Data tables, tallies and frequency tables
A month of weather cards, a basket of lunch choices, a class full of answers about pets. How do you turn a pile of separate answers into a few numbers you can read at a glance?
Where you are 8 in Data and variables
When you reach for it
Every school morning Vieno’s class looks out the window and puts one picture on the class weather chart: a sun for a sunny day, a cloud for a cloudy day, a cloud with rain for a rainy day. After a month the chart is full. Ask the class “what was the weather like this month?” and the chart, as it stands, does not quite answer. It is a list of separate days. To answer, the days have to be sorted by kind of weather and each kind counted.
A collection of answers like this, one for each day or one for each child, is called data. When every answer is one of a few kinds (sunny, cloudy or rainy; dog, cat or fish), the kinds are called categories. This lesson turns data like that into a short table of categories and counts, and then reads the table to answer questions. Here the data always fall into at most three categories.
The lesson takes four days.
- Day 1 sorts the data into categories and counts each one.
- Day 2 keeps the count with tally marks, bundled in fives.
- Day 3 writes the counts in a table, and reads a table someone else made.
- Day 4 answers questions from the table: how many in all, how many in one category, and how many more or fewer in one category than in another.
The child already counts objects (Counting to 100 and beyond) and has met sorting objects into groups by one feature. The questions on Day 4 need only adding and subtracting within twenty and adding a one-digit number to a two-digit number.
The procedure
The running problem for all four days is Vieno’s weather chart. It has 12 sunny days, 7 cloudy days and 5 rainy days. How many school days were there in all? How many more sunny days than rainy days?
Day 1: sort, then count
The chart shows the school days in the order they happened, one row for each week. To find how much of each kind of weather there was, take one category at a time. Point to each sun and count: there are 12. Then each shaded cloud: there are 7. Then each rain cloud: there are 5.
With real cards it is easier to move them. Take each card off the chart and put it on one of three piles, then count each pile. Moving the cards keeps the count honest: a card that has been sorted is no longer on the chart, so no card is counted twice and none is skipped. With a printed chart that cannot be moved, crossing off each picture as it is counted does the same job.
Two things make the sorting work.
Every item fits a category. Every square on the chart holds one of the three pictures, so every day lands on some pile.
No item fits two. The class puts up one picture a day, chosen at morning meeting, so a day that starts sunny and turns rainy still has only one picture. If a day could sit on two piles, it would be counted twice.
Because of those two facts, the three counts share out the whole chart: each school day is in exactly one of the counts. Day 4 turns this into a check.
Something is lost in sorting, too. The piles say how many days were rainy, not which days. Once the cards are in piles, nobody can tell from the piles whether the first Monday was sunny. That is the price of a short summary, and it is fine for every question this lesson asks.
Day 2: tally marks
Counting a finished chart is one way. Often the data arrive one at a time: one picture each morning, one answer from each child as the line goes by. A tally mark keeps the count as the data arrive. For each new item, draw one short upright mark in the row for its category.
The first four marks stand side by side. The fifth mark is drawn slanting across the four, and that closes a bundle of five. The sixth mark starts a new group beside the bundle.
The bundle is there to make reading quick. A row of twelve loose marks has to be counted one at a time, and it is easy to lose your place. In bundles, the eye takes in five at once, and the row is read by counting by fives, then by ones. The slanting mark does two jobs at once: it counts the fifth item, and it shows that the bundle is full, which a fifth upright mark could not show.
Here is Vieno’s chart kept as tallies. Sunny is two bundles and two loose marks: “five, ten,” then “eleven, twelve,” so 12. Cloudy is one bundle and two loose marks: “five,” then “six, seven,” so 7. Rainy is exactly one bundle, so 5.
Two bundles of five make ten. So two bundles and two loose marks are the same as 1 ten and 2 ones, which is how 12 is built in Tens and ones. A child who sees the two bundles as one ten reads a long tally faster.
Day 3: the table
Tally marks are good for keeping a count while it grows. A finished count is shorter written as a number. A data table puts each category in its own row with its count beside it.
Weather this month
| Weather | Number of days |
|---|---|
| sunny | 12 |
| cloudy | 7 |
| rainy | 5 |
Each part of the table has a job. The title says what was counted. The heading of the left column says what the categories are. The heading of the right column says what the numbers count: days, not children or cards. Each row holds one category and its count. A table of categories, each beside the number of times it came up, is what later books call a frequency table, and the count in each row is the frequency of that category.
A table can be read in two directions.
- From a category to its count. How many days were cloudy? Find the cloudy row and read across: 7.
- From a count to its category. Which weather came 5 times? Find 5 in the number column and read back across: rainy.
The rows can go in any order; each count stays with its own category, so moving whole rows changes nothing the table says. Putting the rows from most to fewest (ordering groups by how many) makes “which was most?” easy to see. This table is already in that order, because 12 > 7 > 5 (Comparing numbers and the symbols >, < and =). So sunny days came most often and rainy days least often.
Day 4: asking the table questions
How many in one category is a single reading: find the row, read across.
How many in all. Every school day is in exactly one row, so the number of days in all is the three counts put together:
Three numbers can be added in any order, and either pair can go first (Properties of addition): the commutative property lets two numbers trade places, and the associative property lets either pair be added first. Here the first two go first. 2 + 7 = 9, so
Then 19 + 5. One of the 5 fills up the ten, 19 + 1 = 20, and 4 are left, 20 + 4 = 24 (Adding a one-digit number to a two-digit number by making a ten). So
There were 24 school days. The chart has 24 squares, so the count checks: no day was missed and none was counted twice.
How many more. To compare sunny days with rainy days, line up each rainy day beside a sunny day.
Five pairs can be made, and then the rainy days run out. The sunny days with no partner are the extra ones:
There were 7 more sunny days than rainy days. The same subtraction answers the question the other way around: there were 7 fewer rainy days than sunny days. Nothing is taken away in this story, yet the answer is a subtraction, because the unmatched part is what is left when the matched days are set aside. Subtraction as the difference explains this in full, and how many more and how many fewer works through stories of every kind.
Worked examples
Read each problem aloud. If the child has counters, cards or cubes, let them build it.
Day 1: sorting a pet survey
Ahti asks ten children in his class, “Which pet would you most like: a dog, a cat or a fish?” He writes down each answer in the order he hears it:
dog, cat, dog, fish, dog, cat, dog, fish, cat, dog
How many children chose each pet?
SolutionThere are three categories: dog, cat and fish. Go through the list once for each pet, crossing off each answer as it is counted. Dog is the 1st, 3rd, 5th, 7th and 10th answer, so 5 children chose a dog. Cat is the 2nd, 6th and 9th answer, so 3 chose a cat. Fish is the 4th and 8th answer, so 2 chose a fish.
Check: every answer is crossed off once, and the counts put together give back all ten answers, 5 + 3 = 8 and 8 + 2 = 10.
Day 1: a category nobody chose
Ahti asks the same question of nine grown-ups waiting at pickup. Their answers are:
cat, dog, cat, cat, dog, cat, cat, dog, cat
How many chose each pet?
SolutionCat is the 1st, 3rd, 4th, 6th, 7th and 9th answer, so 6 grown-ups chose a cat. Dog is the 2nd, 5th and 8th, so 3 chose a dog. Nobody chose a fish, so the fish count is 0.
The fish category stays, with 0 beside it. The question offered three pets, and “no one chose a fish” is something the data say. Dropping the fish would make it look as if fish had never been a choice. Check: 6 + 3 + 0 = 9, one for each grown-up.
Day 2: tallying lunch choices
Each morning Sylvi’s class picks a lunch: pizza, tacos or salad. On Monday Sylvi makes a tally mark as each child answers. In the end 9 children chose pizza, 11 chose tacos and 4 chose salad. What do the three rows of her tally look like?
SolutionThink of each count as fives and ones.
Pizza: 9 is 5 and 4 more, so the row is one bundle of five and 4 loose marks. Reading it back: “five,” then “six, seven, eight, nine.”
Tacos: 11 is 5, 5 and 1 more, so the row is two bundles and 1 loose mark: “five, ten, eleven.”
Salad: 4 is less than 5, so the fifth mark never came. The row is 4 upright marks with no bundle at all.
Day 2: reading a long tally
At recess Ahti asks many children the pet question and keeps a tally. How many children chose each pet?
Count each row by fives for the bundles, then by ones for the loose marks.
Dog: three bundles, “five, ten, fifteen,” then one loose mark, “sixteen.” So 16 children chose a dog.
Cat: two bundles, “five, ten,” then three loose marks, “eleven, twelve, thirteen.” So 13 chose a cat.
Fish: one bundle, “five,” then four loose marks, “six, seven, eight, nine.” So 9 chose a fish.
Counting by fives works because every bundle holds exactly five marks, four upright and one across.
Day 3: from a tally to a table
Make a table from Ahti’s recess tally. Then use the table to say which pet 13 children chose.
SolutionThe categories go in the left column and the counts read from the tally in the right column. The numbers count children, so the right heading says so.
Pets children at recess would like
| Pet | Number of children |
|---|---|
| dog | 16 |
| cat | 13 |
| fish | 9 |
To find which pet 13 children chose, find 13 in the number column and read back across: cat. The rows are already in order from most to fewest, because 16 > 13 > 9, so dog was chosen most.
Day 3: reading a table
Sylvi made this table on Friday.
Lunch choices on Friday
| Lunch | Number of children |
|---|---|
| tacos | 8 |
| salad | 6 |
| pizza | 13 |
How many children chose tacos? Which lunch did 6 children choose? Put the three lunches in order from most to fewest.
SolutionFind the tacos row and read across: 8 children chose tacos.
Find 6 in the number column and read back across: salad.
The largest count is 13, then 8, then 6, since 13 > 8 > 6. So the order from most to fewest is pizza, tacos, salad. Sylvi’s rows were not in this order, and that is fine: the table says the same thing either way.
Day 4: in all, and how many more
Use Sylvi’s Friday table. How many children chose a lunch on Friday? How many more children chose pizza than tacos?
SolutionEvery child chose exactly one lunch, so the number of children is the three counts added. Any order gives the same total, so pick an easy one: put 13 and 6 together first.
Then 19 + 8: one of the 8 fills the ten, 19 + 1 = 20, and 7 are left, 20 + 7 = 27.
So 27 children chose a lunch on Friday.
For “how many more”, match each child who chose tacos with a child who chose pizza. After 8 pairs, 13 - 8 pizza children have no partner. Take the 8 away in two steps, first to ten: 13 - 3 = 10, then 10 - 5 = 5.
So 5 more children chose pizza than tacos. Check by adding: 8 + 5 = 13.
Day 4: a missing count
Sylvi asks all 20 children in her class the pet question, and every child picks exactly one pet. Her table shows 9 for dog and 7 for cat, but the fish count got smudged. How many children chose a fish? How many fewer chose a fish than a dog?
SolutionEvery child is in exactly one row, so the three counts must add up to 20:
First 9 + 7: one of the 7 fills the ten, 9 + 1 = 10, and 6 are left, so 9 + 7 = 16. Now 16 + \square = 20. Count up from 16: seventeen, eighteen, nineteen, twenty. That is four steps, so \square = 4.
4 children chose a fish. Check: 9 + 7 + 4 = 20.
For “how many fewer”, match each fish child with a dog child. After 4 pairs, the dog children with no partner are
So 5 fewer children chose a fish than a dog.
Boundaries
- Each item in exactly one category. A table of counts only makes sense when the categories cover every item and no item fits two. If a day could be both sunny and rainy, the class needs a rule (the weather at morning meeting) or the counts will not add up to the number of days.
- The counts add to the whole only for that reason. “How many in all” is the sum of the rows because each item sits in one row. If one child were allowed to choose two pets, the counts would add to more than the number of children.
- A table says how many, not which. The weather table cannot tell which days were rainy, or in what order the weather came. That information was in the chart and was set aside in sorting.
- A count is a whole number, and it can be 0. A category nobody chose stays in the table with 0.
- “How many more” compares two rows of the same table. Both counts must count the same kind of thing; days are compared with days and children with children.
- Three categories here. The same sorting, tallying and table work for four or more categories; this lesson keeps to three.
Where this goes next
The next step is to draw the table. A picture graph shows each count as a row of pictures, one picture per item, and a bar graph shows it as a bar whose length is the count. Later, a picture can stand for several items at once, as in Scaled picture graphs.
Collecting the data on purpose, by asking everyone the same question and writing down every answer, is collecting data with a survey. Data that come as kinds, as here, are set beside data that come as amounts in types of data. Much later, each count is written as a share of the whole in Relative frequency tables.
Exercises
Day 1: sort, then count
- Vieno writes down the weather for two weeks of school days: sunny, rainy, sunny, cloudy, sunny, sunny, rainy, cloudy, sunny, sunny. How many days of each kind were there?
- Sylvi has these lunch slips: tacos, salad, salad, pizza, tacos, salad, salad, pizza, salad, tacos, salad. How many slips are there for each lunch?
- Look at the top row of Vieno’s weather chart (the first week). How many of those five days were sunny? How many were rainy?
- Ahti asks seven grown-ups the pet question. They answer dog, cat, cat, dog, cat, cat, dog. How many chose each pet?
- Sylvi had 12 lunch slips. After sorting, her three piles hold 5, 4 and 2 slips. Was every slip sorted?
Day 2: tally marks
- Describe the tally for 8.
- Describe the tally for 14.
- A row of tally marks has three bundles and two loose marks. What number is that?
- A row of tally marks has four bundles and no loose marks. What number is that?
- Vieno’s sunny row has one bundle and four loose marks. The next day is sunny. What does she draw, and what does the row show now?
Day 3: the table
Use this table for exercises 11 to 14.
Lunch choices on Tuesday
| Lunch | Number of children |
|---|---|
| pizza | 10 |
| tacos | 4 |
| salad | 9 |
- How many children chose salad?
- Which lunch did 4 children choose?
- Which lunch was chosen most?
- Write the three lunches in order from most to fewest.
- Ahti’s tally shows dog: two bundles and no loose marks; cat: one bundle and one loose mark; fish: three loose marks. Make a table from it.
Day 4: asking the table questions
Use Vieno’s weather table (12 sunny, 7 cloudy, 5 rainy) for exercises 16 to 18.
- How many more cloudy days were there than rainy days?
- How many days were not sunny?
- How many fewer cloudy days were there than sunny days?
Use the Tuesday lunch table for exercises 19 to 21.
- How many children chose a lunch on Tuesday?
- How many more children chose pizza than tacos?
- How many fewer children chose tacos than salad?
- Vieno keeps the weather for 15 school days. There were 6 sunny days and 5 rainy days, and the rest were cloudy. How many days were cloudy?
Explain
- Why is the fifth tally mark drawn across the other four instead of standing up beside them?
- Sylvi’s table lists pizza, tacos, salad. Ahti copies the same counts but lists salad, pizza, tacos. Do the two tables say the same thing? Why?
- Sylvi sorts all the lunch slips into three piles and adds the three counts. Why must the total be the number of slips? What would make it come out different?
Answers
- Sunny 6, rainy 2, cloudy 2. Sunny is the 1st, 3rd, 5th, 6th, 9th and 10th day; rainy the 2nd and 7th; cloudy the 4th and 8th. Check: 6 + 2 + 2 = 10 days.
- Salad 6, tacos 3, pizza 2. Salad is the 2nd, 3rd, 6th, 7th, 9th and 11th slip; tacos the 1st, 5th and 10th; pizza the 4th and 8th. Check: 6 + 3 + 2 = 11 slips.
- 3 sunny and 1 rainy. The first week reads sun, sun, cloud, rain, sun, so 3 suns and 1 rain cloud (and 1 cloudy day).
- Dog 3, cat 4, fish 0. Dog is the 1st, 4th and 7th answer; cat the other four; nobody chose a fish, and the fish row stays with 0. Check: 3 + 4 + 0 = 7.
- No. 5 + 4 = 9 and 9 + 2 = 11, so the piles hold 11 slips, one fewer than 12. One slip was not sorted, or was lost.
- One bundle of five and 3 loose marks, because 8 is 5 and 3 more: “five, six, seven, eight.”
- Two bundles of five and 4 loose marks, because 14 is 5, 5 and 4 more: “five, ten, eleven, twelve, thirteen, fourteen.”
- 17. “Five, ten, fifteen,” then “sixteen, seventeen.”
- 20. “Five, ten, fifteen, twenty.”
- She draws the tenth mark slanting across the four loose marks, which closes a second bundle. The row now shows two bundles and no loose marks, which is 10.
- 9. Find the salad row and read across.
- Tacos. Find 4 in the number column and read back across.
- Pizza, with 10, since 10 > 9 and 10 > 4.
- Pizza, salad, tacos, because 10 > 9 > 4.
- A table with a title such as Pets Ahti asked about, a Pet column and a Number of children column: dog 10 (two bundles, “five, ten”), cat 6 (one bundle and one mark), fish 3.
- 2 more. Match the 5 rainy days with 5 cloudy days; 7 - 5 = 2 cloudy days have no partner.
- 12 days. The days that were not sunny are the cloudy and rainy days: 7 + 5 = 12.
- 5 fewer. Match the 7 cloudy days with 7 sunny days; 12 - 7 = 5 sunny days are left over, so there were 5 fewer cloudy days.
- 23 children. 10 + 9 = 19, then 19 + 1 = 20 and 20 + 3 = 23, so 10 + 9 + 4 = 23.
- 6 more. 10 - 4 = 6. Check: 4 + 6 = 10.
- 5 fewer. 9 - 4 = 5. Check: 4 + 5 = 9.
- 4 cloudy days. 6 + 5 = 11, so 11 + \square = 15, and counting up from 11 to 15 is four steps: \square = 4. Check: 6 + 5 + 4 = 15.
- A fifth upright mark would look like the others, and the reader would have to count every mark. The slanting mark counts the fifth item and also shows that the group is full, so each bundle is seen at once as five and the row can be read by counting by fives.
- Yes. Each count stays beside its own lunch, so every lunch still has the same number. Only the order of the rows changed, and the order of the rows is not part of what the table says.
- Every slip goes on exactly one pile, so adding the piles counts each slip once. The total would come out different if a slip were left out of every pile (too small), or if a slip could be put on two piles or counted twice (too large).