Original exam-style questions with full mark schemes. Use the filters to choose level and difficulty, and tick questions to build a printable test.
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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
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1IGMC-049-00-01Averages, range and frequency tablesPaper CEasy6 marks
No calculator
Ten students recorded the number of hours they spent reading in one week. 7, 3, 9, 4, 7, 12, 5, 7, 8, 6
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Find the range.
(1)
(d)
Calculate the mean.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 7 hours
✓ 1
Part (b)
M1 ordering 3, 4, 5, 6, 7, 7, 7, 8, 9, 12 and taking the 5th and 6th values ; A1 7 hours
✓ 2
Part (c)
B1 12 − 3 = 9 hours
✓ 1
Part (d)
M1 total 68 ÷ 10 ; A1 6.8 hours
✓ 2
Exam tip: Always write the data in order before finding the median. With an even number of values, the median is halfway between the two middle values.
2IGMC-049-00-02Averages, range and frequency tablesPaper CEasy6 marks
Calculator
25 households were asked how many pets they own. The results are shown below. Number of pets: 0, 1, 2, 3, 4 Frequency: 5, 8, 7, 4, 1
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Calculate the mean number of pets per household.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 1 pet
✓ 1
Part (b)
M1 the median is the 13th value, or cumulative frequencies 5, 13, … ; A1 1 pet
Common error: Dividing by 5, the number of categories, instead of 25, the total frequency. The mean must lie between the smallest and largest values, 0 and 4.
3IGMC-049-00-03Averages, range and frequency tablesPaper CEasy4 marks
No calculator
The mean of five numbers is 14. Four of the numbers are 10, 12, 15 and 18.
(a)
Find the fifth number.
(2)
(b)
A sixth number is added. The mean of all six numbers is now 15. Find the sixth number.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5 × 14 = 70 ; A1 70 − 55 = 15
✓ 2
Part (b)
M1 6 × 15 = 90 ; A1 90 − 70 = 20
✓ 2
Exam tip: When a mean and the number of values are given, multiply them first to find the total. Most missing-value problems start from this total.
4IGMC-049-00-04Averages, range and frequency tablesPaper CEasy5 marks
No calculator
Ana and Ben each played five basketball games. Their points in each game are shown below. Ana: 12, 18, 15, 20, 10 Ben: 14, 16, 15, 17, 13
(a)
Calculate the mean number of points for each player.
(2)
(b)
Find the range of points for each player.
(2)
(c)
Use your answers to compare the performances of the two players.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 Ana 75 ÷ 5 = 15 ; B1 Ben 75 ÷ 5 = 15
✓ 2
Part (b)
B1 Ana 20 − 10 = 10 ; B1 Ben 17 − 13 = 4
✓ 2
Part (c)
B1 Both players have the same mean, but Ben's range is smaller, so Ben's scoring is more consistent
✓ 1
Exam tip: A good comparison uses one average and one measure of spread, and says what each means in the context of the question.
5IGMC-049-00-05Averages, range and frequency tablesPaper CEasy5 marks
Calculator
The annual salaries, in dollars, of the seven workers at a small company are shown below. 21 000, 23 000, 22 000, 24 000, 21 000, 25 000, 98 000
(a)
Find the median salary.
(2)
(b)
Calculate the mean salary. Give your answer correct to 3 significant figures.
(2)
(c)
Explain why the median is a better average than the mean for these salaries.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 ordering the salaries (21 000, 21 000, 22 000, 23 000, 24 000, 25 000, 98 000) and selecting the 4th value ; A1 23 000 dollars
✓ 2
Part (b)
M1 234 000 ÷ 7 (= 33428.57…) ; A1 33 400 dollars
✓ 2
Part (c)
B1 The mean is pulled up by the one very large value of 98 000 dollars (it is higher than six of the seven salaries), while the median is not affected by it
✓ 1
Common error: Choosing the mean just because it uses every value. An extreme value (outlier) pulls the mean away from most of the data, so the median is more representative here.
6IGMC-049-01-01Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy4 marks
No calculator
In parts (a) to (c), state the type of correlation you would expect between each pair of variables. Then answer part (d).
(a)
The daily maximum temperature at a beach and the number of ice lollies sold that day.
(1)
(b)
The average outside temperature in a month and the cost of heating a house that month.
(1)
(c)
The house number of an adult and their height in centimetres.
(1)
(d)
A survey of children aged 5 to 11 shows strong positive correlation between shoe size and reading score. Mina says, "Having bigger feet makes children better at reading." Explain why Mina is wrong.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 Positive (correlation)
✓ 1
Part (b)
B1 Negative (correlation)
✓ 1
Part (c)
B1 No correlation
✓ 1
Part (d)
B1 One valid explanation. Either state that correlation does not show that one variable causes the other, or state that a third factor (age) makes both increase, e.g. older children have bigger feet and also read better
✓ 1
Common error: A strong correlation never proves that one variable causes the other. Look for a third factor, such as age, that affects both.
7IGMC-049-01-02Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy6 marks
No calculator
The table shows the number of hours, h, that 8 students spent revising and their test mark, m.
Write down the age and value of the car that does not fit the general trend.
(1)
(b)
Ignoring this car, what type of correlation is shown?
(1)
(c)
Describe the relationship between the age and the value of these cars.
(1)
(d)
Suggest one reason why the car in part (a) might have a higher value than expected.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 Age 5 years, value $19 800
✓ 1
Part (b)
B1 Negative
✓ 1
Part (c)
B1 The older the car, the lower its value
✓ 1
Part (d)
B1 Any sensible reason, e.g. very low mileage, excellent condition, rare extras
✓ 1
Exam tip: When asked to describe a relationship, write a sentence in context using both variables, not just the word 'negative'.
9IGMC-049-00-06Averages, range and frequency tablesPaper CMedium7 marks
Calculator
The table shows the times, t minutes, taken by 40 workers to travel to work. 0 < t ≤ 10: 4 10 < t ≤ 20: 11 20 < t ≤ 30: 15 30 < t ≤ 40: 8 40 < t ≤ 60: 2
(a)
Write down the modal class.
(1)
(b)
Write down the class interval that contains the median.
(1)
(c)
Calculate an estimate of the mean time.
(4)
(d)
Explain why your answer to the previous part is only an estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 20 < t ≤ 30
✓ 1
Part (b)
B1 20 < t ≤ 30 (cumulative frequencies 4, 15, 30 show that the 20th and 21st values lie in this class)
B1 The exact times are not known; the midpoint is used to represent every value in each class
✓ 1
Common error: Using the upper class boundary or the class width instead of the midpoint. The last class is wider than the others, so its midpoint is 50, not 45.
10IGMC-049-00-08Averages, range and frequency tablesPaper CMedium5 marks
Calculator
Class A has 18 students and their mean test mark is 64. Class B has 12 students and their mean test mark is 74.
(a)
Calculate the mean mark for all 30 students.
(3)
(b)
One student's mark in Class A was recorded as 47 instead of the correct mark of 74. Calculate the correct mean mark for Class A.
Common error: Giving the range as 23 − 2 = 21 by using the frequencies. The range is worked out from the data values (number of people), not from the frequencies.
12IGMC-049-00-12Averages, range and frequency tablesPaper CMedium4 marks
Calculator
A chess club has 20 members with a mean age of 31 years. Two members, aged 45 and 52, leave the club. Three new members join, and their mean age is 24 years.
(a)
Find the total of the ages of the original 20 members.
A line of best fit is drawn through the points (500, 18) and (2500, 6).
(a)
Describe the correlation shown by the data.
(1)
(b)
Calculate the gradient of the line of best fit.
(2)
(c)
Interpret the gradient in the context of the question.
(1)
(d)
Find the equation of the line of best fit in the form T = ah + b.
(1)
(e)
Use your equation to estimate the temperature at an altitude of 1800 m.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 (Strong) negative
✓ 1
Part (b)
M1 (6 − 18) ÷ (2500 − 500)
✓ 1
A1 −0.006
✓ 1
Part (c)
B1 Temperature falls by 0.006 °C for each metre of altitude (6 °C per 1000 m)
✓ 1
Part (d)
B1 T = −0.006h + 21
✓ 1
Part (e)
B1 10.2 °C
✓ 1
Common error: Keep the minus sign in the gradient. A falling line must have a negative gradient, and the y-intercept follows from substituting one point.
14IGMC-049-01-05Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium6 marks
Calculator
A shop records the rainfall, r mm, and the number of umbrellas sold, u, on 6 days.
r: 2, 5, 8, 12, 15, 18 u: 9, 14, 20, 27, 31, 37
A line of best fit passes through the mean point and the point (0, 4).
(a)
Calculate the coordinates of the mean point.
(2)
(b)
Find the gradient of the line of best fit.
(2)
(c)
Use the line of best fit to estimate the number of umbrellas sold on a day with 14 mm of rainfall. Give your answer to the nearest whole number.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 mean r = 60 ÷ 6 = 10
✓ 1
B1 mean u = 138 ÷ 6 = 23
✓ 1
Part (b)
M1 (23 − 4) ÷ (10 − 0)
✓ 1
A1 1.9
✓ 1
Part (c)
M1 4 + 1.9 × 14 (= 30.6)
✓ 1
A1 31
✓ 1
Exam tip: The mean point is found by averaging each variable separately. Plotting it first gives a reliable anchor for the line of best fit.
15IGMC-049-01-13Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium5 marks
Calculator
A student tested a phone battery. The line of best fit for the remaining charge, C%, against the screen-on time, t hours, has equation C = 100 − 12.5t. The data were collected for values of t from 0 to 7.
(a)
What type of correlation does this line show?
(1)
(b)
Explain what the number 12.5 represents.
(1)
(c)
Estimate the screen-on time when the remaining charge is 30%.
(2)
(d)
Explain why the equation should not be used to predict the charge after 10 hours.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 Negative
✓ 1
Part (b)
B1 The charge falls by 12.5 percentage points for each hour of screen-on time
✓ 1
Part (c)
M1 100 − 12.5t = 30 or 70 ÷ 12.5
✓ 1
A1 5.6 hours
✓ 1
Part (d)
B1 10 is outside the data range / gives −25%, which is impossible
✓ 1
Exam tip: When the output value is given, substitute it into the equation and solve for the input, rather than guessing.
16IGMC-049-00-09Averages, range and frequency tablesPaper CHard5 marks
No calculator
Four positive integers have a mode of 6, a median of 7 and a mean of 8.
(a)
Find the four integers. Show your reasoning.
(4)
(b)
Write down the range of the four integers.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 total = 4 × 8 = 32 ; B1 in order a ≤ b ≤ c ≤ d, the median gives b + c = 14, so c cannot be 6 and the two 6s (needed for the mode) must be a and b ; M1 c = 14 − 6 = 8 and d = 32 − 6 − 6 − 8 ; A1 6, 6, 8, 12
✓ 4
Part (b)
B1 12 − 6 = 6
✓ 1
Exam tip: Write the unknown values in order as boxes: _ ≤ _ ≤ _ ≤ _. Fill in what each average tells you, one condition at a time, then check all three averages at the end.
17IGMC-049-01-14Scatter diagrams, correlation, cumulative frequency and histogramsPaper CHard7 marks
Calculator
A gym records the daily maximum temperature, x °C, and the number of bottles of water sold, y, on 6 days.
x: 3, 6, 8, k, 13, 16 y: 20, 26, 31, 35, 44, 48
The mean temperature is 9.5 °C.
(a)
Find the value of k.
(2)
(b)
Calculate the mean number of bottles sold.
(1)
(c)
A line of best fit passes through the mean point and (0, 15). Find its equation in the form y = mx + c.
(2)
(d)
A seventh day, with x = 20 and y = 60, is added to the data. Find the new mean point, giving the coordinates correct to 3 significant figures where necessary.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 6 × 9.5 = 57
✓ 1
A1 k = 11
✓ 1
Part (b)
B1 34
✓ 1
Part (c)
M1 (34 − 15) ÷ 9.5
✓ 1
A1 y = 2x + 15
✓ 1
Part (d)
M1 (57 + 20) ÷ 7 or (204 + 60) ÷ 7
✓ 1
A1 (11, 37.7)
✓ 1
Exam tip: Use mean × number of values to get the total, then subtract the known values to find a missing one.
18IGMC-049-01-06Scatter diagrams, correlation, cumulative frequency and histogramsPaper EEasy4 marks
No calculator
The times, t minutes, that 50 people took to complete a puzzle are shown.
0 < t ≤ 10: 5 10 < t ≤ 20: 12 20 < t ≤ 30: 18 30 < t ≤ 40: 10 40 < t ≤ 50: 5
(a)
Complete a cumulative frequency table for t ≤ 10, t ≤ 20, t ≤ 30, t ≤ 40 and t ≤ 50.
(2)
(b)
Write down the class interval that contains the median time.
(1)
(c)
How many people took more than 30 minutes?
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 5, 17, 35
✓ 1
B1 45, 50
✓ 1
Part (b)
B1 20 < t ≤ 30
✓ 1
Part (c)
B1 15
✓ 1
Common error: Cumulative frequencies are plotted at the upper class boundary (10, 20, 30, …), not at the midpoint of each class.
19IGMC-049-01-09Scatter diagrams, correlation, cumulative frequency and histogramsPaper EEasy3 marks
No calculator
The heights, h cm, of 83 seedlings are shown.
0 < h ≤ 10: 12 10 < h ≤ 15: 18 15 < h ≤ 20: 25 20 < h ≤ 30: 20 30 < h ≤ 50: 8
The data are to be shown in a histogram.
(a)
Calculate the frequency density for each class.
(2)
(b)
Write down the class with the tallest bar.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 1.2, 3.6, 5
✓ 1
B1 2, 0.4
✓ 1
Part (b)
B1 15 < h ≤ 20
✓ 1
Common error: The tallest bar is the class with the greatest frequency density, not necessarily the greatest frequency.
20IGMC-049-00-07Averages, range and frequency tablesPaper EMedium7 marks
No calculator
The table shows the number of goals scored by a team in each match of a season. Goals: 0, 1, 2, 3, 4 Frequency: 4, x, 6, 5, 2 The mean number of goals per match is 1.75.
(a)
Find the value of x.
(4)
(b)
Write down the mode.
(1)
(c)
Find the median.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Σfx = x + 35 or total frequency = 17 + x ; M1 (x + 35)/(17 + x) = 1.75 ; M1 x + 35 = 29.75 + 1.75x leading to 0.75x = 5.25 ; A1 x = 7
✓ 4
Part (b)
B1 1 goal
✓ 1
Part (c)
M1 24 matches, so the median is between the 12th and 13th values; cumulative frequencies 4, 11, 17 ; A1 2 goals
✓ 2
Exam tip: Write separate expressions for the total frequency and for Σfx in terms of x, then form one equation. Check that x comes out as a whole number, because it is a frequency.
21IGMC-049-00-11Averages, range and frequency tablesPaper EMedium5 marks
No calculator
Five numbers are x, x + 4, 2x, 2x + 3, 3x − 1 The mean of the five numbers is 12.
(a)
Find the value of x.
(3)
(b)
Find the median of the five numbers.
(1)
(c)
Find the range of the five numbers.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum = 9x + 6 ; M1 (9x + 6)/5 = 12 or 9x + 6 = 60 ; A1 x = 6
✓ 3
Part (b)
B1 12 (from 6, 10, 12, 15, 17)
✓ 1
Part (c)
B1 17 − 6 = 11
✓ 1
Exam tip: After finding x, work out all five actual numbers and put them in order. The algebraic order is not always the numerical order.
22IGMC-049-01-07Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium6 marks
Calculator
The reaction times of 80 students in Year 9 and 80 students in Year 11 were measured. Information read from the cumulative frequency curves is given below.
Year 9: median 0.31 s, lower quartile 0.26 s, upper quartile 0.38 s Year 11: median 0.27 s, lower quartile 0.24 s, upper quartile 0.29 s
(a)
Find the interquartile range for each year group.
(2)
(b)
Make two comparisons between the reaction times of the two year groups.
(2)
(c)
For Year 9, the cumulative frequency for a time of 0.35 s is 52. Calculate the percentage of Year 9 students with a reaction time greater than 0.35 s.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 Year 9: 0.12 s
✓ 1
B1 Year 11: 0.05 s
✓ 1
Part (b)
B1 Year 11 have faster reaction times on average (lower median)
✓ 1
B1 Year 11 times are more consistent (smaller IQR)
✓ 1
Part (c)
M1 (80 − 52) ÷ 80 × 100
✓ 1
A1 35%
✓ 1
Exam tip: For comparisons, use one average (median) and one measure of spread (IQR), and say what each means in context.
23IGMC-049-01-08Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The cumulative frequency table shows the masses, m grams, of 120 apples.
m ≤ 100: 8 m ≤ 120: 30 m ≤ 140: 72 m ≤ 160: 104 m ≤ 180: 120
Assume the masses are evenly spread within each class.
(a)
How many apples have a mass greater than 120 g but not more than 160 g?
(1)
(b)
Estimate the median mass.
(3)
(c)
The heaviest 20% of apples are labelled "large". Estimate the minimum mass of a large apple.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 74
✓ 1
Part (b)
M1 uses 60th apple, in class 120 < m ≤ 140
✓ 1
M1 120 + (60 − 30) ÷ 42 × 20
✓ 1
A1 134 g (134.3 g)
✓ 1
Part (c)
M1 0.8 × 120 = 96th apple
✓ 1
M1 140 + (96 − 72) ÷ 32 × 20
✓ 1
A1 155 g
✓ 1
Common error: The heaviest 20% start at the 80th percentile, so use 96 apples, not 24.
24IGMC-049-01-10Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
A histogram shows the ages, a years, of visitors to a museum one morning. The frequency densities are:
0 < a ≤ 10: 1.5 10 < a ≤ 25: 4 25 < a ≤ 40: 3.2 40 < a ≤ 70: 1.0
(a)
Find the frequency for each class.
(2)
(b)
Estimate the number of visitors aged over 50.
(1)
(c)
Calculate an estimate of the mean age of the visitors.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 15 and 60
✓ 1
B1 48 and 30
✓ 1
Part (b)
B1 20
✓ 1
Part (c)
M1 midpoints 5, 17.5, 32.5, 55
✓ 1
M1 Σfx = 75 + 1050 + 1560 + 1650 = 4335
✓ 1
M1 ÷ 153 (dependent on previous M1)
✓ 1
A1 28.3 years
✓ 1
Exam tip: Frequency = frequency density × class width. Divide Σfx by the total frequency, not by the number of classes.
25IGMC-049-00-13Averages, range and frequency tablesPaper EHard6 marks
Calculator
The heights, h cm, of 50 plants are summarised below. 0 < h ≤ 10: 5 10 < h ≤ 20: p 20 < h ≤ 30: 18 30 < h ≤ 40: q 40 < h ≤ 50: 4 An estimate of the mean height is 25.6 cm.
(a)
Find the values of p and q.
(5)
(b)
Write down the class interval that contains the median height.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 p + q = 23 ; M1 using midpoints: 5×5 + 15p + 25×18 + 35q + 45×4 ; M1 655 + 15p + 35q = 50 × 25.6 = 1280, so 15p + 35q = 625 ; M1 solving the simultaneous equations correctly, e.g. 15(23 − q) + 35q = 625 giving 20q = 280 ; A1 p = 9 and q = 14
✓ 5
Part (b)
B1 20 < h ≤ 30
✓ 1
Exam tip: Two unknowns need two equations: one from the total frequency and one from the estimated mean. Check by substituting: 25 + 135 + 450 + 490 + 180 = 1280.
26IGMC-049-00-14Averages, range and frequency tablesPaper EHard8 marks
No calculator
Five positive integers have • a mean of 10 • a median of 10 • a mode of 13 (the only mode) • a range of 8.
(a)
Find the five integers.
(4)
(b)
A sixth positive integer, n, is added to the list. The median of the six integers is 11. Find n.
(2)
(c)
Find the mean of the six integers. Give your answer as a mixed number.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 total = 5 × 10 = 50 ; B1 middle value is 10 and the two largest values are 13 and 13 ; M1 smallest value = 13 − 8 = 5 and second value = 50 − 5 − 10 − 13 − 13 ; A1 5, 9, 10, 13, 13
✓ 4
Part (b)
M1 median is the mean of the 3rd and 4th values, so n lies between 10 and 13 and (10 + n)/2 = 11 ; A1 n = 12
✓ 2
Part (c)
M1 (50 + 12) ÷ 6 ; A1 \(10\frac{1}{3}\)
✓ 2
Exam tip: Check your final list against every condition. For example, if the second value had come out as 10, both 10 and 13 would appear twice, and 13 would no longer be the only mode.
27IGMC-049-01-11Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard8 marks
Calculator
The speeds, v km/h, of 120 cars were recorded.
30 < v ≤ 40: 14 40 < v ≤ 50: p 50 < v ≤ 55: 26 55 < v ≤ 60: q 60 < v ≤ 80: 12
In a histogram of the data, the bar for 40 < v ≤ 50 has frequency density 3.8.
(a)
Find p.
(1)
(b)
Find q and the frequency density for the class 55 < v ≤ 60.
(3)
(c)
Estimate the number of cars travelling faster than 52 km/h.
(3)
(d)
Express your answer to part (c) as a percentage of all the cars.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 p = 3.8 × 10 = 38
✓ 1
Part (b)
M1 120 − (14 + 38 + 26 + 12)
✓ 1
A1 q = 30
✓ 1
B1 frequency density 6 (follow through q ÷ 5)
✓ 1
Part (c)
M1 3/5 × 26 = 15.6
✓ 1
M1 15.6 + 30 + 12 (follow through their q)
✓ 1
A1 57.6 (accept 58)
✓ 1
Part (d)
B1 48% (accept 48.3% from using 58; follow through their (c) ÷ 120 × 100)
✓ 1
Exam tip: To estimate part of a class, take the fraction of the class width you need and multiply it by that class's frequency.
28IGMC-049-01-12Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard7 marks
Calculator
The cumulative frequency table shows the distances, d km, that 200 workers travel to work.
d ≤ 2: 24 d ≤ 5: 84 d ≤ 10: 164 d ≤ 20: 192 d ≤ 40: 200
Assume the distances are evenly spread within each class.
(a)
Calculate the frequency densities needed to draw a histogram, for classes 0 < d ≤ 2, 2 < d ≤ 5, 5 < d ≤ 10, 10 < d ≤ 20 and 20 < d ≤ 40.
(3)
(b)
Estimate the interquartile range of the distances.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequencies 24, 60, 80, 28, 8
✓ 1
M1 frequency ÷ class width
✓ 1
A1 12, 20, 16, 2.8, 0.4
✓ 1
Part (b)
M1 uses 50th and 150th values
✓ 1
M1 LQ = 2 + (50 − 24) ÷ 60 × 3 = 3.3
✓ 1
M1 UQ = 5 + (150 − 84) ÷ 80 × 5 = 9.125
✓ 1
A1 5.83 km
✓ 1
Common error: Do not use cumulative frequencies as heights in a histogram. First find each class frequency by subtracting, then divide by the class width.
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