TMUA Paper 1: Statistics and probability, set 2

• Answer all questions. No calculator.
• Each question has exactly one correct answer.
• This is a demanding set: several questions need multi-step working, and the prompts give less signposting than the real exam.

1. 11 mark

   A company employs 200 workers across two shifts, and every worker recently sat the same safety test. The day shift has 120 workers and the night shift has 80. Across both shifts, 130 workers passed the test in total.

   The number of day-shift workers who FAILED the test is exactly half the number of night-shift workers who PASSED it.

   Work out how many night-shift workers failed the test.

   1. A 20
   2. B 10
   3. C 60
   4. D 28
2. 21 mark

   A histogram shows the mass, in kg, of 90 parcels handled by a courier depot in one day, using unequal class widths.

   The bar for the interval 0 <= m < 10 has a frequency of 20. The bar for the interval 10 <= m < 30 has a frequency of 30. All remaining parcels fall in the third and final interval, which starts at 30 kg.

   The frequency density of the third bar is exactly half the frequency density of the first bar. Work out the width of the third class interval.

   1. A 20
   2. B 70
   3. C 90
   4. D 40
3. 31 mark

   A cumulative frequency graph shows the amount of time, in minutes, that 84 shoppers spent in a supermarket. The graph passes through the points (0, 0), (10, 12), (20, 30), (40, 70), (60, 82) and (70, 84).

   Using linear interpolation between the two plotted points on either side of the median position, estimate the median time spent in the supermarket.

   1. A 26
   2. B 30
   3. C 34
   4. D 40
4. 41 mark

   A cumulative frequency graph shows the delivery times, in minutes, of 200 parcels sent by a delivery company. The graph passes through the points (0, 0), (10, 20), (20, 60), (30, 140), (40, 180) and (50, 200).

   Using linear interpolation, estimate the interquartile range of the delivery times.

   1. A 15
   2. B 32.5
   3. C 7.5
   4. D 20
5. 51 mark

   Two classes sat the same maths test, out of 100 marks. Class A's results have a median score of 62 and an interquartile range of 10. Class B's results have a median score of 58 and an interquartile range of 22.

   Which of the following is a valid comparison of the two classes' performance, based only on these summary statistics?

   1. A Class A's median score is higher, so Class A's students consistently outperformed Class B's on every question in the test.
   2. B Because Class B's interquartile range is more than double Class A's, Class B's median score must also be higher, since a wider spread always corresponds to a higher typical score.
   3. C The two classes cannot be compared at all, because the median and interquartile range only apply to grouped, continuous data taken from a cumulative frequency graph, not to a class's raw test scores.
   4. D Class A had a higher median and a smaller interquartile range, so on average it did better, and its scores were more consistently clustered than Class B's.
6. 61 mark

   A gardener records the height of a young plant, in cm, once a week for its first 12 weeks. A line of best fit is drawn through this data, with equation y = 4x + 3, where x is the plant's age in weeks and y is its height in cm.

   A student uses this equation to predict the plant's height after 5 years, that is, at x = 260 weeks, and calculates y = 4 x 260 + 3 = 1043 cm.

   Which of the following is the best comment on this prediction?

   1. A The calculation is arithmetically correct, so the prediction of 1043 cm is reliable regardless of the plant's age.
   2. B The prediction is unreliable because a line of best fit can only ever be used to interpolate between two data points that were actually plotted, and it can never be used to predict a height at any other age, even one within the original 12 weeks.
   3. C The prediction is unreliable because it extrapolates the line of best fit far beyond the range of ages, 0 to 12 weeks, that it was fitted to, and there is no reason to expect the same linear trend to continue growing at this rate for 260 weeks.
   4. D The prediction is unreliable because the gradient of 4 and the intercept of 3 must have been calculated incorrectly, since no real plant could have a height described by a straight-line equation.
7. 71 mark

   A frequency tree records data for 150 members of a climbing club. First, it splits them by age group: 60% of the 150 members are adults, and the rest are juniors.

   Each age group is then split by which kind of climbing they do. Of the adults, 5/9 climb outdoors and the rest climb indoors. Of the juniors, 30% climb outdoors and the rest climb indoors.

   Work out the total number of the 150 members, combining both adults and juniors, who climb indoors.

   1. A 82
   2. B 40
   3. C 68
   4. D 42
8. 81 mark

   Every one of 90 students at a sixth-form college takes at least one of three activities: Drama, Music and Sport. 40 students take Drama, 35 take Music, and 47 take Sport.

   12 students take both Drama and Music, 15 take both Music and Sport, and 10 take both Drama and Sport (each of these three figures includes anyone who takes all three activities). 5 students take all three activities.

   By reasoning systematically about the situation as a Venn diagram of the three activities, work out how many of the 90 students take EXACTLY ONE of the three activities.

   1. A 48
   2. B 63
   3. C 85
   4. D 22
9. 91 mark

   In a survey of 60 people, 24 own a car, 18 own a bicycle, and 10 own both a car and a bicycle.

   A person is chosen at random from those who own AT LEAST ONE of a car or a bicycle. Find the probability that this person owns BOTH a car and a bicycle.

   1. A 1/6
   2. B 5/21
   3. C 5/16
   4. D 5/7
10. 101 mark

    A game uses two fair spinners. Spinner X has four equal sections, numbered 1, 2, 2 and 3 (the number 2 appears on two separate sections). Spinner Y has three equal sections, numbered 1, 3 and 5.

    Both spinners are spun once, and the two numbers they land on are multiplied together.

    By constructing a possibility space of all equally likely outcomes, find the probability that the product is an odd number.

    1. A 1/2
    2. B 2/3
    3. C 1
    4. D 1/4
11. 111 mark

    A bag contains 5 green counters and 3 yellow counters, 8 counters in total. Three counters are removed from the bag at random, one after another, without replacement.

    Using a tree diagram, find the probability that exactly two of the three counters removed are green (and the other is yellow).

    1. A 5/28
    2. B 225/512
    3. C 5/14
    4. D 15/28
12. 121 mark

    At a school, the probability that a randomly chosen student walks to school is 1/4, and the probability that they travel some other way is 3/4.

    Of the students who walk to school, 1/5 also play a sport after school. Of the students who travel some other way, 2/3 also play a sport after school.

    Using a tree diagram, find the probability that a student who plays a sport after school is one who walks to school.

    1. A 1/20
    2. B 1/11
    3. C 1/4
    4. D 1/5

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

1. Question 1Answer: C

   1. Let p_n be the number of night-shift workers who passed. The day-shift failures equal half of this, so day-shift failures = 0.5 x p_n.
   2. The day-shift total is 120, so day-shift passes = 120 - 0.5 x p_n.
   3. The two shifts' passes must add to the overall total of 130 passes, so (120 - 0.5 x p_n) + p_n = 130, which simplifies to 120 + 0.5 x p_n = 130.
   4. So 0.5 x p_n = 10, giving p_n = 20 night-shift passes.
   5. The night shift has 80 workers in total, so the number who failed is 80 - 20 = 60.
   6. The answer is C.
   • Why not A: 20 is the number of night-shift workers who PASSED the test, correctly found along the way as part of solving the simultaneous equations, but it answers a different question from the one asked, which wants the number of night-shift workers who FAILED.
   • Why not B: 10 is the number of day-shift workers who FAILED the test, found correctly from the given ratio, but this is the wrong shift: the question asks about the night shift, not the day shift.
   • Why not D: 28 comes from assuming every shift has the same pass rate as the company overall (130 out of 200, or 65%), applying that overall rate directly to the night shift's total of 80 (0.65 x 80 = 52 passes, so 80 - 52 = 28 fails) instead of using the specific relationship stated between day-shift failures and night-shift passes.

2. Question 2Answer: D

   1. The first bar has frequency 20 and width 10 - 0 = 10, so its frequency density is 20 / 10 = 2.
   2. The third bar's frequency density is half of this: 2 / 2 = 1.
   3. The third bar's frequency is what remains once the first two bars are accounted for: 90 - 20 - 30 = 40.
   4. Frequency = width x frequency density, so width = frequency / density = 40 / 1 = 40.
   5. The answer is D.
   • Why not A: This uses the first bar's frequency density itself (20 / 10 = 2) rather than half of it, giving a width of 40 / 2 = 20 instead of using the required density of 1.
   • Why not B: This finds the third bar's frequency incorrectly using 90 - 20 = 70, forgetting to also subtract the second bar's frequency of 30, and then divides by the correct density of 1 to get 70.
   • Why not C: This treats the whole 90 parcels as belonging to the third bar, forgetting to subtract the 20 and 30 parcels already accounted for by the first two bars, before dividing by the correct density of 1.

3. Question 3Answer: A

   1. There are 84 shoppers, so the median position is at cumulative frequency 84 / 2 = 42.
   2. This lies between the plotted points (20, 30) and (40, 70), since the cumulative frequency rises from 30 to 70 across that interval.
   3. Within that interval, 42 is (42 - 30) / (70 - 30) = 12 / 40 = 0.3 of the way from 30 to 70.
   4. The time axis spans from 20 to 40 minutes across this interval, a width of 20, so the estimated median is 20 + 0.3 x 20 = 20 + 6 = 26 minutes.
   5. The answer is A.
   • Why not B: This assumes the median falls exactly halfway through the class interval containing it (20 + (40 - 20) / 2 = 30), rather than using the actual cumulative frequencies either side of the median position to weight where within the interval it falls.
   • Why not C: This measures the interpolation fraction from the wrong end of the interval, using how far the median position is BELOW the upper cumulative frequency (70 - 42 = 28, out of the interval's span of 40) instead of how far it is ABOVE the lower one.
   • Why not D: This reads off the upper boundary of the class interval containing the median (40 minutes) without interpolating within it at all.

4. Question 4Answer: A

   1. There are 200 parcels, so the lower quartile position is 200 / 4 = 50 and the upper quartile position is 3 x 200 / 4 = 150.
   2. The lower quartile position of 50 lies between (10, 20) and (20, 60): it is (50 - 20) / (60 - 20) = 30 / 40 = 0.75 of the way across, giving a lower quartile of 10 + 0.75 x 10 = 17.5 minutes.
   3. The upper quartile position of 150 lies between (30, 140) and (40, 180): it is (150 - 140) / (180 - 140) = 10 / 40 = 0.25 of the way across, giving an upper quartile of 30 + 0.25 x 10 = 32.5 minutes.
   4. The interquartile range is the upper quartile minus the lower quartile: 32.5 - 17.5 = 15.
   5. The answer is A.
   • Why not B: 32.5 is the estimated upper quartile itself, correctly found by interpolating between (30, 140) and (40, 180), but the working stops there rather than going on to subtract the lower quartile to get the interquartile range.
   • Why not C: 7.5 is half of the correct interquartile range: this reports the semi-interquartile range rather than the interquartile range the question asks for.
   • Why not D: 20 comes from measuring the upper quartile's interpolation fraction from the wrong end of its interval (using (180 - 150) / (180 - 140) instead of (150 - 140) / (180 - 140)), which gives an upper quartile of 37.5 instead of 32.5, and so an interquartile range of 37.5 - 17.5 = 20.

5. Question 5Answer: D

   1. The median measures the typical, central score. Class A's median of 62 is higher than Class B's 58, so on average Class A's students scored better.
   2. The interquartile range measures how spread out the middle 50 per cent of scores are. Class A's interquartile range of 10 is smaller than Class B's 22, so Class A's scores were more tightly clustered around its median, that is, more consistent.
   3. Comparing like-for-like summary values, median against median and interquartile range against interquartile range, is the valid way to compare two distributions.
   4. The answer is D.
   • Why not A: This correctly notes Class A's higher median but wrongly stretches a comparison of typical (central) scores into a claim about every individual question, which a single median score cannot support.
   • Why not B: This has the relationship backwards: the interquartile range measures spread, not typical score, so a bigger spread does not imply a higher median. In fact Class B's median (58) is the lower of the two.
   • Why not C: The median and interquartile range can be found for any numerical data set, including a class's raw test scores; a cumulative frequency graph is only one way of estimating them from grouped data, not a requirement for using them at all.

6. Question 6Answer: C

   1. A line of best fit is only trustworthy for describing the relationship over the range of values it was fitted to, here ages 0 to 12 weeks.
   2. Extrapolating it out to x = 260, more than twenty times beyond that range, assumes the same linear growth rate continues indefinitely, which the original 12 weeks of data cannot support.
   3. A real plant's growth would be very unlikely to remain linear over five years, so a numerically correct calculation can still give a meaningless prediction once extrapolated this far.
   4. The answer is C.
   • Why not A: Being arithmetically correct is not the issue: a correctly evaluated equation can still give an unreliable real-world prediction if it is applied far outside the range of data it was built from.
   • Why not B: This overstates the limitation. Interpolating within the range the line was fitted to, 0 to 12 weeks, is a legitimate use of a line of best fit; the actual problem here is specifically extrapolating to 260 weeks, well beyond that range, not using the line at all.
   • Why not D: This wrongly blames the equation itself. The line may fit the original 12 weeks of growth data very well; the problem is extending a relationship that far beyond the range of ages it describes, not an error in finding it.

7. Question 7Answer: A

   1. 60% of the 150 members are adults: 0.6 x 150 = 90 adults, leaving 150 - 90 = 60 juniors.
   2. Of the 90 adults, 5/9 climb outdoors: 90 x 5/9 = 50 outdoor adults, so 90 - 50 = 40 climb indoors.
   3. Of the 60 juniors, 30% climb outdoors: 0.3 x 60 = 18 outdoor juniors, so 60 - 18 = 42 climb indoors.
   4. Combining both groups' indoor climbers: 40 + 42 = 82.
   5. The answer is A.
   • Why not B: 40 is the number of ADULTS who climb indoors alone: this correctly follows the adult branch of the tree but forgets to also add the juniors who climb indoors.
   • Why not C: 68 correctly totals both branches of the tree, but adds up the OUTDOOR climbers rather than the indoor ones the question asks for.
   • Why not D: 42 is the number of JUNIORS who climb indoors alone: this correctly follows the junior branch of the tree but forgets to also add the adults who climb indoors.

8. Question 8Answer: B

   1. For each activity, the number taking exactly that one activity is its total minus BOTH pair-overlaps involving it, plus the all-three group added back on (since it was subtracted twice).
   2. Drama only: 40 - 12 - 10 + 5 = 23. Music only: 35 - 12 - 15 + 5 = 13. Sport only: 47 - 10 - 15 + 5 = 27.
   3. Adding these three regions gives the number who take exactly one activity: 23 + 13 + 27 = 63.
   4. As a check, exactly two activities accounts for (12 - 5) + (15 - 5) + (10 - 5) = 7 + 10 + 5 = 22 students, and all three accounts for 5, and 63 + 22 + 5 = 90, matching the total.
   5. The answer is B.
   • Why not A: 48 comes from subtracting each pair-overlap figure from its activity's total but forgetting to add the all-three group back on for each activity (for example, treating 'Drama only' as 40 - 12 - 10 = 18 rather than 40 - 12 - 10 + 5 = 23), which double-subtracts the students who take all three activities.
   • Why not C: 85 comes from subtracting only the all-three group from the total of 90 (90 - 5 = 85), ignoring the pair overlaps entirely rather than reasoning about the diagram's regions properly.
   • Why not D: 22 is the number of students who take EXACTLY TWO of the three activities, not exactly one: it answers a different, though related, question about the diagram.

9. Question 9Answer: C

   1. The number who own at least one of a car or a bicycle is found by adding the two totals and subtracting those counted twice: 24 + 18 - 10 = 32.
   2. The number who own both is given directly as 10.
   3. The probability that a person chosen from those owning at least one item owns both is 10 / 32, which simplifies to 5 / 16.
   4. The answer is C.
   • Why not A: 1/6 uses all 60 people surveyed as the denominator (10/60), rather than restricting attention to only those who own at least one of the two items, as the question specifies.
   • Why not B: 5/21 comes from adding the car and bicycle totals without subtracting the overlap (24 + 18 = 42 'at least one', when this double-counts the 10 who own both), giving 10/42 instead of the correct 10/32.
   • Why not D: 5/7 uses the number who own a car ONLY (24 - 10 = 14) as the denominator, rather than the number who own at least one of the two items, giving 10/14 instead of 10/32.

10. Question 10Answer: A

    1. There are 4 x 3 = 12 equally likely outcomes in total, treating each of spinner X's four physical sections, and each of spinner Y's three, as equally likely.
    2. Spinner Y's sections are 1, 3 and 5, which are all odd, so the product's parity depends entirely on spinner X's section.
    3. Spinner X's sections are 1, 2, 2 and 3: two of the four sections show an odd number (1 and 3), and two show an even number (the two sections showing 2).
    4. The product is odd exactly when spinner X lands on an odd section, which happens on 2 of its 4 equally likely sections, giving a probability of 2/4 = 1/2.
    5. The answer is A.
    • Why not B: 2/3 treats spinner X's four sections as only three distinct outcomes, 1, 2 and 3, each equally likely, collapsing the two separate sections both showing 2 into a single outcome. This undercounts the possibility space: the two sections showing 2 are two separate, equally likely physical outcomes, not one.
    • Why not C: 1 assumes that, because spinner Y always lands on an odd number, the product must always be odd regardless of what spinner X shows. This forgets that an even number (from either of X's two sections showing 2) multiplied by an odd number still gives an even product.
    • Why not D: 1/4 wrongly assumes spinner Y has an equal chance of landing on an odd or even number, giving it a probability of 1/2 of being odd, and multiplies this by spinner X's correct probability of landing on an odd section (1/2), rather than recognising that all three of spinner Y's sections, 1, 3 and 5, are odd.

11. Question 11Answer: D

    1. There are three orders in which exactly two green counters and one yellow counter can be drawn: green-green-yellow, green-yellow-green, and yellow-green-green.
    2. For green-green-yellow: P = (5/8) x (4/7) x (3/6) = 5/28. The same value, 5/28, is obtained for each of the other two orders, since without replacement the same three counters (two green, one yellow) are simply removed in a different sequence.
    3. Adding the three equal branches gives the total probability: 5/28 + 5/28 + 5/28 = 15/28.
    4. The answer is D.
    • Why not A: 5/28 is the probability of just ONE specific order of colours, green then green then yellow. It correctly multiplies the changing probabilities along that one branch of the tree, but forgets that 'exactly two green and one yellow' can also occur as green-yellow-green or yellow-green-green, which must be added on as well.
    • Why not B: 225/512 comes from using the formula for drawing WITH replacement (treating each draw as an independent 5/8 chance of green), then multiplying by 3 for the three orders. Because the counters are not replaced, the probabilities along each branch actually change after every draw, so this overstates the true probability.
    • Why not C: 5/14 comes from correctly finding the probability of one order (5/28) but then multiplying by 2 rather than 3, having missed one of the three possible orders in which exactly two greens and one yellow can occur.

12. Question 12Answer: B

    1. P(walks AND plays sport) = 1/4 x 1/5 = 1/20.
    2. P(does not walk AND plays sport) = 3/4 x 2/3 = 1/2 = 10/20.
    3. The overall probability of playing a sport, combining both branches, is 1/20 + 10/20 = 11/20.
    4. The probability that a student who plays a sport walks to school is the walking-and-sport branch divided by the overall sport probability: (1/20) / (11/20) = 1/11.
    5. The answer is B.
    • Why not A: 1/20 is the probability that a student both walks to school AND plays a sport, found correctly as 1/4 x 1/5, but this stops one step early: it has not been divided by the overall probability of playing a sport, so it is not yet a conditional probability.
    • Why not C: 1/4 simply reports the ORIGINAL probability of walking to school, ignoring the information that the student plays a sport after school entirely.
    • Why not D: 1/5 answers the reverse conditional probability, the probability that a student plays a sport GIVEN that they walk to school, rather than the probability that a student walks to school given that they play a sport.