Tally and Frequency Tables

Count in gates of five; the table keeps score.

Counting as the data arrive

A team writes down how many goals it scored in each of its 14 matches: 1, 0, 2, 1, 1, 0, 1, 2, 0, 1, 1, 2, 0, 1. The list is hard to read as it stands. A tally chart sorts it.

Make one row for each value, 0, 1 and 2. Then go through the list once, in order, and for each number draw one stroke, called a tally mark, in its row. Tick each number off the list as it is tallied, so that none is missed and none is counted twice.

Tally marks are grouped in fives. The first four strokes stand upright, and the fifth is drawn across them, which closes the group like a gate. A row of tally marks can then be counted in fives at a glance.

goalstallyfrequency041723

The row for 1 goal holds a gate of five and two more strokes: 7 matches with 1 goal.

The frequency column

The number of times a value occurs is its frequency. The last column of the table records it: count each row of tally marks and write the number. The row for 1 goal has a gate of five and two more strokes, so its frequency is 5 + 2 = 7. The row for 0 goals has four strokes, a frequency of 4, and the row for 2 goals has a frequency of 3.

Every match was tallied exactly once, so the frequencies add up to the number of matches: 4 + 7 + 3 = 14. This is a check on the tally. A total that differs from the length of the list means a value was missed or tallied twice.

goalstallyfrequency041723total14

The frequencies 4, 7 and 3 add up to 14, one for each match.

Grouping values into classes

Sixteen students take a test on which a score can be any whole number from 0 to 29. Their scores are 12, 25, 7, 18, 14, 21, 3, 16, 19, 27, 11, 9, 15, 23, 13 and 20. A row for every possible score would need 30 rows, most of them holding one tally mark or none, and the table would show no pattern.

Instead, group the scores into classes: 0–9, 10–19 and 20–29. Each row now counts a whole interval of scores. Tally the list into the three classes as before, and the frequencies are 3, 8 and 5, which add up to 16.

The classes must not overlap, and together they must cover every possible score, so that each score belongs to exactly one class: 19 goes in 10–19, and 20 goes in 20–29. The table is now easier to read, but it has lost detail. It says that 5 students scored from 20 to 29, but no longer which scores they were.

scorefrequency0–9310–19820–295

Grouped into classes of ten, the 16 scores have frequencies 3, 8 and 5.

From a frequency table to a total

A frequency table also gives the total of all the values. In the goals table, 7 matches had 1 goal each, which is 1 × 7 = 7 goals, and 3 matches had 2 goals each, which is 2 × 3 = 6 goals. Multiply each value by its frequency and add: 0 × 4 + 1 × 7 + 2 × 3 = 0 + 7 + 6 = 13 goals in the 14 matches. The mean is 13 ÷ 14 ≈ 0.93 goals a match.

This works backward as well. The mean is the total divided by the number of values, so the total is the mean times the number of values. A team with a mean of 1.5 goals over 10 matches scored 1.5 × 10 = 15 goals.

So when a frequency is missing from a table, two facts can find it: the frequencies add up to the number of values, and the values times their frequencies add up to the total.

Worked example: Frequency Distribution Table with Hidden Frequencies

Question A teacher surveyed 40 students on the number of books they read over the holidays. The table recorded: 4 students read 0 books; 12 students read 1 book; k students read 2 books; 8 students read 3 books; and m students read 4 books. The average number of books read across the 40 students was 2.05. (a) How many students read 2 books? (b) How many students read 4 books?

  1. 1.Known students = 4 + 12 + 8 = 24. Leftover students for 2 and 4 books = 40 − 24 = 16 students.

    0 books41 book122 booksk3 books84 booksmk + m = 16
    0 books41 book122 booksk3 books84 booksmk + m = 16
    24 students are known; k + m = 16.
  2. 2.Total books = 40 × 2.05 = 82. Known books = 12 + 24 = 36. Leftover books = 82 − 36 = 46.

    0 books41 book122 booksk3 books84 booksmk + m = 16Booksknown 3682 books
    0 books41 book122 booksk3 books84 booksmk + m = 16Booksknown 3682 books
    82 books in all; 36 are known, so the 16 read 46.
  3. 3.Assume all 16 students read 2 books: Total = 16 × 2 = 32 books.

    0 books41 book122 booksk3 books84 booksmk + m = 16Booksknown 3646 for the 1682 booksIf all 216 × 2 = 32
    0 books41 book122 booksk3 books84 booksmk + m = 16Booksknown 3646 for the 1682 booksIf all 216 × 2 = 32
    Suppose the 16 all read 2: 32 books.
  4. 4.Shortfall: 46 − 32 = 14 books.

    0 books41 book122 booksk3 books84 booksmk + m = 16Booksknown 3646 for the 1682 booksIf all 216 × 2 = 3214 short
    0 books41 book122 booksk3 books84 booksmk + m = 16Booksknown 3646 for the 1682 booksIf all 216 × 2 = 3214 short
    14 short; each 4-book reader adds 2 more than a 2-book reader.
  5. 5.Difference between 4 books and 2 books: 4 − 2 = 2 books per student.

  6. 6.Number of students who read 4 books (m): 14 ÷ 2 = 7 students.

    0 books41 book122 booksk3 books84 booksmk + m = 16Booksknown 3646 for the 1682 booksIf all 216 × 2 = 3214 short14 ÷ 2 = 7 read 4
    0 books41 book122 booksk3 books84 booksmk + m = 16Booksknown 3646 for the 1682 booksIf all 216 × 2 = 3214 short14 ÷ 2 = 7 read 4
    (b) m = 7.
  7. 7.(b) m = 7.

    0 books41 book122 booksk3 books84 books7k + m = 16Booksknown 3646 for the 1682 booksIf all 216 × 2 = 3214 short14 ÷ 2 = 7 read 4
    0 books41 book122 booksk3 books84 books7k + m = 16Booksknown 3646 for the 1682 booksIf all 216 × 2 = 3214 short14 ÷ 2 = 7 read 4
    (b) m = 7.
  8. 8.Number of students who read 2 books (k): 16 − 7 = 9 students.

    0 books41 book122 booksk3 books84 books7k + m = 16Booksknown 3646 for the 1682 booksIf all 216 × 2 = 3214 short14 ÷ 2 = 7 read 4
    0 books41 book122 booksk3 books84 books7k + m = 16Booksknown 3646 for the 1682 booksIf all 216 × 2 = 3214 short14 ÷ 2 = 7 read 4
    (a) k = 16 − 7 = 9.
  9. 9.(a) k = 9.

    0 books41 book122 books93 books84 books7k + m = 16Booksknown 3646 for the 1682 booksIf all 216 × 2 = 3214 short14 ÷ 2 = 7 read 4
    0 books41 book122 books93 books84 books7k + m = 16Booksknown 3646 for the 1682 booksIf all 216 × 2 = 3214 short14 ÷ 2 = 7 read 4
    (a) k = 16 − 7 = 9.

Answer: (a) 9 students; (b) 7 students

Common mistakes

  • Forgetting to multiply the books by their frequency (adding 2 + 4 = 6 books instead of 2k + 4m).
  • Including the 0-book students in the books sum as 4 books instead of 0.

More averages and charts problems, worked step by step →

Practice Tally and Frequency Tables in the app