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Math 8 · 8.SP.A.4 · S.CP.A.4

Conditional relative frequencies

Of the museum visitors who took a guided tour, what share bought something at the shop? And is that the same number as the share of shoppers who took a tour?

Where you are 3 in Two-way tables

When you reach for it

A two-way table counts one group of people or things by two questions at once. A made-up museum asked 80 visitors on their way out whether they took a guided tour and whether they bought something at the shop.

Table 1bought somethingbought nothingTotal
took a tour24832
no tour64248
Total305080

The four inner counts are the joint frequencies and the totals round the edge are the marginal frequencies (Joint and marginal frequencies). Now ask: “of the visitors who took a tour, what share bought something?” The answer is not 24 out of 80. The question’s first words narrow the whole down to the 32 tour visitors, and the answer is a share of them.

You reach for a conditional relative frequency whenever a question names a group and asks what share of that group has some other property, and whenever groups of different sizes have to be compared.

The procedure

Part 1: a cell over its row total or its column total

A conditional relative frequency is the count in one cell of a two-way table divided by the total of the row, or of the column, that the question picks out. That row or column is the condition: the group the question has narrowed down to.

\text{share} = \frac{\text{count in the cell}}{\text{total of the group}}

This is the same part over whole as any share (Finding what percent one number is of another); what is new is only that the whole is a row or column total rather than the grand total. For the tour visitors, the group is the “took a tour” row, total 32, and those who also bought something are the cell 24:

\frac{24}{32} = \frac{3}{4} = 0.75

So 0.75, or 75\%, of the tour visitors bought something.

24
8
bought something bought nothing 32 on a tour
\displaystyle \frac{24}{32} = \frac{3}{4} = 0.75
Figure 1The words “of the visitors who took a tour” make the 32 tour visitors the whole. The share who bought something is 24 out of that 32.

Which total: read it from the wording. A conditional question always names its group, and the group’s total is what you divide by. Words such as “of”, “among”, “out of” and “what share of the” usually come just before the group.

  • “Of the visitors who took a tour, what share bought something?” The group is the tour visitors, so divide by 32.
  • “What share of the visitors who bought something took a tour?” The group is the buyers, so divide by 30.
  • “Of those who play chess, what share are in the choir?” divides by the chess players, while “of those in the choir, what share play chess?” divides by the choir.

The cell is always the members of the group who also have the property asked about. When a question names no group, as in “what share of all the visitors took a tour and bought something?”, the whole is the grand total: \displaystyle \frac{24}{80} = 0.3. That is a share of the whole table, not a conditional one.

Shares may be written as fractions, decimals or percents. When a share does not come out exactly, this lesson rounds it to three decimal places and writes \approx.

Row-relative tables. Dividing every cell by its own row total gives a row-relative table, in which each row describes one group on its own scale. The tour row is \displaystyle \frac{24}{32} = 0.75 and \displaystyle \frac{8}{32} = 0.25; the no-tour row is \displaystyle \frac{6}{48} = 0.125 and \displaystyle \frac{42}{48} = 0.875; the Total row is \displaystyle \frac{30}{80} = 0.375 and \displaystyle \frac{50}{80} = 0.625.

bought somethingbought nothingTotal
took a tour0.750.251
no tour0.1250.8751
Total0.3750.6251

Every row adds to 1, because the cells of a row add up to the row total. If a row holds counts a and b with total r = a + b, then

\frac{a}{r} + \frac{b}{r} = \frac{a + b}{r} = \frac{r}{r} = 1

and the same works with three or more cells. In words: every tour visitor either bought something or did not. The columns of a row-relative table need not add to 1: 0.75 + 0.125 adds shares of two different groups.

Column-relative tables. Dividing every cell by its column total instead makes each column a group: the buyers and the non-buyers. The buyers’ column is \displaystyle \frac{24}{30} = 0.8 and \displaystyle \frac{6}{30} = 0.2; the non-buyers’ column is \displaystyle \frac{8}{50} = 0.16 and \displaystyle \frac{42}{50} = 0.84; the Total column is \displaystyle \frac{32}{80} = 0.4 and \displaystyle \frac{48}{80} = 0.6.

bought somethingbought nothingTotal
took a tour0.80.160.4
no tour0.20.840.6
Total111

Now each column adds to 1, for the same reason. The entry 0.8 says that of the visitors who bought something, 80\% took a tour.

Part 2: two directions, fair comparisons, and working back

The two directions give different numbers. The cell “took a tour and bought something” holds 24 visitors. Over its row total it gives the share of tour visitors who bought something, \displaystyle \frac{24}{32} = 0.75. Over its column total it gives the share of buyers who took a tour, \displaystyle \frac{24}{30} = 0.8. The same 24 people, two different wholes, and two answers to two different questions.

24
8
24
6
toured and bought 32 on a tour 30 who bought
\displaystyle \frac{24}{32} = 0.75
\displaystyle \frac{24}{30} = 0.8
Figure 2The same 24 visitors, measured against two different wholes, give two different shares.

The two agree only when the row total equals the column total, and they can be far apart. If a museum had only 4 tour visitors, all of whom bought something, and 100 buyers in all, then every tour visitor bought something, \displaystyle \frac{4}{4} = 1, while only \displaystyle \frac{4}{100} = 0.04 of the buyers took a tour. So a sentence has to keep its group. “75\% of tour visitors bought something” and “75\% of buyers took a tour” are different claims, and for Table 1 only the first is true.

Comparing groups of different sizes. A count mixes how common something is inside a group with how big the group is. In Table 1, 24 tour visitors and 6 others bought something, but out of wholes of 32 and 48. The shares 0.75 and 0.125 each answer “what share of this group bought something?”, so they can be set side by side fairly. The groups being compared decide the direction: to compare tour visitors with the others, divide by their row totals. Whether shares that differ from group to group mean the two questions are linked is the subject of Association in a two way table and segmented bar graphs.

Working back from shares. Since a share is the cell divided by the group’s total, multiplying both sides by that total gives the cell back:

\text{cell} = \text{share} \times \text{group total}

For instance 0.75 \times 32 = 24. So from the shares and the group totals the counts can be filled in, and from the counts every other share follows, including the other direction. The totals are needed: doubling every count leaves every share the same, so shares alone fix the counts only up to scale.

Worked examples

Examples 1 to 4 go with Part 1, and Examples 5 to 7 with Part 2.

Example 1

Single shares in a park count

A made-up count in a park recorded each cyclist who passed, by age group and by whether they wore a cycle helmet.

helmetno helmetTotal
child (under 12)451560
teen (12 to 17)182240
adult (18 and over)6238100
Total12575200

(a) Of the children counted, what share wore a helmet? (b) Among the adults, what share wore no helmet?

Solution

(a) The group is the children, so the whole is the child row total, 60. The children who wore a helmet are the cell 45:

\frac{45}{60} = \frac{3}{4} = 0.75

So 75\% of the children wore a helmet.

(b) “Among the adults” makes the adult row, total 100, the whole. The adults with no helmet are the cell 38, and \displaystyle \frac{38}{100} = 0.38, so 38\% of the adults wore no helmet. As a check, \displaystyle \frac{62}{100} = 0.62 of the adults did wear one, and 0.62 + 0.38 = 1.

Example 2

A row-relative table for a cinema

A made-up cinema sorted one evening’s 300 tickets by ticket type and film genre.

childadultTotal
animation5644100
comedy1696112
adventure88088
Total80220300

Make the row-relative table, and say what its comedy row tells you.

Solution

Each cell is divided by its own row total. For animation, \displaystyle \frac{56}{100} = 0.56 and \displaystyle \frac{44}{100} = 0.44. For comedy, \displaystyle \frac{16}{112} \approx 0.143 and \displaystyle \frac{96}{112} \approx 0.857. For adventure, \displaystyle \frac{8}{88} \approx 0.091 and \displaystyle \frac{80}{88} \approx 0.909. For the Total row, \displaystyle \frac{80}{300} \approx 0.267 and \displaystyle \frac{220}{300} \approx 0.733.

childadultTotal
animation0.560.441
comedy0.1430.8571
adventure0.0910.9091
Total0.2670.7331

Every row adds to 1, for instance 0.143 + 0.857 = 1. The comedy row describes the comedy tickets on their own: about 14.3\% of them were child tickets and about 85.7\% adult tickets.

Example 3

A column-relative table for the park count

Make the column-relative table for the park count in Example 1, and say in words what the entry for teens in the helmet column means.

Solution

Now each column is the group: the helmet wearers (total 125), the cyclists with no helmet (total 75), and everyone (total 200). In the helmet column, \displaystyle \frac{45}{125} = 0.36, \displaystyle \frac{18}{125} = 0.144 and \displaystyle \frac{62}{125} = 0.496. In the no helmet column, \displaystyle \frac{15}{75} = 0.2, \displaystyle \frac{22}{75} \approx 0.293 and \displaystyle \frac{38}{75} \approx 0.507. In the Total column, \displaystyle \frac{60}{200} = 0.3, \displaystyle \frac{40}{200} = 0.2 and \displaystyle \frac{100}{200} = 0.5.

helmetno helmetTotal
child0.360.20.3
teen0.1440.2930.2
adult0.4960.5070.5
Total111

The columns check: 0.36 + 0.144 + 0.496 = 1, 0.2 + 0.293 + 0.507 = 1 and 0.3 + 0.2 + 0.5 = 1.

The teen entry in the helmet column, 0.144, says that of the cyclists wearing a helmet, 14.4\% were teens. The share of teens who wore a helmet is a different question: it divides the same cell by the teen row, \displaystyle \frac{18}{40} = 0.45.

Example 4

Choosing the total from the wording

For each question about the cinema in Example 2, name the total to divide by, then find the share. (a) Among the child tickets, what share were for adventure? (b) What share of the adult tickets were for comedy? (c) What share of all the tickets were adult tickets for a comedy? (d) Giedre says, “20\% of the adult tickets were for animation.” Is she right?

Solution

(a) “Among the child tickets” names the child column, total 80. The cell is the 8 child adventure tickets, and \displaystyle \frac{8}{80} = 0.1.

(b) “Of the adult tickets” names the adult column, total 220. The cell is the 96 adult comedy tickets, and \displaystyle \frac{96}{220} = \frac{24}{55} \approx 0.436.

(c) “Of all the tickets” narrows nothing, so the whole is the grand total, 300. The cell is the same 96, and \displaystyle \frac{96}{300} = 0.32: a share of the whole table, not a conditional one, which is why it is smaller than the answer to (b).

(d) Her group is the adult column, total 220. The cell is the 44 adult animation tickets, and \displaystyle \frac{44}{220} = 0.2 = 20\%, so she is right. The same 44 over the animation row total, \displaystyle \frac{44}{100} = 0.44, would be the share of animation tickets that were adult tickets, a different claim.

Example 5

Two directions on the cinema table

Use the cinema table in Example 2. (a) Compare the share of child tickets that were for animation with the share of animation tickets that were child tickets. (b) Do the same for adult tickets and adventure.

Solution

(a) Child tickets that were for animation: the group is the child column, total 80, so the share is \displaystyle \frac{56}{80} = 0.7. Animation tickets that were child tickets: the group is the animation row, total 100, so the share is \displaystyle \frac{56}{100} = 0.56. The same 56 tickets over wholes of 80 and 100; the first share is larger because its whole is smaller.

(b) Adult tickets that were for adventure: the adult column, \displaystyle \frac{80}{220} = \frac{4}{11} \approx 0.364. Adventure tickets that were adult tickets: the adventure row, \displaystyle \frac{80}{88} = \frac{10}{11} \approx 0.909. Only about a third of the adult tickets were for adventure, yet nearly every adventure ticket was an adult’s, because the 88 adventure tickets are a much smaller whole than the 220 adult tickets. “Most adventure tickets were adult tickets” is true; “most adult tickets were for adventure” is false.

Example 6

Comparing groups of different sizes

Jareth reads the park count in Example 1 and says, “Adults are the most likely to wear a helmet: 62 adults wore one, more than any other age group.” Use conditional shares to compare the age groups fairly.

Solution

The count 62 is the largest, but 100 adults were counted against 60 children and 40 teens, so more adults could wear helmets simply because more adults passed by. “How likely is a member of this group to wear a helmet?” makes each age group its own whole, so each helmet count is divided by its row total:

\begin{gathered} \text{children: } \frac{45}{60} = 0.75 \\ \text{teens: } \frac{18}{40} = 0.45 \\ \text{adults: } \frac{62}{100} = 0.62 \end{gathered}

Compared fairly, the children had the largest share wearing a helmet, then the adults, then the teens. Jareth’s count is right, but his claim does not follow from it. The helmet column shares of Example 3 (0.36 for children, 0.496 for adults) say how the helmet wearers split among the age groups; they still carry the group sizes and do not answer his question.

Example 7

Filling a table back from shares

One afternoon a made-up cinema sold 250 tickets, and 40\% of them were child tickets. Of the child tickets, \displaystyle \frac{3}{4} were for animation; of the adult tickets, 20\% were for animation. Every ticket not for animation counts as “another genre”. What share of the animation tickets were child tickets?

Solution

The totals come first: 0.4 \times 250 = 100 child tickets, so 250 - 100 = 150 adult tickets. Each cell is a share times its group’s total. The child animation tickets are \displaystyle \frac{3}{4} \times 100 = 75, leaving 25 child tickets for another genre. The adult animation tickets are 0.2 \times 150 = 30, leaving 120.

animationanother genreTotal
child7525100
adult30120150
Total105145250

The columns add to 105 + 145 = 250, which checks the table. The question names the animation tickets as its group, total 105:

\frac{75}{105} = \frac{5}{7} \approx 0.714

The shares given were of the rows and the question asked about a column; the counts made that turn possible.

Boundaries

  • The group must have members. A row or column whose total is 0 has no shares at all: “of the visitors who arrived by boat, what share bought something?” has no answer when nobody did; the answer is not 0.
  • The cell must sit inside the group. Dividing a cell by the total of a row or column it is not in gives a number with no meaning.
  • A row of a row-relative table adds to exactly 1 because each member of the group falls in exactly one cell of that row. If a survey let one person be counted in two columns, the shares of a row could add to more than 1. Rounded shares may add to 0.999 or 1.001; the exact fractions always add to 1.
  • Shares alone do not give counts, and a row-relative table alone does not give column shares. Both need the group totals.

Where this goes next

The next step sets conditional shares side by side across a whole table. If the share of children wearing a helmet is very different from the share of adults, age group and helmet look connected; if the shares are about the same from row to row, they do not. That is Association in a two way table and segmented bar graphs, where each row of a row-relative table becomes one bar of a segmented bar graph.

Later the same division becomes Conditional probability: the chance of one event given that another happened. The two directions differ there for the same reason they differ here: one cell measured against two different wholes.

Exercises

Part 1: one share at a time

Table 2 is a made-up survey of 120 visitors at a second museum.

Table 2bought somethingbought nothingTotal
took a tour28735
no tour127385
Total4080120
  1. Table 2: Of the visitors who took a tour, what share bought something?
  2. Table 2: Of the visitors who bought nothing, what share took no tour?

Part 1: row and column tables

Table 3 is a made-up cinema’s Saturday matinee, 150 tickets in all.

Table 3animationmusicalTotal
child362460
adult276390
Total6387150
  1. Make the row-relative table for Table 3, and check that each row adds to 1.
  2. Example 1 found two shares from the park count. Complete the row-relative table for that count: the child, teen, adult and Total rows.

Part 1: which total?

For each question, name the total you divide by, then find the share.

  1. Table 3: Among the musical tickets, what share were child tickets?
  2. Table 3: What share of the animation tickets were adult tickets?
  3. Table 2: Faolan writes, “70\% of the visitors who bought something took a tour.” Which total is he dividing by, and is he right?

Part 2: two directions

  1. Table 2: Huxley writes, “One in five tour visitors bought nothing, so one in five visitors who bought nothing took a tour.” Check each half of his sentence.
  2. Table 3: Find the share of adult tickets that were for a musical, and the share of musical tickets that were adult tickets. Which is larger, and why?

Part 2: comparing groups fairly

  1. On a made-up Saturday, 42 of the 56 tour visitors at a museum bought something. On Sunday, 50 of the 75 tour visitors did. Which day had more tour visitors who bought something, and which had the larger share?
  2. Table 3: (a) Is a child ticket or an adult ticket more likely to be for animation? (b) Is an animation ticket or a musical ticket more likely to be a child ticket? Compare shares.
  3. On a second made-up day in the park, 28 of the 35 children and 54 of the 90 adults wore a helmet. Ines-Ruth says more adults than children wore helmets, so the adults were more careful. Compare the two groups fairly.

Part 2: filling a table back

  1. A made-up museum had 200 visitors, and 80 of them took a tour. Of the tour visitors, 90\% bought something; of the others, 10\% did. Fill in the two-way table, then find the share of the buyers who took a tour.
  2. A made-up cinema sold 400 tickets, 35\% of them child tickets. Half of the child tickets and 15\% of the adult tickets were for animation. What share of the animation tickets were child tickets?
  3. In a made-up park count, 90 cyclists wore a helmet and 60 did not. Of the helmet wearers, 40\% were children; of those without a helmet, 25\% were children. What share of the children wore a helmet?

Reasoning

  1. Kalliope adds the helmet column of the row-relative table from Exercise 4 and gets 0.75 + 0.45 + 0.62 = 1.82. She thinks something is wrong because it is not 1. Explain why nothing is wrong, and say which numbers in that table do have to add to 1.
  2. A cell holds a people, with a > 0. Its row total is r and its column total is c. Show that the share of the row in that cell equals the share of the column in that cell exactly when r = c.
  3. Two made-up museums have the same row-relative table as Table 1: 0.75 and 0.25 for the tour visitors, 0.125 and 0.875 for the others. The first is Table 1 itself. The second has 32 tour visitors and 16 other visitors. Find the share of the buyers who took a tour at each, and say what this shows about reading column shares from a row-relative table.
  4. Odalys says, “Of the chess players, 40\% are in the choir, so 40\% of the choir play chess.” Without a table, explain why the second half does not follow from the first, and say what extra counts would settle it.
  5. Explain why a table of shares alone, with no totals, can never be filled back into a table of counts.
Answers
  1. The group is the tour visitors, row total 35: \displaystyle \frac{28}{35} = \frac{4}{5} = 0.8.
  2. The group is the visitors who bought nothing, column total 80; 73 of them took no tour: \displaystyle \frac{73}{80} = 0.9125.
  3. Child: \displaystyle \frac{36}{60} = 0.6 and \displaystyle \frac{24}{60} = 0.4. Adult: \displaystyle \frac{27}{90} = 0.3 and \displaystyle \frac{63}{90} = 0.7. Total: \displaystyle \frac{63}{150} = 0.42 and \displaystyle \frac{87}{150} = 0.58. Each pair adds to 1.
  4. Child: 0.75 and 0.25. Teen: \displaystyle \frac{18}{40} = 0.45 and \displaystyle \frac{22}{40} = 0.55. Adult: 0.62 and 0.38. Total: \displaystyle \frac{125}{200} = 0.625 and \displaystyle \frac{75}{200} = 0.375. Each row’s Total entry is 1.
  5. The musical column, 87: \displaystyle \frac{24}{87} = \frac{8}{29} \approx 0.276.
  6. The animation column, 63: \displaystyle \frac{27}{63} = \frac{3}{7} \approx 0.429.
  7. The buyers’ column, 40: \displaystyle \frac{28}{40} = 0.7 = 70\%. He is right.
  8. First half true: \displaystyle \frac{7}{35} = \frac{1}{5} of the tour visitors bought nothing. Second half false: of the 80 who bought nothing, 7 took a tour, \displaystyle \frac{7}{80} = 0.0875, fewer than one in eleven. Same cell, different whole.
  9. \displaystyle \frac{63}{90} = 0.7 of adult tickets were for a musical; \displaystyle \frac{63}{87} = \frac{21}{29} \approx 0.724 of musical tickets were adult. The second is larger: the same 63 over a smaller whole.
  10. Sunday had more, 50 against 42. Saturday had the larger share: \displaystyle \frac{42}{56} = 0.75 against \displaystyle \frac{50}{75} = \frac{2}{3} \approx 0.667.
  11. (a) A child ticket: \displaystyle \frac{36}{60} = 0.6 against \displaystyle \frac{27}{90} = 0.3. (b) An animation ticket: \displaystyle \frac{36}{63} = \frac{4}{7} \approx 0.571 against \displaystyle \frac{24}{87} \approx 0.276.
  12. Children \displaystyle \frac{28}{35} = 0.8, adults \displaystyle \frac{54}{90} = 0.6. The adult count is larger only because more adults were counted; a larger share of the children wore a helmet.
  13. Tour: 0.9 \times 80 = 72 bought, 8 did not. Others: 0.1 \times 120 = 12 bought, 108 did not. Buyers 84, non-buyers 116, and 84 + 116 = 200. Share of buyers who took a tour: \displaystyle \frac{72}{84} = \frac{6}{7} \approx 0.857.
  14. Child tickets 0.35 \times 400 = 140, adult 260. Child animation 70, adult animation 0.15 \times 260 = 39, animation total 109. Share: \displaystyle \frac{70}{109} \approx 0.642.
  15. Children with a helmet 0.4 \times 90 = 36, without 0.25 \times 60 = 15, so 51 children. Share: \displaystyle \frac{36}{51} = \frac{12}{17} \approx 0.706.
  16. The three numbers are shares of three different groups (children, teens, adults), so they have no single whole to fill. Each row must add to 1, such as 0.75 + 0.25 for the children.
  17. The shares are \displaystyle \frac{a}{r} and \displaystyle \frac{a}{c}. If r = c they are equal. If \displaystyle \frac{a}{r} = \frac{a}{c} with a > 0, cross-multiplying gives ac = ar, and dividing by a gives c = r.
  18. First: \displaystyle \frac{24}{30} = 0.8. Second: 0.75 \times 32 = 24 tour visitors and 0.125 \times 16 = 2 others bought something, so \displaystyle \frac{24}{26} = \frac{12}{13} \approx 0.923. Same row-relative table, different column shares: column shares need the row totals.
  19. The 40\% divides the chess players in the choir by the number of chess players; the second claim divides the same people by the number in the choir. They agree only if those two numbers are equal. The number of chess players and the number in the choir would settle it.
  20. Multiplying every count by the same number, say 2, multiplies each cell and each total by 2 and leaves every share the same. So one table of shares fits many tables of counts; only the totals fix the scale.
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