Printable · GCSE Foundation · ages 14-16
Frequency of outcomes, tables and frequency trees worksheet — GCSE Foundation
Fifteen questions on "frequency of outcomes, tables and frequency trees" — DfE statement P1. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Frequency of outcomes, tables and frequency trees worksheet — GCSE Foundation
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- 1.180 gardeners at an allotment were asked whether they grow vegetables organically. 100 of the gardeners are on the north plots. 64 of the north-plot gardeners grow organically. 30 of the south-plot gardeners do not grow organically. Work out how many south-plot gardeners grow organically.
- 2.A cinema sold 150 tickets for a Saturday screening. A frequency tree records the sales: 90 of the tickets were bought by adults and the rest by children. Of the adult tickets, 54 were for the 3D showing and the rest were for the standard showing. Work out how many adult tickets were for the standard showing.
- 3.A frequency tree records the results of 160 patients who took a new medicine. It splits them into those who reported side effects and those who did not. 15% of the patients reported side effects. Work out how many of the 160 patients did not report side effects.
- 4.A frequency tree records how 90 pupils travel to school and whether they were late. The first pair of branches splits the 90 pupils into 50 who walk and 40 who do not walk. On the walking branch, 6 were late and 44 were not. On the other branch, 10 were late and 30 were not. One of the 90 pupils is chosen at random. Work out the probability that the pupil walks to school and was late.
- 5.A two-way table records how 80 students travel to school. 45 of the students are boys and the rest are girls. 27 of the boys walk to school. Work out the probability, as a fraction, that a student chosen at random from the boys walks to school.
- 6.150 people at a gym were asked whether they prefer weight training or cardio; each person chose exactly one. 84 of the people are women. 50 of the women prefer cardio. 40 of the men prefer weight training. Work out the probability that a person, chosen at random from the 150, prefers weight training. Give your answer as a fraction in its simplest form.
- 7.A survey of 160 employees at a company recorded whether they cycle to work, using a frequency tree. The first branch splits them into 90 who work full-time and 70 who work part-time. Of the full-time employees, 27 cycle to work. Of the part-time employees, 14 cycle to work. Work out the probability that an employee, chosen at random from the 160, cycles to work. Give your answer as a fraction in its simplest form.
- 8.A garden centre recorded 240 plant sales using a frequency tree. The first stage splits the sales into three types: shrubs, bedding plants and trees. 96 sales were shrubs, 114 were bedding plants and the rest were trees. Work out the probability that a sale chosen at random from these 240 was a tree.
- 9.A clinic recorded 300 booked appointments using a frequency tree. The first branch splits them into 210 adult appointments and the rest child appointments. Of the adult appointments, 189 were attended and the rest were missed. Of the child appointments, 81 were attended. Work out the probability that a booked appointment, chosen at random from the 300, was missed. Give your answer as a fraction in its simplest form.
- 10.A council surveyed 140 households about recycling. 85 of the households are in Zone A. 52 of the Zone A households recycle glass. 21 of the Zone B households do not recycle glass. Work out the probability that a household, chosen at random from the 140, recycles glass. Give your answer as a fraction in its simplest form.
- 11.180 people entered a fun run. A frequency tree splits them into 108 adults and the rest children. It then splits each of those groups into those who finished and those who did not. 99 of the adults finished, and 153 of the 180 people finished altogether. Work out how many children did not finish.
- 12.A weather station records whether it rains each day for 40 days: it rains on 9 of the days and does not rain on the rest. A local forecaster claims that the probability of rain on any day is 0.3. Work out the relative frequency of rain from the recorded data.
- 13.A table shows how two Year 10 classes did in a spelling test. In class A, 18 out of 24 pupils passed. In class B, 21 out of 30 pupils passed. Which class had the greater proportion of pupils passing?
- 14.A two-way table records 200 members of a gym. 80 of the members are women and the rest are men. 60 of the women attend yoga classes. Work out the percentage of the women who attend yoga classes.
- 15.A frequency tree records 320 calls to a helpline. The calls first split into calls about billing and calls about technical support. Of the technical support calls, the probability that a call was resolved on the first contact is 0.75. If 180 technical support calls were resolved on the first contact, work out how many calls were about billing.
Answer key
- (b) 50 — There are 180 − 100 = 80 south-plot gardeners. 30 of them do not grow organically, so the rest do: 80 − 30 = 50. Writing 64 is wrong because that is the number of NORTH-plot gardeners who grow organically, not south. Writing 30 again is wrong because that is the number of south-plot gardeners who do NOT grow organically — the question asks for those who do. Writing 80 is wrong because that is the whole south-plot total, without subtracting the 30 who do not grow organically. The answer is 50.
- (c) 36 — On the adult branch there are 90 tickets in total, and 54 of them are for the 3D showing, so the standard-showing branch is 90 − 54 = 36. Subtracting 54 from the overall total of 150 gives 96, but 150 is the total for ALL tickets, not just the adult branch, so 96 is wrong. Writing 60 is wrong because that is the number of CHILD tickets (150 − 90 = 60), not adult standard tickets. Writing 54 again is wrong because that is the number of adult 3D tickets, not the number of adult standard tickets. The adult standard-showing branch is 36.
- (d) 136 — 15% of 160 = 0.15 × 160 = 24 patients reported side effects, so 160 − 24 = 136 did not. Stopping after finding the number who reported side effects, 24, answers the wrong question — it is not the number who did NOT report them. Misreading '15%' as a raw count of 15 patients, rather than a percentage, gives 160 − 15 = 145. Subtracting 15% of 160 twice, 160 − 24 − 24 = 112, double-counts the side-effect group.
- (d) 1/15 — Method: a probability read from a frequency tree is the count at the end of the branch you want, divided by the total number in the whole experiment. Working: the branch for walking followed by the branch for being late ends with 6 pupils, and the experiment covers all 90 pupils, so the probability is 6/90. Dividing the top and the bottom by 6 gives 1/15. Answer: the probability is 1/15. The distractors: 3/25 is 6/50 and comes from dividing the 6 by the 50 walkers rather than by the whole group, which answers a different question about walkers only; 8/45 is 16/90 and comes from counting every late pupil, the 6 walkers and the 10 others together, instead of only the late walkers; 1/9 is 10/90 and comes from reading the late count on the branch for pupils who do not walk.
- (d) 3/5 — 'Chosen from the boys' restricts the sample space to the 45 boys, of whom 27 walk. So the probability is 27/45 = 3/5. Using all 80 students as the denominator instead of just the boys gives 27/80, which is the probability that a student chosen from everyone is a boy who walks — not what was asked. Using the 18 boys who do NOT walk (45 − 27) as the numerator instead of the 27 who do gives 18/45 = 2/5. Dividing the 27 walking boys by the number of girls (80 − 45 = 35) instead of by the number of boys gives 27/35.
- (b) 37/75 — There are 150 − 84 = 66 men. 84 − 50 = 34 women prefer weight training, and 40 men prefer weight training, so 34 + 40 = 74 people in total prefer weight training, out of 150: 74/150 = 37/75. Writing 4/15 is wrong because it only counts the men who prefer weight training (40/150, simplified), leaving out the 34 women. Writing 17/75 is wrong because it only counts the women who prefer weight training (34/150, simplified), leaving out the 40 men. Writing 37/42 is wrong because it uses the number of women (84) as the denominator instead of the whole gym (150) — 74/84 simplifies to 37/42, but that is not a probability out of the whole group. The probability is 37/75.
- (b) 41/160 — In total, 27 + 14 = 41 of the 160 employees cycle to work, so the probability is 41/160 (41 and 160 share no common factor, so this is already in its simplest form). Writing 27/160 is wrong because it only counts the full-time cyclists and leaves out the 14 part-time cyclists. Writing 41/90 is wrong because it uses the full-time total (90) as the denominator instead of the whole survey (160). Writing 1/5 is wrong because it only uses the part-time branch, simplifying 14/70 to 1/5 and ignoring the full-time cyclists completely. The probability is 41/160.
- (b) 1/8 — Shrubs, bedding plants and trees are the three branches at the first stage of the tree, so they must total 240: tree sales = 240 − 96 − 114 = 30. So P(tree) = 30/240 = 1/8. Using the shrub count instead, 96/240 = 2/5, is the probability of a shrub sale, not a tree sale. Using the bedding-plant count instead, 114/240 = 19/40, is the probability of a bedding-plant sale. Subtracting the bedding count from the shrub count (114 − 96 = 18) instead of subtracting both from 240 gives 18/240 = 3/40, which is not the number of tree sales at all.
- (c) 1/10 — On the adult branch, 210 − 189 = 21 appointments were missed. There are 300 − 210 = 90 child appointments, and 90 − 81 = 9 of those were missed. In total, 21 + 9 = 30 appointments were missed, out of 300: 30/300 = 1/10. Writing 7/100 is wrong because 21/300 simplifies to 7/100, and 21 only counts the adult branch, leaving out the 9 missed child appointments. Writing 3/100 is wrong because 9/300 simplifies to 3/100, and 9 only counts the child branch, leaving out the 21 missed adult appointments. Writing 1/9 is wrong because it divides the 30 missed appointments by the 270 that were attended (300 − 30) instead of by the whole 300 booked. The probability is 1/10.
- (a) 43/70 — There are 140 − 85 = 55 Zone B households, and 55 − 21 = 34 of them recycle glass. In total, 52 + 34 = 86 households recycle glass, out of 140: 86/140 = 43/70. Writing 13/35 is wrong because 52/140 simplifies to 13/35, and 52 only counts Zone A, leaving out the 34 Zone B recyclers. Writing 17/70 is wrong because 34/140 simplifies to 17/70, and 34 only counts Zone B, leaving out the 52 Zone A recyclers. Writing 73/140 is wrong because it adds the 52 Zone A recyclers to the 21 Zone B households that do NOT recycle, mixing up two different groups instead of adding the two recycling groups. The probability is 43/70.
- (d) 18 — Method: a frequency tree can be filled from either end, because a pair of branches adds up to the number above it and the matching end branches add up to the total for that outcome. Working: the children number 180 − 108 = 72. Of the 153 finishers, 99 were adults, so the children who finished number 153 − 99 = 54. Along the children's branch, 72 − 54 = 18 children did not finish. Answer: 18 children did not finish. The distractors: 27 is 180 − 153 and counts everyone who did not finish, adults included; 9 is 108 − 99 and counts only the adults who did not finish; 54 is the number of children who did finish, which is the branch next to the one asked for.
- (b) 0.225 — The relative frequency of rain is the number of rainy days out of all days recorded: 9 ÷ 40 = 0.225, which is noticeably less than the forecaster's claimed 0.3. Using the number of dry days, 40 − 9 = 31, as the denominator instead of the total of 40 gives 9 ÷ 31 = 0.29 (2 d.p.). Simply reporting the forecaster's claimed value, 0.3, without calculating anything from the data at all, ignores the recorded results completely. Misplacing the decimal point, treating 9 out of 40 as 9%, gives 0.09 instead of 0.225.
- (a) Class A — Class A: 18/24 = 0.75 = 75%. Class B: 21/30 = 0.7 = 70%. Since 75% > 70%, class A had the greater proportion passing, even though fewer pupils passed there in total. Choosing class B compares the raw numbers of pupils who passed (21 > 18) rather than the proportions. The two proportions are not equal — 0.75 and 0.7 are different values, so the two classes did not have the same pass rate. The class sizes being different does not prevent a comparison: converting each to a proportion makes the two classes directly comparable, so the answer can be determined.
- (a) 75% — 'Percentage of the women' restricts the group to the 80 women, of whom 60 attend yoga: 60/80 = 0.75 = 75%. Dividing by the number of men (200 − 80 = 120) instead of the number of women gives 60/120 = 0.5 = 50%. Dividing by all 200 members instead of just the 80 women gives 60/200 = 0.3 = 30%. Using the 20 women who do NOT attend yoga (80 − 60) as the numerator instead of the 60 who do gives 20/80 = 0.25 = 25%.
- (c) 80 — Since 180 calls are 0.75 of all the technical support calls, the technical support total is 180 ÷ 0.75 = 240. The billing calls make up the rest of the 320 calls, so 320 − 240 = 80. Choosing 240 comes from stopping after finding the technical support total and forgetting the question asks for the billing calls, which are the rest. Choosing 185 comes from multiplying 180 × 0.75 = 135 instead of dividing, then working out 320 − 135 = 185. Choosing 140 comes from using 180 directly as the whole technical support total, ignoring the probability altogether, then working out 320 − 180 = 140.
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