumps, shown as ?. She remembers that the median was 125 cm. What was the smudged leaf?

Figure 6The question mark is the smudged leaf.

Solution

Twelve is even, so the median is halfway between the sixth and seventh values. Rows 10 and 11 hold four values, so row 12 holds the fifth, sixth and seventh: 122, the smudged value, and 128. The smudged value is the sixth.

A median of 125 means the two middle values add to 2 \times 125 = 250. So the sixth value is 250 - 128 = 122, and the smudged leaf is 2.

This fits the plot. The leaves in a row are in order, so the smudged leaf had to be between 2 and 8, and 2 is allowed: two jumpers cleared 122 cm. The middle values are then 122 and 128, and \displaystyle \frac{122 + 128}{2} = 125.

Boundaries

Numbers only. The plot splits values into digits, so it needs numerical data. Colors, names and other categories have no stem or leaf; a bar graph shows them.
One digit per leaf, every value to the same place. If Dalton recorded one pod as 7.25 cm and the rest to tenths, the values must first be written to one place. Rounding 7.25 to fit means the plot no longer keeps that exact value.
Small data sets. Every value is written out, so hundreds of values make rows too long to read. A histogram, which only counts, handles large sets.
A useful number of stems. If every value shares one or two stems, the plot is one or two long rows and shows no shape. If the values spread over dozens of stems, most rows are empty. Either way the leaf’s place does not suit the data.
Back to back means two groups. One column of stems has only two sides. Both groups must use the same unit and the same place for the leaf, or a shared row would mix different things. Three or more groups need another display, such as dot plots or Box plots on one scale.

Where this goes next

A stem and leaf plot is sorted, so anything that depends on order, such as the quartiles behind a box plot, can be read straight off it. A histogram gives up the single values in exchange for handling far larger data sets. Comparing two groups by center and spread, which a back-to-back plot invites, is Comparing two distributions. Picking the display that answers a given question is Choosing a display for a question.

Exercises

Make a plot

Chioma counts the grapes on twelve bunches from a second vine: 54, 61, 47, 58, 66, 52, 49, 63, 58, 70, 55, 61. Make a stem and leaf plot with a key, and say which row is longest.
A younger group’s high jump heights, in centimeters, are 126, 131, 118, 140, 129, 122, 135, 119, 133. Make a stem and leaf plot with a key, and find the median.
Dalton measures ten more pea pods, in centimeters: 8.3, 7.6, 9.1, 8.0, 7.6, 8.7, 6.9, 8.5, 9.4, 8.2. Make a stem and leaf plot with a key.
Chioma counts the bunches on each of ten vines: 23, 31, 28, 52, 35, 24, 37, 30, 26, 33. Make a stem and leaf plot, and explain why it needs a row for stem 4.
Make a stem and leaf plot of the data set 7, 15, 3, 22, 18, 9, 26, 14.

Read a plot

Exercises 6 to 9 use this plot of a junior group’s high jumps.

Figure 7A junior group’s jumps, for Exercises 6 to 9.

How many jumpers are there? What are the lowest and highest jumps?
Find the median jump.
How many jumpers cleared at least 100 cm?
Which height was cleared by two jumpers? Which row is longest, and what does that say about the jumps?

Exercises 10 and 11 use this plot of pea pod lengths.

Figure 8Eight pea pods, for Exercises 10 and 11.

List the values the plot stands for, in order, and find their median.
Suppose the key instead said 6 \mid 4 means 64 grapes. What would the median be, and what about the picture would change?

Back-to-back plots

Chioma counts grapes on eight bunches from an old vine, 82, 75, 91, 68, 87, 79, 94, 85, and eight from a young vine, 56, 63, 71, 59, 66, 62, 74, 68. Draw a back-to-back plot with the old vine on the left.
For the two vines in Exercise 12, find each vine’s median. Which vine’s bunches hold more grapes in general?

Exercises 14 to 16 use this plot of one group’s high jumps in spring and in autumn.

Figure 9One group’s jumps in spring and in autumn, for Exercises 14 to 16.

Write out the spring heights in order from smallest to largest.
Give the lowest and highest jump in each season.
Find the median for each season, and say in one sentence what the plot shows.

Reasoning

Exercise 17 uses this plot of grapes per bunch.

Figure 10One smudged leaf, for Exercise 17.

One leaf is smudged. Which digits could it be? Does the median depend on which one it is?
Explain why a plot keeps a row for a stem with no leaves, and why its leaves are spaced evenly.
A stem and leaf plot and a histogram of the same data, with bins ten wide, have rows and bars of the same lengths. What can the stem and leaf plot tell you that the histogram cannot?
On the left side of a back-to-back plot the leaves grow from right to left. Explain why they are written that way.
Dalton measures one pod as 7.25 cm and the rest to the nearest tenth of a centimeter. Explain why 7.25 cannot go on his plot as it stands, and what he could do.

Answers

Rows: 4 \mid 7\ \ 9; 5 \mid 2\ \ 4\ \ 5\ \ 8\ \ 8; 6 \mid 1\ \ 1\ \ 3\ \ 6; 7 \mid 0. Key: 5 \mid 2 means 52 grapes. The longest row is stem 5, with five leaves: more bunches held fifty-something grapes than any other stretch of ten.
Rows: 11 \mid 8\ \ 9; 12 \mid 2\ \ 6\ \ 9; 13 \mid 1\ \ 3\ \ 5; 14 \mid 0. Key: 12 \mid 6 means 126 cm. Nine values, so the median is the fifth: counting 118, 119, 122, 126, 129 gives a median of 129 cm.
Rows: 6 \mid 9; 7 \mid 6\ \ 6; 8 \mid 0\ \ 2\ \ 3\ \ 5\ \ 7; 9 \mid 1\ \ 4. Key: 8 \mid 3 means 8.3 cm. The pod of 8.0 cm gives the leaf 0, which must be written.
Rows: 2 \mid 3\ \ 4\ \ 6\ \ 8; 3 \mid 0\ \ 1\ \ 3\ \ 5\ \ 7; 4 \mid (empty); 5 \mid 2. Key: 3 \mid 0 means 30 bunches. No vine had forty-something bunches. Dropping row 4 would put 52 straight under the 30s and hide the gap that sets that vine apart.
Rows: 0 \mid 3\ \ 7\ \ 9; 1 \mid 4\ \ 5\ \ 8; 2 \mid 2\ \ 6. Key: 1 \mid 4 means 14. The one-digit values 3, 7 and 9 have stem 0.
1 + 5 + 4 + 2 + 1 = 13 jumpers. The lowest jump is 88 cm and the highest 125 cm.
Thirteen values, so the median is the seventh. Rows 8 and 9 hold six, so the seventh is the first leaf of row 10: the median is 101 cm.
The leaves in rows 10, 11 and 12: 4 + 2 + 1 = 7 jumpers.
95 cm, since row 9 has two leaves of 5. Row 9 is longest, with five leaves: more jumps fell between 90 and 99 cm than in any other stretch of ten.
5.5, 5.9, 6.0, 6.4, 6.4, 6.8, 7.2, 7.7, in centimeters. Eight values, so the median is halfway between the fourth and fifth, 6.4 and 6.4: the median is 6.4 cm.
The values become 55, 59, 60, 64, 64, 68, 72, 77 grapes, and the median 64 grapes. The picture does not change at all; only the key does, which is why a plot needs one.
Each row is written old vine’s leaves, stem, young vine’s leaves, and a side with no leaves is left blank: \mid 5 \mid 6\ \ 9; 8 \mid 6 \mid 2\ \ 3\ \ 6\ \ 8; 9\ \ 5 \mid 7 \mid 1\ \ 4; 7\ \ 5\ \ 2 \mid 8 \mid; 4\ \ 1 \mid 9 \mid. Key: 5 \mid 7 on the left means 75 grapes, and 7 \mid 1 on the right means 71 grapes.
Old vine in order: 68, 75, 79, 82, 85, 87, 91, 94, median \displaystyle \frac{82 + 85}{2} = 83.5 grapes. Young vine: 56, 59, 62, 63, 66, 68, 71, 74, median \displaystyle \frac{63 + 66}{2} = 64.5 grapes. The old vine’s bunches hold more grapes in general.
Reading each left leaf with its stem, from the top row down: 103, 107, 111, 115, 116, 118, 120, 124, 129, 132 cm.
Spring: lowest 103 cm, highest 132 cm. Autumn: lowest 112 cm, highest 140 cm.
Spring has ten values, so its median is \displaystyle \frac{116 + 118}{2} = 117 cm. Autumn has eleven, so its median is the sixth, 128 cm. The group jumped higher in autumn: the median rose by 11 cm and the leaves moved down the plot.
The leaves in a row are in order, so the smudged leaf lies between 1 and 6: it could be any of 1, 2, 3, 4, 5, 6. The smudged value is the fourth of ten, and the median is halfway between the fifth and sixth, 56 and 60, whatever the smudge is. The median is 58 grapes in every case.
The rows stand for equal stretches of ten, like evenly spaced marks on a number line. Leaving out an empty row would make the values on either side of a gap look like neighbors. Even spacing makes a row’s length count its leaves, so longer rows really hold more values.
The histogram says only how many values fall in each bin. The plot keeps every value, so it shows which values are in each row, whether any repeat, and the exact smallest value, largest value and median.
Writing outward from the shared stem makes the two sides mirror images: on each side, the closer a leaf is to the stem, the smaller its value. Each row can then be compared across at a glance.
Every leaf is one digit in the same place. On a plot with tenths as leaves, 7.25 would need two digits after the stem. Dalton could round it to the nearest tenth, 7.3 cm, and accept that the plot no longer keeps its exact value, or measure every pod to the hundredth and make the hundredths digit the leaf.