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
MathsUKwww.geekhero.co.uk
- 1.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.
- 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 heights, h cm, of 90 seedlings, the class 12 ≤ h < 18 contains 36 seedlings. Work out the frequency density for this class.
- 4.The times, t minutes, taken by 80 people to travel to work are grouped like this: 0 ≤ t < 10, 6 people; 10 ≤ t < 20, 14 people; 20 ≤ t < 30, 25 people; 30 ≤ t < 40, 20 people; 40 ≤ t < 50, 15 people. Work out the cumulative frequency for t < 30.
- 5.The masses, m kg, of 80 sacks of grain are summarised by these cumulative frequencies: m < 10, 6 sacks; m < 20, 22 sacks; m < 30, 58 sacks; m < 40, 74 sacks; m < 50, 80 sacks. Use interpolation to estimate the median mass.
- 6.A gym draws a histogram of the times, t minutes, that its members spend on one machine. The bar for 0 ≤ t < 10 has a frequency density of 1.8 per minute, the bar for 10 ≤ t < 25 has a frequency density of 3.2 per minute, and the bar for 25 ≤ t < 55 has a frequency density of 0.9 per minute. Members who spend 10 minutes or more on the machine pay an extra charge. Work out the number of members who pay the extra charge.
- 7.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.
- 8.Forty pupils in class P and forty pupils in class Q each solved a puzzle. The times, in seconds, were summarised using cumulative frequency. For class P the lower quartile is 24, the median is 38 and the upper quartile is 46. For class Q the lower quartile is 30, the median is 35 and the upper quartile is 44. Write down the statement that correctly compares the two classes.
- 9.The times, t minutes, taken by 120 runners to finish a fun run are summarised by these cumulative frequencies: t < 20, 8 runners; t < 30, 26 runners; t < 40, 74 runners; t < 50, 110 runners; t < 60, 120 runners. Work out the number of runners who took 40 minutes or longer to finish.
- 10.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.
- 11.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.
- 12.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.
- 13.The times, t minutes, of 70 visits to a website are grouped into two classes: 0 ≤ t < 4, which contains 30 visits, and 4 ≤ t < 20, which contains 40 visits. A histogram is drawn. Work out how many times taller the bar for 0 ≤ t < 4 is than the bar for 4 ≤ t < 20.
- 14.A histogram shows the speeds, v mph, of 100 vehicles passing a checkpoint. The bar for 0 ≤ v < 20 has a frequency density of 1 vehicle per mph, the bar for 20 ≤ v < 30 has a frequency density of 3 vehicles per mph, the bar for 30 ≤ v < 50 has a frequency density of 2 vehicles per mph, and the bar for 50 ≤ v < 70 has a frequency density of 0.5 vehicles per mph. Estimate the mean speed of the vehicles.
- 15.A histogram shows the ages, in years, of 250 members of a running club. The bar for the class 10 ≤ age < 20 has a frequency density of 4.5 members per year, the bar for 20 ≤ age < 35 has a frequency density of 6 members per year, and the bar for 50 ≤ age < 70 has a frequency density of 2.75 members per year. Work out the frequency of the remaining class, 35 ≤ age < 50.
Answer key
- (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) 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.
- (b) 6 — Method: frequency density = frequency ÷ class width. Working: the class 12 ≤ h < 18 has width 18 − 12 = 6, so frequency density = 36 ÷ 6 = 6. Answer: the frequency density is 6 seedlings per cm. Watch which numbers you use: taking the lower bound, 12, as the width instead of 18 − 12 = 6 gives 36 ÷ 12 = 3; dividing the total number of seedlings, 90, rather than this class's frequency, 36, by the width gives 90 ÷ 6 = 15, a density that belongs to no single class; and multiplying instead of dividing gives 36 × 6 = 216, far too large a density for so narrow a class.
- (d) 45 — Method: a cumulative frequency is a running total — it counts everybody in every class up to and including the one that ends at the value given. Working: the classes that lie wholly below 30 minutes are 0 ≤ t < 10, 10 ≤ t < 20 and 20 ≤ t < 30, with frequencies 6, 14 and 25, so the running total is 6 + 14 = 20 and then 20 + 25 = 45. Answer: 45 people took less than 30 minutes. The distractors: 25 comes from quoting the frequency of the class 20 ≤ t < 30 on its own instead of the running total; 65 comes from accumulating one class too many and including 30 ≤ t < 40, which is 45 + 20; 35 comes from accumulating from the top downwards, 15 + 20, which counts the people who took 30 minutes or more rather than fewer.
- (d) 25 — Method: the median is estimated at position n ÷ 2 in the cumulative frequency table, then interpolated across the class it falls in: lower boundary, plus the fraction of the way through the class, times the class width. Working: there are 80 sacks, so the median sits at position 80 ÷ 2 = 40. Before the class 20 ≤ m < 30 the cumulative frequency is 22, and by the end of it, it is 58, so this class holds the 40th sack; its frequency is 58 − 22 = 36 and its width is 30 − 20 = 10. The extra distance needed into the class is 40 − 22 = 18, and 18 ÷ 36 × 10 = 5, so the median is 20 + 5 = 25. Answer: the estimated median mass is 25 kg. Watch which numbers the interpolation uses: reading off just the lower boundary of the median class, 20, ignores how far into that class the 40th sack actually falls; treating n ÷ 2 = 40 itself as the median mass mistakes a position in the list for a mass in kilograms; and using the target position, 40, as the extra distance into the class instead of subtracting the sacks already counted changes the calculation to 20 + 40 ÷ 36 × 10. That comes to 20 + 11.1 = 31.1, overshooting the class because it never subtracts the 22 sacks already counted before it.
- (c) 75 — Method: the number in a class is the area of its bar, frequency density × class width, so work out the frequency of each class that lies at or above 10 minutes and add them. Working: the class 10 ≤ t < 25 is 15 minutes wide with a frequency density of 3.2, giving 3.2 × 15 = 48 members; the class 25 ≤ t < 55 is 30 minutes wide with a frequency density of 0.9, giving 0.9 × 30 = 27 members; the total charged is 48 + 27 = 75. Answer: 75 members pay the extra charge. The distractors: 4.1 comes from adding the two frequency densities, 3.2 + 0.9, as though each height were a count; 93 comes from including the class 0 ≤ t < 10 as well, 1.8 × 10 = 18 added to 48 and 27, which charges every member; 27 comes from using only the class 25 ≤ t < 55 and forgetting that 10 ≤ t < 25 is also at or above 10 minutes.
- (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.
- (b) Q was faster on average and more consistent — Method: compare the medians for the average and the interquartile ranges for the spread, remembering that a shorter time is faster and a smaller interquartile range means more consistent. Working: the median for class Q is 35 seconds against 38 seconds for class P, so class Q was faster on average; the interquartile range for class P is 46 − 24 = 22 seconds and for class Q it is 44 − 30 = 14 seconds, so class Q's times are more tightly grouped. Answer: class Q was faster on average and more consistent. The distractors: calling Q slower comes from comparing the lower quartiles, 30 against 24, as though a quartile were the average; calling Q less consistent comes from using the gap between the median and the upper quartile as the spread, 44 − 35 = 9 against 46 − 38 = 8, instead of the full interquartile range; the statement that Q was both slower and less consistent comes from making both of those mistakes together.
- (c) 46 — Method: the cumulative frequency table gives the number of runners below each time; to find the number at or above a time, subtract that cumulative frequency from the total. Working: the cumulative frequency for t < 40 is 74, so 120 runners in total take away the 74 who finished in under 40 minutes: 120 − 74 = 46. Answer: 46 runners took 40 minutes or longer. Watch which boundary and which subtraction you use: reading off t < 50 instead of t < 40 and subtracting, 120 − 110 = 10, answers a different question, '50 minutes or longer'; giving 74 itself as the answer reports how many finished below 40 minutes, the opposite of what was asked; and subtracting the two nearby cumulative frequencies, 110 − 74 = 36, finds how many took between 40 and 50 minutes, not everyone from 40 minutes upward.
- (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.
- (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) 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) 3 — Method: the height of a bar is its frequency density, frequency ÷ class width, so work out both heights and divide one by the other. Working: the class 0 ≤ t < 4 is 4 minutes wide and holds 30 visits, so its frequency density is 30 ÷ 4 = 7.5 per minute; the class 4 ≤ t < 20 is 16 minutes wide and holds 40 visits, so its frequency density is 40 ÷ 16 = 2.5 per minute; dividing the heights, 7.5 ÷ 2.5 = 3. Answer: the first bar is 3 times as tall. The distractors: 0.75 comes from comparing the frequencies, 30 ÷ 40, as though the frequencies were the heights, which is the mistake the unequal widths are there to expose; 4 comes from comparing the class widths, 16 ÷ 4, instead of the heights; 5 comes from subtracting the two frequency densities, 7.5 − 2.5, which answers how much taller rather than how many times taller.
- (a) 31.5 — Method: to estimate the mean from a histogram, first turn each bar into a frequency (frequency density × class width), then use mean = Σ(frequency × midpoint) ÷ Σfrequency, with the midpoint standing in for every value in that class. Working: the four classes have widths 20, 10, 20 and 20, so their frequencies are 1 × 20 = 20, 3 × 10 = 30, 2 × 20 = 40 and 0.5 × 20 = 10, which do add to the 100 vehicles stated. Their midpoints are 10, 25, 40 and 60, so Σfx = 20 × 10 + 30 × 25 + 40 × 40 + 10 × 60 = 200 + 750 + 1600 + 600 = 3150, and the mean is 3150 ÷ 100 = 31.5. Answer: the estimated mean speed is 31.5 mph. Watch which numbers you treat as the frequencies and which as the values: using the frequency densities themselves as the frequencies, without multiplying by the class widths first, gives 1 × 10 + 3 × 25 + 2 × 40 + 0.5 × 60 = 195 spread over 1 + 3 + 2 + 0.5 = 6.5, and 195 ÷ 6.5 = 30, a mean built from the wrong 'frequencies' altogether; averaging the four midpoints on their own, (10 + 25 + 40 + 60) ÷ 4 = 33.75, ignores how many vehicles are actually in each class; and using each class's lower boundary in place of its midpoint, 20 × 0 + 30 × 20 + 40 × 30 + 10 × 50 = 2300 and 2300 ÷ 100 = 23, systematically underestimates every class by roughly half its width.
- (d) 60 — Method: on a histogram the frequency of a class is its frequency density × its class width, and the frequencies of all the classes add up to the total, so turn each labelled bar into a frequency and subtract their total from 250. Working: 10 ≤ age < 20 has width 20 − 10 = 10, so its frequency is 4.5 × 10 = 45; 20 ≤ age < 35 has width 35 − 20 = 15, so its frequency is 6 × 15 = 90; 50 ≤ age < 70 has width 70 − 50 = 20, so its frequency is 2.75 × 20 = 55. Those three come to 45 + 90 + 55 = 190, and the total is 250, so the missing frequency is 250 − 190 = 60. Answer: the class 35 ≤ age < 50 has 60 members. Watch what you do with the total and the three frequencies you have found: giving the total, 250, as the answer forgets that three bars have already accounted for some of the members; giving 190, the total of the other three classes, reports how many members are not in this class rather than how many are; and leaving one of the three out of the subtraction, for example 45 + 90 = 135 and 250 − 135 = 115, still owes the class at 50 ≤ age < 70 its 55 members.
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