Printable · GCSE Higher · ages 14-16
Histograms and cumulative frequency graphs worksheet — GCSE Higher
Fifteen questions on "histograms and cumulative frequency graphs" — DfE statement S3. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Higher only
Histograms and cumulative frequency graphs worksheet — GCSE Higher
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- 1.The masses, m kg, of 150 boxes are summarised by these cumulative frequencies: m < 5, 18 boxes; m < 10, 52 boxes; m < 20, 96 boxes; m < 35, 130 boxes; m < 60, 150 boxes. Work out the number of boxes with a mass in the class 10 ≤ m < 20.
- 2.The distances, d km, cycled by 180 riders in a charity sportive are summarised by these cumulative frequencies: d < 30, 20 riders; d < 60, 60 riders; d < 80, 120 riders; d < 100, 160 riders; d < 130, 180 riders. Use interpolation to estimate the median distance cycled.
- 3.In a histogram of the lengths, x cm, of some rods, the bar for 10 ≤ x < 30 has a frequency density of 3 per cm. The bar for 30 ≤ x < 45 is twice as tall as the bar for 10 ≤ x < 30. Work out the number of rods with a length in the class 30 ≤ x < 45.
- 4.In a histogram of the masses, m grams, of some pebbles, the bar for the class 50 ≤ m < 80 has a frequency density of 2.4 per gram. Work out the number of pebbles in this class.
- 5.A vet records the masses, m kg, of the dogs seen in one week as a histogram. The bar for 0 ≤ m < 5 has a frequency density of 4 per kg, the bar for 5 ≤ m < 15 has a frequency density of 2.6 per kg, and the bar for 15 ≤ m < 40 has a frequency density of 1.2 per kg. Work out the total number of dogs seen that week.
- 6.A cumulative frequency graph for the diameters, d mm, of 320 ball bearings is plotted from these points (upper class boundary, cumulative frequency): (10, 30), (20, 90), (30, 190), (40, 280), (50, 320). Estimate the diameter below which 90% of the ball bearings measure.
- 7.A bus company records the delay, d minutes, of 250 buses: 0 ≤ d < 2, 60 buses; 2 ≤ d < 5, 90 buses; 5 ≤ d < 10, 75 buses; 10 ≤ d < 20, 25 buses. The company refunds the fare whenever a bus is more than 8 minutes late. Estimate the number of refunds it must pay.
- 8.Marta is drawing a cumulative frequency diagram for the times, t seconds, of 100 telephone calls. The grouped frequencies are: 0 ≤ t < 10, 7 calls; 10 ≤ t < 20, 19 calls; 20 ≤ t < 30, 34 calls; 30 ≤ t < 40, 40 calls. Write down the coordinates of the point Marta should plot for the class 20 ≤ t < 30.
- 9.The heights, h cm, of 80 plants are grouped like this: 0 ≤ h < 20, 14 plants; 20 ≤ h < 40, 22 plants; 40 ≤ h < 50, 16 plants; 50 ≤ h < 80, 28 plants. Write down the class interval that contains the lower quartile.
- 10.In a histogram of the distances, d metres, thrown by some athletes, the bar covering 20 ≤ d < 60 has a constant frequency density of 1.8 per metre. Estimate the number of throws of at least 20 metres but less than 35 metres.
- 11.A histogram shows the times, t minutes, taken by 120 visitors to complete an escape room. The bar for 0 ≤ t < 10 has a frequency density of 5 visitors per minute, the bar for 10 ≤ t < 20 has a frequency density of 2 visitors per minute, the bar for 20 ≤ t < 40 has a frequency density of 1.5 visitors per minute, and the bar for 40 ≤ t < 60 has a frequency density of 1 visitor per minute. Work out which class contains the median time.
- 12.The times, t seconds, taken by 142 competitors to complete a lap are grouped like this: 0 ≤ t < 10, 20 competitors; 10 ≤ t < 25, 12 competitors; 25 ≤ t < 45, 50 competitors; 45 ≤ t < 75, 60 competitors. A histogram is drawn. Write down the class whose bar is the tallest.
- 13.In a histogram of the times, t minutes, taken by some people to complete a task, the class 15 ≤ t < 30 contains 24 people. Work out the frequency density for this class.
- 14.The times, t minutes, of 80 journeys are summarised by these cumulative frequencies: t < 10, 8 journeys; t < 20, 28 journeys; t < 30, 52 journeys; t < 40, 72 journeys; t < 50, 80 journeys. Estimate the interquartile range.
- 15.The masses, m grams, of 100 apples are grouped like this: 100 ≤ m < 120, 10 apples; 120 ≤ m < 140, 30 apples; 140 ≤ m < 160, 40 apples; 160 ≤ m < 200, 20 apples. Estimate the median mass.
Answer key
- (d) 44 — Method: a cumulative frequency counts everything below a value, so the frequency of a class is the running total at the top of the class minus the running total at the bottom of it. Working: the running total below 20 kg is 96 and the running total below 10 kg is 52, so the number of boxes in the class 10 ≤ m < 20 is 96 − 52 = 44. Answer: 44 boxes. The distractors: 96 comes from quoting the running total at 20 kg itself, which counts every box below 20 kg rather than only those in this class; 34 comes from subtracting the wrong pair, 52 − 18, which gives the class 5 ≤ m < 10 instead; 54 comes from subtracting from the grand total, 150 − 96, which gives the boxes of 20 kg or more.
- (c) 70 — Method: estimate the median from the cumulative frequency table by interpolation: find its position, n ÷ 2, locate the class it falls in, then add the fraction of the way through that class (adjusted for the cumulative frequency reached before it) to the class's lower boundary. Working: there are 180 riders, so the median is at position 180 ÷ 2 = 90. Before the class 60 ≤ d < 80 the cumulative frequency is 60, and by the end of it, 120, so the 90th rider falls in this class; its frequency is 120 − 60 = 60 and its width is 80 − 60 = 20. The extra distance needed into the class is 90 − 60 = 30, and 30 ÷ 60 × 20 = 10, so the median is 60 + 10 = 70. Answer: the estimated median distance is 70 km. Watch which numbers the interpolation actually uses: reading off just the class's lower boundary, 60, ignores how far into the class the 90th rider falls; using the target position, 90, as the extra distance instead of subtracting the 60 riders already counted before the class gives 90 ÷ 60 × 20 = 30, so 60 + 30 = 90, overshooting by treating the whole position as if none of it had already been counted; and using the total number of riders, 180, instead of half of it as the target position lands in the very last class, giving an estimate of 130 km — further than any rider is known to have ridden by that point in the table.
- (c) 90 — Method: the height of a bar is its frequency density, so twice as tall means twice the frequency density — not twice the frequency, because the two classes have different widths. Then frequency = frequency density × class width. Working: the first bar has frequency density 3 per cm, so the second has frequency density 2 × 3 = 6 per cm; the class 30 ≤ x < 45 is 45 − 30 = 15 cm wide, so its frequency is 6 × 15 = 90. Answer: 90 rods. The distractors: 120 comes from doubling the first bar's frequency instead of its height — the first class holds 3 × 20 = 60 rods, and doubling that ignores the fact that the second class is narrower; 45 comes from using the first bar's frequency density, 3, for the second bar, 3 × 15, and so never using the information that it is twice as tall; 6 comes from stopping at the frequency density of the taller bar and quoting a height as though it were a count.
- (c) 72 — Method: on a histogram the frequency of a class is the area of its bar, so frequency = frequency density × class width. Working: the class 50 ≤ m < 80 has width 80 − 50 = 30 grams and a frequency density of 2.4 per gram, so the frequency is 2.4 × 30 = 72. Answer: 72 pebbles. The distractors: 192 comes from using the upper class boundary, 80, as the width, giving 2.4 × 80; 12.5 comes from dividing the width by the density, 30 ÷ 2.4, which reverses the area rule; 2.4 comes from reading the height of the bar as the frequency itself, the commonest mistake on histograms, where a height is a density and only an area is a count.
- (c) 76 — Method: the frequency of each class is the area of its bar, frequency density × class width, so work out all three frequencies and add them. Working: the widths are 5, 10 and 25 kg, so the frequencies are 4 × 5 = 20, 2.6 × 10 = 26 and 1.2 × 25 = 30; the total is 20 + 26 + 30 = 76. Answer: 76 dogs. The distractors: 7.8 comes from adding the three frequency densities, 4 + 2.6 + 1.2, treating each height as though it were a count; 107 comes from using each upper class boundary as the width, giving 4 × 5, 2.6 × 15 and 1.2 × 40; 39 comes from using the first class width, 5, for every bar, which ignores the unequal intervals and gives 20 + 13 + 6.
- (c) 42 — Method: find the target cumulative frequency, 90% of the total, locate the class it falls in from the plotted points, then interpolate: lower boundary, plus the extra distance needed into the class divided by the class's frequency, times its width. Working: 90% of 320 is 0.9 × 320 = 288. The plotted points show a cumulative frequency of 280 at d = 40 and 320 at d = 50, so the class 40 ≤ d < 50 has frequency 320 − 280 = 40 and width 50 − 40 = 10, and 288 falls inside it. The extra distance needed into the class is 288 − 280 = 8, and 8 ÷ 40 × 10 = 2, so the diameter is 40 + 2 = 42. Answer: the estimated diameter is 42 mm. Watch which point and which class the interpolation actually uses: reading off d = 40, the plotted point just below the target, instead of interpolating the extra 8 ball bearings into the next 10 mm, stops one step short of the true answer; finding the diameter below which only 10% lie instead of 90% gives a target of 0.1 × 320 = 32, which falls in the class 10 ≤ d < 20 — the extra distance into that class is 32 − 30 = 2, and 2 ÷ 60 × 10 = 0.3, so this route gives 10 + 0.3 = 10.3, the bottom decile rather than the top 90%; and interpolating within the class 30 ≤ d < 40 instead of 40 ≤ d < 50, as though 288 had not yet reached a cumulative frequency of 280, treats the extra distance as 288 − 190 = 98, and 98 ÷ 90 × 10 = 10.9, giving 30 + 10.9 = 40.9, one class too early.
- (b) 55 — Method: count the classes that lie wholly above 8 minutes, then use linear interpolation for the class that 8 cuts through, assuming the delays in that class are spread evenly. Working: the class 10 ≤ d < 20 lies wholly above 8 and holds 25 buses; the value 8 lies in the class 5 ≤ d < 10, which is 5 minutes wide and holds 75 buses, and the part above 8 runs from 8 to 10, a width of 2, so the estimated share is (2 ÷ 5) × 75 = 30 buses; the estimate is 30 + 25 = 55. Answer: about 55 refunds. The distractors: 100 comes from adding the whole of the class 5 ≤ d < 10, 75 + 25, and so refunding buses only 5 minutes late; 25 comes from using only the class 10 ≤ d < 20 and ignoring the part class that 8 minutes cuts through; 70 comes from taking the part of the class from 5 up to 8 instead of from 8 up to 10, giving (3 ÷ 5) × 75 = 45 and then 45 + 25.
- (a) (30, 60) — Method: a cumulative frequency point is plotted at the upper boundary of its class, paired with the running total of all the frequencies up to and including that class. Working: the running totals are 7, then 7 + 19 = 26, then 26 + 34 = 60, then 60 + 40 = 100; the class 20 ≤ t < 30 has upper boundary 30, and the running total there is 60. Answer: the point for that class is plotted at 30 seconds against a cumulative frequency of 60. The distractors: (25, 60) comes from plotting at the class midpoint, which is what a frequency polygon uses and not what a cumulative frequency diagram uses; (30, 34) comes from plotting the class frequency, 34, rather than the running total; (20, 60) comes from plotting at the lower boundary of the class, which would claim that 60 calls took less than 20 seconds when only 26 did.
- (c) 20 ≤ h < 40 — Method: with 80 values the lower quartile is the 80 ÷ 4 = 20th value in order, so build a running total until it first reaches 20. Working: the running totals are 14, then 14 + 22 = 36, then 52, then 80; the 20th plant is past 14 but not past 36, so it lies in the second class. Answer: the lower quartile lies in the class 20 ≤ h < 40. The distractors: 0 ≤ h < 20 comes from believing that the bottom quarter of the data must all sit in the first class, when that class holds only 14 of the 80 plants; 40 ≤ h < 50 comes from using the position 80 ÷ 2 = 40 and so locating the median rather than the lower quartile; 50 ≤ h < 80 comes from counting 20 plants down from the tallest instead of up from the shortest, which locates the upper quartile at the 60th plant.
- (c) 27 — Method: a frequency is the area of the part of the bar being asked about, so frequency = frequency density × the width of that part. Working: the part asked about runs from 20 to 35, so its width is 35 − 20 = 15 metres; the frequency density there is 1.8 per metre, so the estimate is 1.8 × 15 = 27. Answer: about 27 throws. The distractors: 72 comes from taking the whole bar, 1.8 × 40, and so counting every throw from 20 up to 60; 1.8 comes from reading the height of the bar as a frequency, when a height is a density and only an area is a count; 63 comes from using the upper value 35 as the width, 1.8 × 35, instead of the width 35 − 20.
- (c) The class with times from 10 up to 20 — Method: to find the median class from a histogram, first turn each bar's frequency density into a frequency using density × class width, build up the cumulative frequency, and find the first class whose cumulative frequency reaches or passes n ÷ 2. Working: the four classes have widths 10, 10, 20 and 20, so their frequencies are 5 × 10 = 50, 2 × 10 = 20, 1.5 × 20 = 30 and 1 × 20 = 20, which add to the 120 visitors stated. The median sits at position 120 ÷ 2 = 60. The cumulative frequency is 50 after the first class and 50 + 20 = 70 after the second, so the 60th visitor is reached during the second class. Answer: the median lies in the class 10 ≤ t < 20. Watch which class each shortcut lands on: the tallest bar belongs to the first class, with the highest frequency density, 5 — but the tallest bar shows where visitors are packed most densely, not where the middle visitor falls, and picking it lands one class too early, at 0 ≤ t < 10; taking half of the total TIME span instead of half of the total NUMBER of visitors, 60 minutes ÷ 2 = 30 minutes, lands in the class 20 ≤ t < 40, confusing a value on the horizontal axis with a position in the data; and using the full 120 visitors as the target position, rather than 120 ÷ 2 = 60, reaches all the way to the last class, 40 ≤ t < 60, treating the whole data set's size as though it were the position of a single middle value.
- (d) 25 ≤ t < 45 — Method: the height of a bar on a histogram is the frequency density, so work out frequency ÷ class width for every class and compare the four heights. Working: 20 ÷ 10 = 2 for the first class; 12 ÷ 15 = 0.8 for the second; 50 ÷ 20 = 2.5 for the third; 60 ÷ 30 = 2 for the fourth. Answer: the largest of 2, 0.8, 2.5 and 2 is 2.5, so the tallest bar is the one for 25 ≤ t < 45. The distractors: 45 ≤ t < 75 comes from picking the class with the greatest frequency, 60, and treating a frequency as a height — but that class is three times as wide, so its 60 competitors are spread thinly; 10 ≤ t < 25 comes from dividing the class width by the frequency, 15 ÷ 12, and picking the largest of those reversed values; 0 ≤ t < 10 comes from assuming the narrowest class must always give the tallest bar, which is only true when the frequencies are equal.
- (a) 1.6 — Method: on a histogram the height of a bar is the frequency density, and frequency density = frequency ÷ class width. Working: the class 15 ≤ t < 30 runs from 15 to 30, so its width is 30 − 15 = 15 minutes; the frequency is 24, so the frequency density is 24 ÷ 15 = 1.6. Answer: 1.6 people per minute. The distractors: 360 comes from multiplying the frequency by the class width, 24 × 15, which uses the area rule backwards — area gives the frequency, so the frequency must be divided by the width to give the height; 0.625 comes from dividing the class width by the frequency, 15 ÷ 24, reversing the formula; 0.8 comes from dividing by the upper class boundary, 24 ÷ 30, instead of by the width of the class.
- (d) 18 minutes — Method: the lower quartile is the 80 ÷ 4 = 20th value and the upper quartile is the 3 × 80 ÷ 4 = 60th value; locate each inside its class by linear interpolation, then subtract. Working: the 20th value lies between the running totals 8 and 28, so it is in the class 10 ≤ t < 20, which holds 20 journeys across 10 minutes, and it is the 20 − 8 = 12th of them, giving 10 + (12 ÷ 20) × 10 = 16 minutes; the 60th value lies between the running totals 52 and 72, so it is in the class 30 ≤ t < 40, which also holds 20 journeys across 10 minutes, and it is the 60 − 52 = 8th of them, giving 30 + (8 ÷ 20) × 10 = 34 minutes; subtracting, 34 − 16 = 18. Answer: an estimated interquartile range of 18 minutes. The distractors: 20 minutes comes from taking the lower boundaries of the two quartile classes, 30 − 10, which locates the classes but never the values inside them; 40 minutes comes from subtracting the two positions, 60 − 20, instead of the two times; 22 minutes comes from interpolating downwards from each upper boundary rather than upwards from each lower boundary, giving 20 − 6 = 14 and 40 − 4 = 36.
- (c) 145 g — Method: find the position of the median from the total frequency, locate the class that contains it, then use linear interpolation inside that class, assuming the apples in it are spread evenly. Working: the median is the 100 ÷ 2 = 50th apple; the running totals are 10, then 10 + 30 = 40, then 40 + 40 = 80, so the 50th apple lies in the class 140 ≤ m < 160; it is the 50 − 40 = 10th of the 40 apples in that class, and the class is 20 g wide, so the median is 140 + (10 ÷ 40) × 20 = 140 + 5 = 145. Answer: an estimated median of 145 g. The distractors: 150 g comes from giving the midpoint of the class that contains the median instead of interpolating inside it; 140 g comes from stopping at the lower boundary of that class, which locates the class but not the value; 155 g comes from measuring the 5 g step down from the upper boundary, 160 − 5, instead of up from the lower boundary.
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