Splitting a total into groups
A school records how 100 children came to school one morning: 60 came by bus and 40 walked. A frequency tree shows this as a split. Write the total, 100, on the left. Draw one branch for each group, name the group along the branch, and write its count at the end.
Every child came one way or the other, so the two groups share out the total with nothing left over: 60 + 40 = 100.
The 100 children split into two branches: 60 came by bus and 40 walked.
Splitting each group again
A second question splits each group again: was the child late or on time? It is asked of the bus children and of the walkers separately, so each group gets its own pair of branches.
Of the 60 children who came by bus, 12 were late and 48 were on time. Of the 40 who walked, 6 were late and 34 were on time. Each number at the end of a branch counts the children described by both labels on the path to it: the 12 at the top are the children who came by bus and were late.
The path from 100 to bus to late ends at 12: twelve children came by bus and were late.
Every fork adds back
At every fork, the branches share out the number they started from: 12 + 48 = 60 for the bus children, and 6 + 34 = 40 for the walkers. The four numbers at the ends add up to the total at the start, 12 + 48 + 6 + 34 = 100, because each child ends at exactly one of them.
To answer a question about lateness alone, add the ends that match it from both groups. The late children are 12 + 6 = 18, and the children on time are 48 + 34 = 82. Check: 18 + 82 = 100.
Recovering a missing number
Because every fork adds back, a single missing number can be found by subtraction. Suppose the number of bus children who were on time is covered. The bus branch holds 60 and 12 of them were late, so 60 − 12 = 48 were on time.
A missing number further back is found by adding. If the 60 were covered, the two branches that split from it, 12 and 48, give 12 + 48 = 60.
The bus branch holds 60 and 12 of them were late, so the covered number is 60 − 12 = 48.
Following one branch
A frequency tree lets you compare the groups. Of the 60 bus children, 12 were late: %. Of the 40 walkers, 6 were late: %. So a child who came by bus was more likely to be late than a child who walked.
Each of those fractions follows one branch and ignores the other. Turn the question round and the whole changes. What fraction of the late children came by bus? The late children are gathered from two branches, 12 + 6 = 18, and 12 of them came by bus: .
So and describe the same 12 children against two different wholes. Read the question for the group it starts from, and make that group the denominator. Looking ahead: in probability, a chance worked out within one group like this is called a conditional probability.
The same counts in a two-way table
The four ends of the tree are the four cells of a two-way table, with a row for each way of coming to school and a column for each answer about lateness. The tree shows the order of the splits, first by travel and then by lateness. The table shows both splits at once. Its row totals, 60 and 40, are the tree’s first pair of branches, and its column totals, 18 and 82, are the late and on-time counts found above by adding ends from both groups.
The ends of the tree fill the cells: 12, 48, 6 and 34. The totals are 60 and 40 across, 18 and 82 down, and 100 in all.
Worked example: A Frequency Tree for a Screening Test, and the Positive Results That Are Correct
Question 1000 people take a screening test for a condition. 40 of them have the condition, and 36 of those 40 test positive. Of the people who do not have the condition, 48 test positive. (a) Complete a frequency tree and find how many people test negative. (b) A person tests positive. What fraction of the people who test positive have the condition?
1.The first pair of branches splits the 1000 people into those who have the condition and those who do not: 1000 − 40 = 960 people do not have it.
The first pair of branches adds up to 1000, so 1000 − 40 = 960 people do not have the condition. 2.The second pairs split each group by the test result. Of the 40 people with the condition, 40 − 36 = 4 test negative. Of the 960 people without it, 960 − 48 = 912 test negative.
Each second pair adds up to the count before it: 40 − 36 = 4 and 960 − 48 = 912. 3.(a) 4 + 912 = 916 people test negative. Check: 36 + 48 = 84 people test positive, and 84 + 916 = 1000.
(a) 4 + 912 = 916 people test negative. 4.The people who test positive come from two branches, 36 who have the condition and 48 who do not, which is 84 people in all.
The people who test positive are on two branches: 36 + 48 = 84. 5.(b) 3684 = 37 of the people who test positive have the condition. That is less than half, because the 48 positive results come from the very large group of people who do not have it.
(b) 3684 = 37 of the people who test positive have the condition.
Answer: (a) 916 people test negative; (b) 37 of the people who test positive have the condition
Common mistakes
- Answering part (b) with 3640. That is the fraction of the people with the condition who test positive. The question starts from a positive result, so the whole is the 84 people who test positive.
- Adding only the two negative counts that are easy to see and forgetting to check. The four end branches must add up to the 1000 people at the start: 36 + 4 + 48 + 912 = 1000. A tree whose ends do not add up has a wrong subtraction in it.