234 free, high-quality past-paper-style questions on Statistics for Cambridge IGCSE Maths, each with a full mark scheme. Covers: Averages, range and frequency tables; Scatter diagrams, correlation, cumulative frequency and histograms. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.1,501 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
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'.
9IGMC3-9-06d-01Averages, range and frequency tablesPaper CEasy6 marks
No calculator
The delays, in minutes, of eight rocket launch attempts were: 12, 7, 15, 7, 20, 9, 7, 13
(a)
Write down the mode.
(1)
(b)
Find the range.
(1)
(c)
Find the median.
(2)
(d)
Find the mean.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
7
✓ 1
Part (b)
20 − 7 = 13
✓ 1
Part (c)
M1 values in order 7, 7, 7, 9, 12, 13, 15, 20
✓ 1
A1 10.5
✓ 1
Part (d)
M1 sum = 90
✓ 1
A1 90 ÷ 8 = 11.25
✓ 1
Common error: Sort the data before choosing the median; with an even number of values, take the mean of the two middle ones.
10IGMC3-9-06d-02Averages, range and frequency tablesPaper CEasy7 marks
Calculator
A technician inspects 40 solar panels and records the number of faults on each. Number of faults: 0, 1, 2, 3, 4 Number of panels: 9, 14, 10, 5, 2
(a)
Show that 40 panels were inspected.
(1)
(b)
Write down the mode.
(1)
(c)
Find the median number of faults.
(2)
(d)
Calculate the mean number of faults per panel.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
9 + 14 + 10 + 5 + 2 = 40
✓ 1
Part (b)
1
✓ 1
Part (c)
M1 20th and 21st values both lie in the group with cumulative total 23
✓ 1
A1 1
✓ 1
Part (d)
M1 0×9 + 1×14 + 2×10 + 3×5 + 4×2
✓ 1
A1 57
✓ 1
A1 57 ÷ 40 = 1.425
✓ 1
Exam tip: For a frequency table, multiply each value by its frequency, add the products, then divide by the total frequency.
11IGMC3-9-06d-03Averages, range and frequency tablesPaper CEasy5 marks
Calculator
Five athletes ran a sprint. Four of the times, in seconds, were 11.2, 10.8, 11.5 and 10.9. The fifth time was x seconds. The mean of all five times was 11.0 seconds.
Exam tip: Given a mean, find the total by multiplying the mean by the number of values.
12IGMC3-9-06d-04Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy6 marks
Calculator
A scientist records the mean daily temperature, t °C, and the heating energy used, E kWh, in a building on six days. (t, E): (2, 41), (5, 36), (8, 30), (11, 27), (14, 20), (17, 16) A line of best fit for the data has equation E = −1.6t + 44.
Scatter diagram of the data with the line of best fit
(a)
Describe the correlation between temperature and energy used.
(1)
(b)
Find the mean of the six temperatures.
(2)
(c)
The scientist uses the line to estimate the energy used on a day with a mean temperature of 30 °C. Explain why this estimate may be unreliable.
(1)
(d)
Use the line to estimate the energy used on a day with a mean temperature of 10 °C.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Negative (correlation)
✓ 1
Part (b)
M1 sum = 57
✓ 1
A1 9.5
✓ 1
Part (c)
30 °C is far outside the range of the data (extrapolation)
✓ 1
Part (d)
M1 −1.6 × 10 + 44
✓ 1
A1 28
✓ 1
Common error: Estimates inside the data range (interpolation) are more reliable than estimates outside it (extrapolation).
13IGMC3-9-06d-05Averages, range and frequency tablesPaper CEasy5 marks
Calculator
The table shows the delivery times, t minutes, of 50 drone deliveries. 0 < t ≤ 10: 6 deliveries; 10 < t ≤ 20: 14; 20 < t ≤ 30: 18; 30 < t ≤ 40: 10; 40 < t ≤ 50: 2.
(a)
Write down the modal class.
(1)
(b)
Explain why a mean calculated from this table is only an estimate.
(1)
(c)
Calculate an estimate of the mean delivery time.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
20 < t ≤ 30
✓ 1
Part (b)
The exact times are not known, only the class they fall in
✓ 1
Part (c)
M1 midpoints 5, 15, 25, 35, 45
✓ 1
M1 6×5 + 14×15 + 18×25 + 10×35 + 2×45 = 1130
✓ 1
A1 1130 ÷ 50 = 22.6
✓ 1
Exam tip: Use the midpoint of each class as the representative value when estimating a mean from grouped data.
14IGMC3-9-08r7-01Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy4 marks
No calculator
A pharmacist records the dose, in mg, of a painkiller given to seven patients: 250, 500, 250, 750, 500, 250, 1000.
Exam tip: Put the data in order before finding the median.
15IGMC3-9-08r7-02Averages, range and frequency tablesPaper CEasy6 marks
Calculator
A games company records how many levels 30 players completed in one session.
Levels completed
1
2
3
4
5
Frequency
4
7
9
6
4
(a)
Write down the modal number of levels completed.
(1)
(b)
Find the median.
(2)
(c)
Calculate the mean number of levels. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 3
✓ 1
Part (b)
M1 15th and 16th values both lie in the group for 3 levels (cumulative 4, 11, 20)
✓ 1
A1 3
✓ 1
Part (c)
M1 sum of frequency × levels = 4 + 14 + 27 + 24 + 20
✓ 1
A1 89
✓ 1
A1 89 ÷ 30 = 2.97
✓ 1
Common error: The median is the middle value of the data, not the middle of the frequency column.
16IGMC3-9-08r7-03Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy4 marks
Calculator
A water company records the volume of water, in megalitres (ML), pumped to a town on six days: 42, 47, 51, x, 45, 49. The mean volume for the six days is 47 ML.
(a)
Find the value of x.
(2)
(b)
On the seventh day the mean for all seven days becomes 47.5 ML. Find the volume pumped on the seventh day.
Exam tip: Work with totals: total = mean × number of values.
17IGMC3-9-08r7-04Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy4 marks
No calculator
A gamer records the hours played per week (x) and the rank points gained (y) for eight players. The hours range from 1 to 9. The points rise steadily as the hours increase.
(a)
Write down the type of correlation shown.
(1)
(b)
A line of best fit has equation y = 12x + 40. Use it to estimate the points gained by a player who plays for 6.5 hours.
(2)
(c)
Explain why the line should not be used to estimate the points for a player who plays 30 hours.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 positive correlation
✓ 1
Part (b)
M1 12 × 6.5 + 40
✓ 1
A1 118
✓ 1
Part (c)
R1 30 is far outside the range of the data (1 to 9), so the trend may not continue (extrapolation)
✓ 1
Common error: Estimates are only reliable inside the range of the data.
18IGMC3-9-08r13-01Averages, range and frequency tablesPaper CEasy6 marks
No calculator
A pharmacist records the number of tablets left in the packs of ten patients after one week: 4, 7, 2, 7, 5, 9, 7, 3, 6, 10.
(a)
Write down the mode.
(1)
(b)
Find the range.
(1)
(c)
Find the median.
(2)
(d)
Find the mean.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 7
✓ 1
Part (b)
B1 10 − 2 = 8
✓ 1
Part (c)
M1 ordered list 2, 3, 4, 5, 6, 7, 7, 7, 9, 10 or middle values 6 and 7 identified
✓ 1
A1 6.5
✓ 1
Part (d)
M1 sum = 60
✓ 1
A1 60 ÷ 10 = 6
✓ 1
Exam tip: With an even number of values, the median is the mean of the two middle values once the data are in order.
19IGMC3-9-08r13-02Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy5 marks
No calculator
Eight students record the number of hours they game each night (x) and the hours of sleep they get (y). The results (x, y) are: (0.5, 8.8), (1, 8.3), (1.5, 8.1), (2, 7.9), (2.5, 7.4), (3, 7.3), (3.5, 6.8), (4, 6.5). A line of best fit has equation y = 9.0 − 0.6x.
Scatter diagram with the line y = 9.0 − 0.6x
(a)
Describe the correlation between hours gaming and hours of sleep.
(1)
(b)
Use the equation to estimate the hours of sleep for a student who games for 4 hours.
(2)
(c)
Use the equation to estimate the hours of sleep for a student who games for 12 hours.
(1)
(d)
Explain why the estimate in part (c) is unlikely to be reliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 negative (correlation)
✓ 1
Part (b)
M1 9.0 − 0.6 × 4
✓ 1
A1 6.6 hours
✓ 1
Part (c)
B1 1.8 hours
✓ 1
Part (d)
R1 12 hours is outside the range of the data (extrapolation) / the prediction is unrealistic
✓ 1
Common error: Do not use a line of best fit to predict far outside the range of the data. The pattern may not continue.
20IGMC3-9-08r13-03Averages, range and frequency tablesPaper CEasy5 marks
Calculator
A water company records the number of burst pipes in its network each week for 40 weeks. Number of bursts: 0, 1, 2, 3, 4 Frequency: 6, 11, f, 7, 3
(a)
Find the value of f.
(2)
(b)
Find the median number of bursts.
(1)
(c)
Calculate the mean number of bursts per week.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 40 − (6 + 11 + 7 + 3)
✓ 1
A1 13
✓ 1
Part (b)
B1 2 (cumulative frequencies 6, 17, 30, so the 20th and 21st values are both 2)
✓ 1
Part (c)
M1 0×6 + 1×11 + 2×13 + 3×7 + 4×3 = 70
✓ 1
A1 70 ÷ 40 = 1.75
✓ 1
Exam tip: For the mean from a frequency table, multiply each value by its frequency, add them up, and divide by the total frequency (not by the number of rows).
21IGMC3-9-08r13-08Averages, range and frequency tablesPaper CEasy5 marks
Calculator
A household's water use (litres) over six days has mean 214. Five of the readings are 198, 225, 210, 240 and 205.
Common error: The mode is the faults value with the highest frequency (0), not the highest frequency itself (12).
25IGMC3-9-10r14-04Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy4 marks
Calculator
A wind farm records the wind speed and the power output of its turbines. The data show that turbines produce more power as the wind speed rises. A line of best fit drawn by eye passes through the points (4 m/s, 60 kW) and (10 m/s, 300 kW).
(a)
Describe the correlation between wind speed and power output.
(1)
(b)
A manager uses the line of best fit to predict the power at 30 m/s. Give one reason why this prediction may be unreliable.
(1)
(c)
Use the line of best fit to estimate the power output at 8 m/s.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 positive correlation
✓ 1
Part (b)
B1 30 m/s is far outside the range of the data (extrapolation) [or turbines shut down in very strong wind]
✓ 1
Part (c)
M1 gradient (300 − 60) ÷ (10 − 4) = 40 kW per m/s, or equivalent
✓ 1
A1 220 kW
✓ 1
Common error: Correlation describes a relationship in the data; it does not by itself show that one quantity causes the other.
26IGMC3-9-09r16-01Averages, range and frequency tablesPaper CEasy6 marks
Calculator
A space probe records the time, in minutes, taken for a radio signal to reach a ground station on eight separate days: 14 9 12 17 9 13 9 19
(a)
Write down the mode.
(1)
(b)
Work out the range.
(1)
(c)
Find the median.
(2)
(d)
Calculate the mean.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
9
✓ 1
Part (b)
19 − 9 = 10
✓ 1
Part (c)
M1 ordering the data: 9, 9, 9, 12, 13, 14, 17, 19
✓ 1
A1 median = (12 + 13) ÷ 2 = 12.5
✓ 1
Part (d)
M1 sum of data = 102 and divided by 8
✓ 1
A1 12.75
✓ 1
Exam tip: Put the data in order before finding the median, and with an even number of values take the mean of the two middle ones.
27IGMC3-9-09r16-02Averages, range and frequency tablesPaper CEasy5 marks
Calculator
A coffee roaster records the number of batches roasted on each of 25 days.
Batches per day
1
2
3
4
5
Number of days
2
5
9
6
3
(a)
Write down the mode.
(1)
(b)
Find the median number of batches per day.
(2)
(c)
Calculate the mean number of batches per day. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
3
✓ 1
Part (b)
M1 identifying the 13th value (cumulative frequencies 2, 7, 16)
✓ 1
A1 3
✓ 1
Part (c)
M1 sum of (batches × days) = 2 + 10 + 27 + 24 + 15 = 78
✓ 1
A1 78 ÷ 25 = 3.12
✓ 1
Common error: Do not divide the sum of the frequencies by 5 (the number of rows); divide the total of batches × days by the total number of days.
28IGMC3-9-09r16-05Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy5 marks
Calculator
A roaster records the roasting time, t minutes, and the mass lost, m grams per 100 g of beans, for five batches.
t
10
12
14
16
18
m
10.6
12.1
13.4
15.0
16.4
Scatter diagram of the data
(a)
Describe the correlation between t and m.
(1)
(b)
Find the mean point (mean of t, mean of m).
(2)
(c)
A line of best fit passes through the mean point and has gradient 0.7. Use it to estimate the mass lost when the roasting time is 15 minutes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Positive correlation (as time increases, mass lost increases)
✓ 1
Part (b)
A1 mean t = 14
✓ 1
A1 mean m = 67.5 ÷ 5 = 13.5, so (14, 13.5)
✓ 1
Part (c)
M1 13.5 + 0.7 × (15 − 14)
✓ 1
A1 14.2 g
✓ 1
Exam tip: A line of best fit always passes through the mean point, which gives a reliable starting point for estimates.
29IGMC3-9-09r16-09Averages, range and frequency tablesPaper CEasy5 marks
Calculator
Five sacks of green coffee beans have masses 62 kg, 58 kg, x kg, 65 kg and 60 kg. The mean mass is 61 kg.
(a)
Write down the range of the four known masses.
(1)
(b)
Show that the total mass of the five sacks is 305 kg.
(1)
(c)
Find x.
(1)
(d)
A sixth sack of mass 66 kg is added. Calculate the new mean mass, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
65 − 58 = 7 kg
✓ 1
Part (b)
5 × 61 = 305 kg
✓ 1
Part (c)
305 − (62 + 58 + 65 + 60) = 60
✓ 1
Part (d)
M1 (305 + 66) ÷ 6
✓ 1
A1 61.8 kg
✓ 1
Exam tip: When the mean is given, multiply it by the number of values to find the total.
30IGMC3-9-08r19-01Averages, range and frequency tablesPaper CEasy5 marks
No calculator
An archaeologist records the depths, in cm, of seven pottery fragments below the surface: 13, 7, 18, 4, 9, 7, 12.
(a)
Work out the mean depth.
(2)
(b)
Find the median depth.
(1)
(c)
Write down the mode.
(1)
(d)
Work out the range.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum = 70
✓ 1
A1 mean = 10 cm
✓ 1
Part (b)
B1 9 cm
✓ 1
Part (c)
B1 7 cm
✓ 1
Part (d)
B1 18 − 4 = 14 cm
✓ 1
Exam tip: Put the data in order before finding the median.
31IGMC3-9-08r19-02Averages, range and frequency tablesPaper CEasy6 marks
Calculator
A cycling club records the number of punctures each rider had on 40 rides.
Punctures
0
1
2
3
Frequency
12
15
8
5
(a)
Show that the total number of rides is 40.
(1)
(b)
Write down the mode.
(1)
(c)
Find the median.
(2)
(d)
Calculate the mean number of punctures per ride.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 12 + 15 + 8 + 5 = 40
✓ 1
Part (b)
B1 1
✓ 1
Part (c)
M1 20th and 21st values are in the group for 1 puncture (cumulative frequencies 12 then 27)
✓ 1
A1 1
✓ 1
Part (d)
M1 (0×12 + 1×15 + 2×8 + 3×5) / 40
✓ 1
A1 1.15
✓ 1
Common error: The mode is the value with the highest frequency, not the highest frequency itself.
32IGMC3-9-08r19-03Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy4 marks
No calculator
A data centre manager records the outside temperature, x °C, and the energy used by the cooling system, y kWh, on ten days. The temperatures were between 14 °C and 32 °C. The points lie close to a straight line that slopes upwards. The line of best fit is y = 1.8x + 12.
(a)
Describe the correlation between x and y.
(1)
(b)
Use the line of best fit to estimate the energy used on a day when the temperature is 25 °C.
(2)
(c)
The manager uses the line to predict the energy used when the temperature is 50 °C. Explain why this prediction may be unreliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 (strong) positive
✓ 1
Part (b)
M1 1.8 × 25 + 12
✓ 1
A1 57 kWh
✓ 1
Part (c)
B1 50 °C is outside the range of the data (extrapolation)
✓ 1
Exam tip: Comment on reliability by referring to the range of the data collected.
33IGMC3-9-08r19-07Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy4 marks
Calculator
The number of minutes of unplanned downtime on seven days in a data centre was 2, 3, 3, 4, 5, 6, 95.
(a)
Calculate the mean downtime, correct to 3 significant figures.
(2)
(b)
Find the median downtime.
(1)
(c)
The manager wants a typical value for the downtime. Which average is better, the mean or the median? Give a reason.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 118 / 7
✓ 1
A1 16.9 minutes
✓ 1
Part (b)
B1 4 minutes
✓ 1
Part (c)
B1 median, because the mean is distorted by the extreme value 95
✓ 1
Common error: Always check for an extreme value before choosing which average represents the data.
34IGMC3-9-08r25-01Averages, range and frequency tablesPaper CEasy5 marks
No calculator
The delays, in minutes, of five flights leaving an airport were: 8, 2, 12, 6, 2
(a)
Write down the mode.
(1)
(b)
Find the median.
(1)
(c)
Find the range.
(1)
(d)
Calculate the mean delay.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 2
✓ 1
Part (b)
B1 6
✓ 1
Part (c)
B1 10
✓ 1
Part (d)
M1 (8 + 2 + 12 + 6 + 2) ÷ 5 or 30 ÷ 5
✓ 1
A1 6 minutes
✓ 1
Common error: Put the data in order before finding the median, and remember the range is largest minus smallest.
35IGMC3-9-08r25-02Averages, range and frequency tablesPaper CEasy7 marks
Calculator
A music streaming service records the lengths, in minutes, of nine tracks on an album: 3.2, 4.1, 2.8, 3.5, 5.0, 3.5, 4.4, 2.9, 3.6
(a)
Find the median length.
(2)
(b)
Find the range.
(1)
(c)
Calculate the mean length. Give your answer correct to 3 significant figures.
(2)
(d)
A tenth track of length 12.0 minutes is added to the album. Calculate the new mean length.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 values in order: 2.8, 2.9, 3.2, 3.5, 3.5, 3.6, 4.1, 4.4, 5.0, or identifies the 5th value
✓ 1
A1 3.5 minutes
✓ 1
Part (b)
B1 2.2 minutes
✓ 1
Part (c)
M1 33.0 ÷ 9
✓ 1
A1 3.67 minutes
✓ 1
Part (d)
M1 (33.0 + 12.0) ÷ 10
✓ 1
A1 4.5 minutes
✓ 1
Exam tip: To update a mean, add to the total of the data and divide by the new number of values. Use the unrounded total.
36IGMC3-9-08r25-07Averages, range and frequency tablesPaper CEasy6 marks
Calculator
A farmer compares two varieties of wheat, P and Q, over five years. The yields, in tonnes per hectare, are: Variety P: 6.1, 5.8, 6.4, 6.0, 6.2 Variety Q: 5.2, 7.0, 6.5, 5.5, 6.3 The mean yield of variety Q is 6.1 tonnes per hectare.
(a)
Calculate the mean yield of variety P.
(2)
(b)
Find the range of the yields of variety P.
(1)
(c)
Find the range of the yields of variety Q.
(1)
(d)
The farmer wants a variety that gives reliable yields. Which variety should she choose? Use the means and the ranges to explain.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 30.5 ÷ 5
✓ 1
A1 6.1 tonnes per hectare
✓ 1
Part (b)
B1 0.6
✓ 1
Part (c)
B1 1.8
✓ 1
Part (d)
B1 variety P
✓ 1
B1 the means are equal but P has the smaller range, so its yields are more consistent
✓ 1
Exam tip: When comparing two data sets, always comment on both an average and a measure of spread, in context.
37IGMC3-9-12r29-01Averages, range and frequency tablesPaper CEasy6 marks
Calculator
A clinic recorded the number of new infections on seven consecutive days: 12, 15, 9, 15, 20, 11, 16.
(a)
Write down the mode.
(1)
(b)
Find the range.
(1)
(c)
Find the median.
(2)
(d)
Calculate the mean.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 15
✓ 1
Part (b)
B1 11
✓ 1
Part (c)
M1 values in order: 9, 11, 12, 15, 15, 16, 20
✓ 1
A1 15
✓ 1
Part (d)
M1 (12 + 15 + 9 + 15 + 20 + 11 + 16)/7 = 98/7
✓ 1
A1 14
✓ 1
Common error: Put the data in order before choosing the median; the middle value of the list as given is usually wrong.
38IGMC3-9-12r29-02Averages, range and frequency tablesPaper CEasy6 marks
Calculator
A supermarket checked the number of items in the baskets of 50 customers. Items: 1, 2, 3, 4, 5, 6 Customers: 8, 12, 15, 10, 3, 2
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Calculate the mean number of items per basket.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 3
✓ 1
Part (b)
M1 25th and 26th values (cumulative frequencies 8, 20, 35)
✓ 1
A1 3
✓ 1
Part (c)
M1 products: 8, 24, 45, 40, 15, 12
✓ 1
M1 sum of products = 144, divided by 50
✓ 1
A1 2.88
✓ 1
Exam tip: For a frequency table, multiply each value by its frequency and divide by the total frequency, not by the number of rows.
39IGMC3-9-12r29-03Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy7 marks
Calculator
An engineer measured the power output P kW of a wind turbine at different wind speeds v m/s. v: 4, 6, 8, 10, 12 P: 30, 190, 420, 610, 790 A line of best fit is drawn through the points (6, 200) and (10, 600).
Scatter diagram with the line of best fit through (6, 200) and (10, 600)
(a)
Describe the correlation between v and P.
(1)
(b)
Calculate the gradient of the line of best fit.
(2)
(c)
Find the equation of the line of best fit in the form P = mv + c.
(2)
(d)
Use the line to estimate the power output when the wind speed is 8.5 m/s.
(1)
(e)
Explain why the line should not be used to estimate the power output at 25 m/s.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 positive correlation (P increases as v increases)
✓ 1
Part (b)
M1 (600 − 200)/(10 − 6)
✓ 1
A1 100
✓ 1
Part (c)
M1 P − 200 = 100(v − 6)
✓ 1
A1 P = 100v − 400
✓ 1
Part (d)
A1 450 kW (ECF from (c))
✓ 1
Part (e)
B1 25 m/s is outside the range of the data (extrapolation), so the relationship may not continue
✓ 1
Common error: Describe correlation using the direction (positive or negative), not just 'it goes up' or 'it is strong'.
40IGMC3-9-12r29-10Averages, range and frequency tablesPaper CEasy5 marks
Calculator
At a clinic, 12 patients had a mean waiting time of 38 minutes. A second group of 8 patients had a mean waiting time of 43 minutes.
(a)
Calculate the total waiting time of the first group.
(1)
(b)
Calculate the mean waiting time of all 20 patients.
(2)
(c)
A 21st patient joins, and the mean waiting time of all 21 patients is now 41 minutes. Calculate the waiting time of the 21st patient.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 456 minutes
✓ 1
Part (b)
M1 (456 + 344)/20
✓ 1
A1 40 minutes
✓ 1
Part (c)
M1 41 × 21 = 861 (total for 21 patients)
✓ 1
A1 861 − 800 = 61 minutes
✓ 1
Exam tip: To combine groups, work with totals (mean × number), never by averaging the two means.
41IGMC3-9-11r31-01Averages, range and frequency tablesPaper CEasy6 marks
Calculator
A clinic records how many tablets each of 40 patients missed during one week of a treatment course. Number of tablets missed: 0, 1, 2, 3, 4 Number of patients: 12, 15, 9, 3, 1
(a)
Write down the mode.
(1)
(b)
Find the median number of tablets missed.
(2)
(c)
Calculate the mean number of tablets missed per patient.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 1
✓ 1
Part (b)
M1 the 20th and 21st values are needed (or cumulative frequencies 12, 27 shown)
✓ 1
A1 1
✓ 1
Part (c)
M1 attempt at \(\sum fx\) with at least three products correct (15, 18, 9, 4)
✓ 1
M1 divide their total by 40
✓ 1
A1 1.15
✓ 1
Common error: In a frequency table, the mean is the total of (value × frequency) divided by the total frequency, not the total of the values divided by the number of columns.
42IGMC3-9-11r31-02Averages, range and frequency tablesPaper CEasy7 marks
No calculator
Seven players record their best scores (in thousands of points) in a puzzle game: 18, 9, 29, 12, 18, 15, 11.
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Work out the range.
(1)
(d)
An eighth player joins. The mean of all eight scores is 17 thousand points. Find the eighth player's score.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 18
✓ 1
Part (b)
M1 arrange in order (9, 11, 12, 15, 18, 18, 29) or identify the 4th value
✓ 1
A1 15
✓ 1
Part (c)
B1 20
✓ 1
Part (d)
M1 total of first seven scores is 112 (accept mean = 16)
✓ 1
M1 \(8\times17 - 112\)
✓ 1
A1 24
✓ 1
Exam tip: Put the data in order before reading off the median, and use total = mean × number of values when a missing value is needed.
43IGMC3-9-11r31-12Averages, range and frequency tablesPaper CEasy4 marks
No calculator
Five gamers have scores 8, 15, 10, 13 and x. The mean of the five scores is 12.
(a)
Find the value of x.
(2)
(b)
Write down the median of the five scores.
(1)
(c)
Work out the range of the five scores.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total \(= 5\times12 = 60\) and \(60 - (8+15+10+13)\)
✓ 1
A1 14
✓ 1
Part (b)
B1 13
✓ 1
Part (c)
B1 7
✓ 1
Exam tip: Use total = mean × number of values to find a missing data value.
44IGMC3-9-11r31-14Averages, range and frequency tablesPaper CEasy6 marks
Calculator
A survey asks 25 teenagers how many hours of video games they played yesterday. Hours: 0, 1, 2, 3, 4 Number of teenagers: 3, 8, x, 5, 2
(a)
Find the value of x.
(1)
(b)
Find the median number of hours.
(2)
(c)
Calculate the mean number of hours.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 7
✓ 1
Part (b)
M1 the 13th value is needed (cumulative frequencies 3, 11, 18 seen)
Exam tip: Find the missing frequency from the total first, then use cumulative frequencies to locate the median.
45IGMC3-9-12r34-01Averages, range and frequency tablesPaper CEasy7 marks
No calculator
The numbers of books returned to a library on ten days were 14, 18, 11, 18, 22, 15, 18, 12, 20, 12.
(a)
Work out the mean.
(2)
(b)
Find the median.
(2)
(c)
Write down the mode.
(1)
(d)
Work out the range.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum = 160 and divide by 10
✓ 1
A1 16
✓ 1
Part (b)
M1 ordered list 11, 12, 12, 14, 15, 18, 18, 18, 20, 22 with middle values 15 and 18
✓ 1
A1 16.5
✓ 1
Part (c)
B1 18
✓ 1
Part (d)
M1 22 − 11
✓ 1
A1 11
✓ 1
Common error: With an even number of values the median is halfway between the two middle values; it is not just one of them.
46IGMC3-9-12r34-02Averages, range and frequency tablesPaper CEasy5 marks
Calculator
At a swimming club, 12 swimmers in group A have a mean 100 m time of 64.5 seconds. 8 swimmers in group B have a mean 100 m time of 67.0 seconds.
(a)
Work out the total of the times for group A.
(2)
(b)
Work out the mean time for all 20 swimmers.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 12 × 64.5
✓ 1
A1 774 seconds
✓ 1
Part (b)
M1 total for group B = 8 × 67.0 = 536
✓ 1
M1 (774 + 536) ÷ 20 (ECF from (a))
✓ 1
A1 65.5 seconds
✓ 1
Common error: The mean of two means (64.5 + 67) ÷ 2 is only right when the groups are the same size; use totals instead.
47IGMC3-9-12r34-03Averages, range and frequency tablesPaper CEasy6 marks
Calculator
A wind farm has 30 turbines. The table shows the number of faults recorded for each turbine in one year.
Faults
0
1
2
3
4
Turbines
12
9
5
3
1
(a)
Write down the mode.
(1)
(b)
Find the median number of faults.
(2)
(c)
Calculate the mean number of faults per turbine, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0
✓ 1
Part (b)
M1 15th and 16th values (cumulative totals 12, 21)
✓ 1
A1 1
✓ 1
Part (c)
M1 (0×12 + 1×9 + 2×5 + 3×3 + 4×1) = 32
✓ 1
M1 32 ÷ 30
✓ 1
A1 1.07
✓ 1
Exam tip: For a frequency table, multiply each value by its frequency and divide by the total frequency, not by the number of columns.
48IGMC3-9-12r34-04Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy6 marks
Calculator
A scatter diagram shows the power output of a wind turbine (in kW) against wind speed (in m/s). A line of best fit is drawn. It passes through the points (6, 150) and (14, 550).
(a)
Describe the correlation shown by the data.
(1)
(b)
Calculate the gradient of the line of best fit and state its units.
(2)
(c)
Use the line of best fit to estimate the power output when the wind speed is 10 m/s.
(2)
(d)
Explain why the line should not be used to estimate the output at 30 m/s.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 positive correlation (output increases as wind speed increases)
✓ 1
Part (b)
M1 (550 − 150) ÷ (14 − 6)
✓ 1
A1 50 kW per m/s
✓ 1
Part (c)
M1 150 + 4 × 50 (or equivalent)
✓ 1
A1 350 kW
✓ 1
Part (d)
B1 outside the range of the data (extrapolation), so the pattern may not continue
✓ 1
Exam tip: Quote the correlation type (positive, negative or none) and add the context in words to earn the mark.
49IGMC3-9-10r36-01Averages, range and frequency tablesPaper CEasy6 marks
Calculator
An archaeologist records the mass, in grams, of nine pottery fragments found in one trench. 12, 15, 9, 15, 20, 11, 15, 18, 17
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Find the range.
(1)
(d)
Calculate the mean. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 15
✓ 1
Part (b)
M1 puts the values in order (or identifies the 5th value)
✓ 1
A1 15
✓ 1
Part (c)
B1 20 − 9 = 11
✓ 1
Part (d)
M1 132 ÷ 9
✓ 1
A1 14.7
✓ 1
Exam tip: Write the values in order before finding the median; it is easy to pick the middle of the unsorted list by mistake.
50IGMC3-9-10r36-02Averages, range and frequency tablesPaper CEasy6 marks
Calculator
The frequency table shows the number of rides per week made by each of 30 members of a cycling club. Rides per week: 1, 2, 3, 4, 5 Number of members: 4, 7, 9, 6, 4
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Calculate the mean number of rides per week. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 3
✓ 1
Part (b)
M1 15th and 16th values both lie in the group for 3 rides, or cumulative frequencies 4, 11, 20
✓ 1
A1 3
✓ 1
Part (c)
M1 at least 3 products of rides × frequency correct
✓ 1
M1 (4 + 14 + 27 + 24 + 20) ÷ 30
✓ 1
A1 2.97
✓ 1
Common error: To find the mean from a frequency table, divide the total of (value × frequency) by the total frequency, not by the number of rows.
51IGMC3-9-10r36-03Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy6 marks
Calculator
A data centre manager studies how its energy use depends on other factors.
(a)
For each pair of quantities, state the type of correlation you would expect on a scatter diagram. (i) the number of servers installed and the electrical power drawn (ii) the age of a cooling unit and its efficiency (iii) the shoe size of the technicians and the number of server failures each month
(3)
(b)
The manager draws a line of best fit for the daily cooling energy, E MWh, against the outside temperature, T °C. The line has equation E = 2.3T + 18. Use the line to estimate the cooling energy when the outside temperature is 28 °C.
(2)
(c)
The manager uses the line to predict the cooling energy for an outside temperature of 60 °C. Give a reason why this prediction may not be reliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 (i) positive
✓ 1
B1 (ii) negative
✓ 1
B1 (iii) no correlation
✓ 1
Part (b)
M1 2.3 × 28 + 18
✓ 1
A1 82.4
✓ 1
Part (c)
B1 60 °C is outside the range of the data, so this is extrapolation
✓ 1
Common error: Correlation describes a relationship in the data; a line of best fit should only be used inside the range of the data it was drawn for.
52IGMC3-9-10r36-04Averages, range and frequency tablesPaper CEasy5 marks
Calculator
The table shows the lengths, l cm, of 50 animal bones found at an excavation. 10 < l ≤ 20: 6 20 < l ≤ 30: 14 30 < l ≤ 40: 18 40 < l ≤ 50: 9 50 < l ≤ 60: 3
(a)
Write down the modal class.
(1)
(b)
Calculate an estimate of the mean length.
(3)
(c)
Explain why your answer to (b) is only an estimate.
Exam tip: Write the data in order before you look for the median.
54IGMC3-9-11r38-02Averages, range and frequency tablesPaper CEasy6 marks
Calculator
A fishing boat records the number of fish of a protected species found in its net on each of 24 trips. Number of fish: 0, 1, 2, 3, 4 Frequency: 3, 5, 8, 6, 2
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Calculate the mean number of fish per trip. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 2
✓ 1
Part (b)
M1 12th and 13th values both lie in the group for 2 fish (cumulative frequencies 3, 8, 16)
✓ 1
A1 2
✓ 1
Part (c)
M1 products of number and frequency found, total 47
✓ 1
M1 47 ÷ 24
✓ 1
A1 1.96
✓ 1
Common error: For a frequency table, dividing the total of the frequencies by the number of groups instead of using (number × frequency) ÷ total frequency.
55IGMC3-9-11r38-03Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy5 marks
No calculator
A fashion shop records the daily maximum temperature, T °C, and the number of sun hats sold, N, on 12 days. On a scatter diagram the points lie close to a straight line sloping upwards from bottom left to top right. A line of best fit passes through the points (10, 12) and (30, 52).
(a)
Write down the type of correlation shown.
(1)
(b)
Describe what this tells you about hat sales and temperature.
(1)
(c)
Use the line of best fit to estimate N when T = 22.
(2)
(d)
The manager uses the line to estimate sales when T = 45. Give one reason why this estimate may be unreliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 positive (correlation)
✓ 1
Part (b)
B1 on hotter days more sun hats are sold (as T increases, N increases)
✓ 1
Part (c)
M1 gradient (52 − 12) ÷ (30 − 10) = 2 used from the point (10, 12), or equivalent reading at T = 22
✓ 1
A1 36
✓ 1
Part (d)
B1 45 is outside the range of the data (extrapolation), so the trend may not continue
✓ 1
Exam tip: Use the two points you are given to find the gradient; do not rely on reading values by eye.
56IGMC3-9-11r38-04Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy7 marks
Calculator
Five delivery robots have a mean battery life of 8.2 hours. Three older robots have a mean battery life of 7.0 hours.
(a)
Work out the total battery life of the five robots.
(2)
(b)
Find the mean battery life of all eight robots.
(2)
(c)
A ninth robot is added. The mean battery life of all nine robots is exactly 7.8 hours. Find the battery life of the ninth robot.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5 × 8.2
✓ 1
A1 41 hours
✓ 1
Part (b)
M1 (41 + 3 × 7.0) ÷ 8
✓ 1
A1 7.75 hours
✓ 1
Part (c)
M1 total for nine robots = 9 × 7.8 = 70.2
✓ 1
M1 70.2 − 62 (total of the eight robots)
✓ 1
A1 8.2 hours
✓ 1
Common error: Averaging two means directly. Each mean must be weighted by the number of items in its group.
57IGMC3-9-11r43-01Averages, range and frequency tablesPaper CEasy6 marks
No calculator
A pharmacist records the dose, in mg, of a drug given to eight patients. 12 15 9 15 20 11 15 13
(a)
Write down the mode.
(1)
(b)
Work out the range.
(1)
(c)
Find the median.
(2)
(d)
Calculate the mean dose.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 15
✓ 1
Part (b)
B1 11
✓ 1
Part (c)
M1 values put in order 9, 11, 12, 13, 15, 15, 15, 20 or the middle two values 13 and 15 identified
✓ 1
A1 14
✓ 1
Part (d)
M1 sum of the doses = 110, divided by 8
✓ 1
A1 13.75 (accept 13.8)
✓ 1
Common error: With an even number of values the median is halfway between the two middle values, not one of them.
58IGMC3-9-11r43-02Averages, range and frequency tablesPaper CEasy6 marks
Calculator
In a game, 24 players each complete a number of levels in one session. The table shows the results. Levels completed: 0, 1, 2, 3, 4 Number of players: 3, 5, 6, 8, 2
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Calculate the mean number of levels completed. Give your answer correct to 3 significant figures.
Exam tip: For a frequency table, multiply each value by its frequency and divide by the total frequency, not by the number of columns.
59IGMC3-9-11r43-03Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy7 marks
No calculator
A water company records the mean daily temperature (°C) and the volume of water used (megalitres, ML) in a town on 12 days. The temperatures range from 8 °C to 34 °C. The scatter diagram shows points lying close to a straight line sloping upwards from left to right. A line of best fit is drawn through the points (10, 30) and (30, 70).
(a)
Describe the correlation.
(1)
(b)
Find the gradient of the line of best fit and explain what it means in this context.
(3)
(c)
Use the line of best fit to estimate the volume of water used on a day when the mean temperature is 24 °C.
(2)
(d)
Explain why the line should not be used to estimate the water used when the temperature is 45 °C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 positive (correlation)
✓ 1
Part (b)
M1 (70 − 30) ÷ (30 − 10)
✓ 1
A1 2
✓ 1
B1 each 1 °C rise in temperature is linked to 2 ML more water used
✓ 1
Part (c)
M1 30 + 2 × (24 − 10)
✓ 1
A1 58 (ML)
✓ 1
Part (d)
B1 45 °C is outside the range of the data (8 °C to 34 °C), so this is extrapolation and may be unreliable
✓ 1
Common error: Always describe a gradient with units and a ‘for each extra’ statement, otherwise the interpretation mark is lost.
60IGMC3-9-11r45-01Averages, range and frequency tablesPaper CEasy6 marks
Calculator
A beekeeper weighs the honey harvested from each of 8 hives. The masses, in kg, are 12, 15, 9, 18, 15, 11, 20, 16.
(a)
Write down the mode.
(1)
(b)
Find the range.
(1)
(c)
Find the median.
(2)
(d)
Calculate the mean mass per hive.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
15 kg
✓ 1
Part (b)
20 − 9 = 11 kg
✓ 1
Part (c)
M1 orders the data: 9, 11, 12, 15, 15, 16, 18, 20
✓ 1
A1 median = (15 + 15) ÷ 2 = 15 kg
✓ 1
Part (d)
M1 total = 116 and divides by 8
✓ 1
A1 14.5 kg
✓ 1
Exam tip: Always write the data in order before looking for the median, and with an even number of values take the middle two.
61IGMC3-9-11r45-02Averages, range and frequency tablesPaper CEasy6 marks
Calculator
During a race a Formula 1 team records the time taken by 20 pit stops.
Time (s)
2.2
2.4
2.6
2.8
3.0
Frequency
4
7
6
2
1
(a)
Write down the modal time.
(1)
(b)
Find the median time.
(2)
(c)
Calculate the mean time.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
2.4 s
✓ 1
Part (b)
M1 the median lies between the 10th and 11th values (cumulative frequencies 4, 11)
Common error: When finding a mean from a frequency table, divide by the total frequency (20), not by the number of different times (5).
62IGMC3-9-11r45-04Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy7 marks
Calculator
A beekeeper counts the bee flights leaving a hive in 10 minutes at different temperatures.
Temperature (°C)
10
14
18
22
26
30
Flights
27
41
47
62
76
85
A line of best fit is drawn for the data. It passes through the points (12, 30) and (28, 78).Scatter diagram with the line of best fit through (12, 30) and (28, 78)
(a)
Describe the correlation between temperature and the number of flights.
(1)
(b)
Calculate the gradient of the line of best fit and explain what it means in this situation.
(3)
(c)
Use the line of best fit to estimate the number of flights at 20 °C.
(2)
(d)
Explain why the line should not be used to estimate the flights at 40 °C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Positive correlation: as temperature rises the number of flights rises
✓ 1
Part (b)
M1 (78 − 30) ÷ (28 − 12)
✓ 1
A1 3
✓ 1
B1 for every 1 °C rise, about 3 more flights in 10 minutes
✓ 1
Part (c)
M1 30 + 3 × (20 − 12) or equivalent
✓ 1
A1 54
✓ 1
Part (d)
40 °C is outside the range of the data (10 °C to 30 °C), so the estimate is an extrapolation and may be unreliable
✓ 1
Common error: A gradient needs a unit and a meaning: say how much the y-quantity changes for each 1 unit increase in x.
63IGMC3-9-11r45-13Averages, range and frequency tablesPaper CEasy7 marks
Calculator
The annual salaries, in thousands of euros (€), of 8 staff in a small clinic are 24, 25, 25, 26, 27, 28, 30, 95.
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Calculate the mean.
(2)
(d)
The clinic manager wants the average that best represents a typical salary. Which average should be used? Give a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
€25 000
✓ 1
Part (b)
M1 the median is the mean of the 4th and 5th values
✓ 1
A1 (26 + 27) ÷ 2 = €26 500
✓ 1
Part (c)
M1 total = 280 divided by 8
✓ 1
A1 €35 000
✓ 1
Part (d)
B1 the median
✓ 1
B1 the mean is pulled upwards by the very high salary of €95 000
✓ 1
Common error: Do not choose the mean when one extreme value distorts it; the median is not affected by extreme values.
64IGMC3-9-10r46-01Averages, range and frequency tablesPaper CEasy9 marks
Calculator
A beekeeper weighs the honey harvested from each of 9 hives, in kg: 18, 22, 25, 19, 28, 22, 27, 22, 24.
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Work out the range.
(1)
(d)
Calculate the mean yield for the 9 hives.
(2)
(e)
A tenth hive is harvested and the mean for all 10 hives is 24 kg. Find the yield of the tenth hive.
Exam tip: To find a missing value from a mean, work with totals: mean × number of values.
65IGMC3-9-10r46-02Scatter diagrams, correlation, cumulative frequency and histogramsPaper CEasy6 marks
No calculator
The table shows the number of visitors each of 20 patients on a hospital ward received on one day. Visitors | 1 | 2 | 3 | 4 | 5 Frequency | 3 | 7 | 6 | 3 | 1
(a)
Write down the modal number of visitors.
(1)
(b)
Find the median number of visitors.
(2)
(c)
Calculate the mean number of visitors per patient.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 2
✓ 1
Part (b)
M1 identifies the 10th and 11th values as 2 and 3
✓ 1
A1 2.5
✓ 1
Part (c)
M1 attempts Σ(visitors × frequency)
✓ 1
A1 3 + 14 + 18 + 12 + 5 = 52
✓ 1
A1 52 ÷ 20 = 2.6
✓ 1
Common error: For a frequency table, multiply each value by its frequency before adding; do not just add the values 1 to 5 and divide by 5.
66IGMC3-9-10r46-03Averages, range and frequency tablesPaper CEasy6 marks
Calculator
A Formula 1 team records the time taken for each of 20 pit stops, in seconds. Time, t (s) | 2.0 < t ≤ 2.4 | 2.4 < t ≤ 2.8 | 2.8 < t ≤ 3.2 | 3.2 < t ≤ 3.6 | 3.6 < t ≤ 4.0 Frequency | 3 | 8 | 5 | 3 | 1
(a)
Write down the modal class.
(1)
(b)
Calculate an estimate of the mean pit stop time.
(4)
(c)
Explain why your answer to (b) is only an estimate.
B1 the exact times are not known; midpoints are used to represent each group
✓ 1
Exam tip: Write down the midpoints in a column before multiplying by the frequencies.
67IGMC3-9-10r46-13Averages, range and frequency tablesPaper CEasy7 marks
No calculator
The numbers of patients arriving at a hospital emergency department in five consecutive hours were 6, 9, 7, 8 and 10.
(a)
Work out the mean number of patients per hour.
(2)
(b)
In the next hour a coach crash brings 35 patients. Find the new mean for all six hours.
(2)
(c)
Find the new median for all six hours.
(2)
(d)
Explain why the median is a better average than the mean for describing a typical hour now.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (6 + 9 + 7 + 8 + 10) ÷ 5
✓ 1
A1 8
✓ 1
Part (b)
M1 (40 + 35) ÷ 6
✓ 1
A1 12.5
✓ 1
Part (c)
M1 orders data: 6, 7, 8, 9, 10, 35 and takes the middle two, 8 and 9
✓ 1
A1 8.5
✓ 1
Part (d)
B1 the median (8.5) is hardly affected by the extreme value 35 while the mean (12.5) is pulled up by it
✓ 1
Common error: When there is an even number of values, the median is the mean of the two middle values.
68IGMC-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.
69IGMC-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.
71IGMC-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).
Scatter diagram with the line of best fit drawn through (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.
73IGMC-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).
Scatter diagram of the data
(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.
74IGMC-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.
75IGMC3-9-06d-06Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium6 marks
Calculator
A group of 15 sports students has a mean mass of 62 kg. A second group of 25 students has a mean mass of 70 kg.
(a)
Find the total mass of the first group.
(1)
(b)
Find the mean mass of all 40 students.
(3)
(c)
A student of mass 67 kg leaves the second group. Show that the mean mass of the remaining 39 students is still 67 kg.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
15 × 62 = 930 kg
✓ 1
Part (b)
M1 25 × 70 = 1750
✓ 1
M1 (930 + 1750) ÷ 40
✓ 1
A1 67
✓ 1
Part (c)
M1 (2680 − 67) ÷ 39
✓ 1
A1 2613 ÷ 39 = 67
✓ 1
Common error: Do not average two means directly unless the groups are the same size; weight by the number in each group.
76IGMC3-9-06d-07Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium6 marks
Calculator
A botanist records the mass of fertiliser, x g, and the plant height, y cm, for five plants. (x, y): (10, 12), (20, 15), (30, 19), (40, 22), (50, 26) A line of best fit is y = 0.4x + 7.
Scatter diagram of the data with the line of best fit
(a)
Find the coordinates of the mean point (x̄, ȳ).
(2)
(b)
Explain what the gradient 0.4 tells you about the plants.
(1)
(c)
The botanist uses the line to predict the height of a plant given 200 g of fertiliser. Give a reason why this is not sensible.
(1)
(d)
Use the line to estimate the height of a plant given 35 g of fertiliser.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
x̄ = 30
✓ 1
ȳ = 18.8
✓ 1
Part (b)
Each extra gram of fertiliser is associated with 0.4 cm more height
✓ 1
Part (c)
200 g is far outside the data range (extrapolation) so the relationship may not continue
✓ 1
Part (d)
M1 0.4 × 35 + 7
✓ 1
A1 21
✓ 1
Exam tip: Link every interpretation of a gradient to the context and its units.
77IGMC3-9-06d-08Averages, range and frequency tablesPaper CMedium7 marks
Calculator
A data logger records five whole-number particle counts per second. In increasing order they have a median of 6, a unique mode of 5, a range of 8 and a mean of 7.2.
(a)
Find the five counts.
(5)
(b)
A sixth count of 9 is recorded. Find the new median.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 median (third value) is 6
✓ 1
B1 mode 5 means the two smallest are 5 and 5
✓ 1
B1 largest = 5 + 8 = 13
✓ 1
M1 total = 5 × 7.2 = 36, so fourth = 36 − (5 + 5 + 6 + 13)
✓ 1
A1 5, 5, 6, 7, 13
✓ 1
Part (b)
M1 order 5, 5, 6, 7, 9, 13 and use the middle two
✓ 1
A1 6.5
✓ 1
Exam tip: Convert each statistic into an equation or restriction on the ordered data, then fit them together.
78IGMC3-9-08r7-05Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium4 marks
Calculator
A clinic records the number of tablets taken per day by 40 patients in a trial.
Tablets per day
0
1
2
3
4
Number of patients
5
12
x
8
3
(a)
Find the value of x.
(1)
(b)
Calculate the mean number of tablets per day.
(2)
(c)
One patient is chosen at random. Find the probability that this patient takes at least 3 tablets per day.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 40 − (5 + 12 + 8 + 3) = 12
✓ 1
Part (b)
M1 (0 + 12 + 24 + 24 + 12) ÷ 40 = 72 ÷ 40
✓ 1
A1 1.8
✓ 1
Part (c)
B1 11/40 (or 0.275)
✓ 1
Exam tip: Multiply each value by its frequency before adding.
79IGMC3-9-08r7-06Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium5 marks
Calculator
The table shows the rainfall x mm and the volume y ML of water added to a reservoir on six days.
x
10
20
30
40
50
60
y
13
21
34
40
51
57
Scatter diagram with the line y = 0.84x + 6.6
(a)
Find the mean point (x̄, ȳ).
(2)
(b)
A line of best fit passes through the mean point and through the point (60, 57). Calculate the gradient of this line.
(1)
(c)
Use the line y = 0.84x + 6.6 to estimate the volume added when the rainfall is 45 mm.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 x̄ = 210 ÷ 6 = 35
✓ 1
B1 ȳ = 216 ÷ 6 = 36
✓ 1
Part (b)
B1 (57 − 36) ÷ (60 − 35) = 0.84
✓ 1
Part (c)
M1 0.84 × 45 + 6.6
✓ 1
A1 44.4 ML
✓ 1
Exam tip: A line of best fit passes through the mean point of the data.
80IGMC3-9-08r7-09Averages, range and frequency tablesPaper CMedium6 marks
Calculator
The ping times, in milliseconds, of two game servers were recorded over seven games. Server A: 32, 35, 29, 41, 38, 35, 35 Server B: 20, 25, 30, 35, 40, 45, 50
(a)
Calculate the mean ping time of server A.
(1)
(b)
Find the range for each server.
(2)
(c)
Both servers have a mean ping time of 35 ms. Which server is more consistent? Give a reason.
(1)
(d)
An eighth game on server A is added and the mean of all eight games becomes 36 ms. Find the ping time of the eighth game.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 245 ÷ 7 = 35 ms
✓ 1
Part (b)
B1 A: 41 − 29 = 12
✓ 1
B1 B: 50 − 20 = 30
✓ 1
Part (c)
R1 Server A, because its range (12) is smaller than B's range (30)
✓ 1
Part (d)
M1 8 × 36 − 245 = 288 − 245
✓ 1
A1 43 ms
✓ 1
Exam tip: Compare an average and a measure of spread when comparing two sets of data.
81IGMC3-9-08r7-10Averages, range and frequency tablesPaper CMedium4 marks
No calculator
Five positive whole numbers, each a dose in mg, are written in order from smallest to largest. The mode is 5, the median is 5, the mean is 6 and the range is 8. The smallest number is 2.
(a)
Write down the total of the five numbers.
(1)
(b)
Find the largest number.
(1)
(c)
Find the five numbers.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 5 × 6 = 30
✓ 1
Part (b)
B1 2 + 8 = 10
✓ 1
Part (c)
M1 two of the numbers are 5 (mode) and the middle is 5, remaining sum 30 − 2 − 10 − 5 − 5 = 8
✓ 1
A1 2, 5, 5, 8, 10
✓ 1
Exam tip: Turn each statistic into an equation or a fact about the ordered list.
82IGMC3-9-08r13-04Averages, range and frequency tablesPaper CMedium7 marks
Calculator
In a trial of a painkiller, the time t minutes taken for 50 patients to feel relief is shown. 0 < t ≤ 10: 5 patients 10 < t ≤ 20: 14 patients 20 < t ≤ 30: 18 patients 30 < t ≤ 40: 9 patients 40 < t ≤ 50: 4 patients
(a)
Write down the modal class.
(1)
(b)
Find the percentage of patients who took more than 30 minutes.
(2)
(c)
Calculate an estimate of the mean time.
(3)
(d)
Explain why your answer to part (c) is only an estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 20 < t ≤ 30
✓ 1
Part (b)
M1 (9 + 4) ÷ 50
✓ 1
A1 26%
✓ 1
Part (c)
M1 midpoints 5, 15, 25, 35, 45 used
✓ 1
M1 5×5 + 14×15 + 18×25 + 9×35 + 4×45 = 1180
✓ 1
A1 1180 ÷ 50 = 23.6 minutes
✓ 1
Part (d)
R1 the exact times are not known, so the midpoint of each class is assumed
✓ 1
Common error: When estimating a mean from grouped data, use the midpoint of each class, not the upper or lower boundary.
83IGMC3-9-08r13-05Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium6 marks
Calculator
A city records the mean daily temperature x °C and the water used y megalitres (ML) on six days. x: 15, 18, 20, 24, 26, 29 y: 312, 330, 336, 360, 366, 396
Scatter diagram of the data
(a)
Find the mean of x and the mean of y.
(2)
(b)
Describe the correlation between temperature and water use.
(1)
(c)
A line of best fit has gradient 6 and passes through the point (22, 350). Find its equation in the form y = mx + c.
(2)
(d)
Use your equation to estimate the water use on a day when the mean temperature is 24 °C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 22
✓ 1
B1 350
✓ 1
Part (b)
B1 positive (correlation)
✓ 1
Part (c)
M1 y − 350 = 6(x − 22) or 350 = 6×22 + c
✓ 1
A1 y = 6x + 218 (ECF from (a))
✓ 1
Part (d)
A1 362 ML (ECF from (c))
✓ 1
Exam tip: A line of best fit should pass through the mean point (mean of x, mean of y), so use it to find the intercept.
84IGMC3-9-08r13-11Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium5 marks
Calculator
A pharmacologist records the dose x mg of a drug and the time y minutes to reach peak effect for seven patients: (5, 62), (10, 55), (15, 49), (20, 41), (25, 70), (30, 33), (35, 28).
(a)
Describe the correlation, ignoring any anomalous point.
(1)
(b)
Write down the anomalous point.
(1)
(c)
A line of best fit passes through (10, 55) and (30, 33). Calculate its gradient.
(2)
(d)
Interpret the gradient in the context of the question.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 negative (correlation)
✓ 1
Part (b)
B1 (25, 70)
✓ 1
Part (c)
M1 (33 − 55) ÷ (30 − 10)
✓ 1
A1 −1.1
✓ 1
Part (d)
R1 each extra 1 mg of dose reduces the time to peak effect by 1.1 minutes (on average)
✓ 1
Common error: The gradient is change in y divided by change in x, taken in the same order for both. A negative gradient matches negative correlation.
85IGMC3-9-08r13-14Averages, range and frequency tablesPaper CMedium7 marks
Calculator
Two gamers, Ana and Ben, each play eight games. Ana's scores: 62, 75, 68, 71, 64, 70, 66, 76. Ben's scores: 50, 90, 71, 84, 58, 77, 66, 56.
(a)
Show that both gamers have a mean score of 69.
(2)
(b)
Find the range of each gamer's scores.
(2)
(c)
Which gamer is more consistent? Give a reason.
(1)
(d)
Ana scores 78 in a ninth game. Find her new mean score.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of each set of scores = 552
✓ 1
A1 552 ÷ 8 = 69 for both
✓ 1
Part (b)
B1 Ana: 76 − 62 = 14
✓ 1
B1 Ben: 90 − 50 = 40
✓ 1
Part (c)
R1 Ana, because her range is smaller (ECF from (b))
✓ 1
Part (d)
M1 (552 + 78) ÷ 9
✓ 1
A1 70
✓ 1
Exam tip: When asked to compare data sets, use one measure of average and one measure of spread, and say what each tells you.
86IGMC3-9-10r14-05Averages, range and frequency tablesPaper CMedium5 marks
Calculator
The mean time of five swimmers over 100 m is 62.4 s. Four of the times are 60.1 s, 63.5 s, 61.8 s and 66.0 s.
Exam tip: When a mean and a count are given, multiply to get the total first, then subtract the known values.
87IGMC3-9-10r14-06Averages, range and frequency tablesPaper CMedium5 marks
Calculator
A library lists the overdue fines paid by seven readers, in pounds: 0.40, 0.60, 0.40, 1.20, 0.80, 0.40, 9.60.
(a)
Find the median fine.
(2)
(b)
Explain why the median is a better measure of the typical fine than the mean.
(1)
(c)
Calculate the mean fine, correct to the nearest penny.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 ordered list 0.40, 0.40, 0.40, 0.60, 0.80, 1.20, 9.60
✓ 1
A1 £0.60
✓ 1
Part (b)
B1 the mean is pulled upward by the extreme value £9.60, but the median is not affected
✓ 1
Part (c)
M1 total 13.40 ÷ 7
✓ 1
A1 £1.91
✓ 1
Common error: Do not choose the mean by default: a single extreme value can make it unrepresentative of most of the data.
88IGMC3-9-10r14-07Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium6 marks
Calculator
Two swimmers, A and B, record five 50 m times in seconds. Swimmer A: 31.2, 30.8, 31.5, 30.9, 31.1 Swimmer B: 30.1, 32.6, 29.8, 31.9, 31.1
(a)
Show that the mean time of swimmer A is 31.1 s.
(2)
(b)
The mean time of swimmer B is also 31.1 s. Swimmer A is more consistent. Use the ranges to justify this statement.
(2)
(c)
Calculate how much greater the range of B is than the range of A.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (31.2 + 30.8 + 31.5 + 30.9 + 31.1) ÷ 5
✓ 1
A1 155.5 ÷ 5 = 31.1
✓ 1
Part (b)
B1 range of A = 31.5 − 30.8 = 0.7 s
✓ 1
B1 range of B = 32.6 − 29.8 = 2.8 s and the smaller range shows A is more consistent
✓ 1
Part (c)
M1 2.8 − 0.7
✓ 1
A1 2.1 s
✓ 1
Exam tip: When two data sets have the same mean, compare their spread (range) to comment on consistency.
89IGMC3-9-10r14-08Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium4 marks
Calculator
The table shows the hub height of six wind turbines and the energy each produced last year.
Hub height (m)
60
70
80
90
100
110
Energy (GWh)
5.1
5.9
6.6
2.0
8.4
9.2
(a)
The turbine with hub height 90 m was shut down for repairs for several months. Explain how the table shows that its result does not fit the pattern.
(1)
(b)
Describe the correlation shown by the other five turbines.
(1)
(c)
Calculate the mean energy produced by the other five turbines.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 2.0 GWh is much lower than the other values, which rise steadily with height
✓ 1
Part (b)
B1 positive correlation
✓ 1
Part (c)
M1 (5.1 + 5.9 + 6.6 + 8.4 + 9.2) ÷ 5 = 35.2 ÷ 5
✓ 1
A1 7.04 GWh
✓ 1
Exam tip: An outlier that has a known cause can be left out when describing the trend; say why it was excluded.
90IGMC3-9-09r16-03Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium6 marks
Calculator
The expansion of a bridge joint, y mm, is measured at six air temperatures, x °C.
x
5
10
15
20
30
35
y
3.2
4.0
5.7
6.5
9.3
10.0
A scatter diagram of these data is drawn. A line of best fit is drawn through the points (0, 1.8) and (40, 11.4).Scatter diagram with the line of best fit through (0, 1.8) and (40, 11.4)
(a)
Describe the correlation between x and y.
(1)
(b)
Work out the gradient of the line of best fit.
(2)
(c)
Use the line of best fit to estimate the expansion at 25 °C.
(2)
(d)
Explain why the line should not be used to estimate the expansion at 120 °C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Positive correlation
✓ 1
Part (b)
M1 (11.4 − 1.8) ÷ (40 − 0)
✓ 1
A1 0.24
✓ 1
Part (c)
M1 1.8 + 0.24 × 25 or reading the line at x = 25
✓ 1
A1 7.8 mm (accept 7.7 to 7.9)
✓ 1
Part (d)
120 °C is far outside the range of the data (5 to 35 °C), so the relationship may not continue (extrapolation)
✓ 1
Common error: Estimating far beyond the range of the data (extrapolation) is unreliable; say that the point lies outside the data range.
91IGMC3-9-09r16-04Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium5 marks
Calculator
A probe collects five rock samples with a mean mass of 8.4 g. A sixth sample is collected and the mean mass of all six samples is 9.0 g.
(a)
Find the total mass of the first five samples.
(1)
(b)
Find the mass of the sixth sample.
(2)
(c)
A seventh sample of mass 6.0 g is collected. Calculate the mean mass of all seven samples, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
5 × 8.4 = 42 g
✓ 1
Part (b)
M1 6 × 9.0 = 54
✓ 1
A1 54 − 42 = 12 g
✓ 1
Part (c)
M1 (54 + 6) ÷ 7
✓ 1
A1 8.57 g
✓ 1
Exam tip: Use mean × number of values to turn a mean back into a total, then add or remove values.
92IGMC3-9-08r19-04Averages, range and frequency tablesPaper CMedium5 marks
Calculator
A club records the number of rest stops made by riders in a sportive. The mean number of stops is 2.5.
Rest stops
1
2
3
4
Frequency
5
9
x
6
(a)
Find the value of x.
(3)
(b)
Find the median number of rest stops.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total of stops = 5 + 18 + 3x + 24
✓ 1
M1 (47 + 3x) / (20 + x) = 2.5
✓ 1
A1 x = 6
✓ 1
Part (b)
M1 total frequency 26, so 13th and 14th values (cumulative frequencies 5 then 14)
✓ 1
A1 2
✓ 1
Exam tip: Write the total frequency as an expression in x before forming the equation.
93IGMC3-9-08r19-05Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium5 marks
Calculator
The mean mass of 9 pottery fragments is 124 g. A tenth fragment is added and the mean mass of the 10 fragments becomes 130 g.
(a)
Find the mass of the tenth fragment.
(2)
(b)
A different box holds 15 fragments with a mean mass of 118 g. Find the mean mass of all 25 fragments.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 130 × 10 − 124 × 9
✓ 1
A1 184 g
✓ 1
Part (b)
B1 total mass of the 15 fragments = 1770 g
✓ 1
M1 (1300 + 1770) / 25
✓ 1
A1 122.8 g
✓ 1
Common error: Do not average the two means; use the total masses.
94IGMC3-9-08r19-06Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium7 marks
Calculator
A cyclist records the weekly training time, x hours, and the time, y minutes, taken to ride 20 km, for six weeks.
x
2
4
5
7
8
10
y
62
58
55
50
46
41
Scatter diagram of the data
(a)
Describe the correlation between x and y.
(1)
(b)
Find the mean point (x̄, ȳ).
(2)
(c)
A line of best fit passes through the mean point and the point (0, 70). Find its equation in the form y = mx + c.
(2)
(d)
Use your equation to estimate the time taken after 9 hours of training in a week.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 negative
✓ 1
Part (b)
B1 x̄ = 6
✓ 1
B1 ȳ = 52
✓ 1
Part (c)
M1 gradient = (52 − 70) / 6
✓ 1
A1 y = −3x + 70
✓ 1
Part (d)
M1 −3 × 9 + 70
✓ 1
A1 43 minutes (ECF from (c))
✓ 1
Exam tip: A line of best fit always passes through the mean point, so use it as one of your two points.
95IGMC3-9-08r19-08Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium5 marks
Calculator
Five riders from team A and five riders from team B each completed a time trial. Team A's times, in minutes, have mean 41 and range 6. Team B's times, in minutes, are 30, 37, 40, 44 and 44.
(a)
Calculate the mean time for team B.
(2)
(b)
Find the range of team B's times.
(1)
(c)
Compare the two teams, using an average and the range.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (30 + 37 + 40 + 44 + 44) / 5
✓ 1
A1 39 minutes
✓ 1
Part (b)
B1 44 − 30 = 14 minutes
✓ 1
Part (c)
B1 team B is faster on average (lower mean)
✓ 1
B1 team A is more consistent (smaller range)
✓ 1
Exam tip: A comparison needs one comment on the average and one on the spread, both in context.
96IGMC3-9-08r25-03Averages, range and frequency tablesPaper CMedium6 marks
Calculator
The mean yield of wheat on eight fields on a farm is 4.5 tonnes per hectare.
(a)
A ninth field has a yield of 5.4 tonnes per hectare. Calculate the mean yield of all nine fields.
(3)
(b)
A tenth field is added and the mean yield of all ten fields is 4.7 tonnes per hectare. Find the yield of the tenth field.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 8 × 4.5 = 36
✓ 1
M1 (36 + 5.4) ÷ 9
✓ 1
A1 4.6 tonnes per hectare
✓ 1
Part (b)
M1 10 × 4.7 = 47
✓ 1
M1 47 − 41.4
✓ 1
A1 5.6 tonnes per hectare
✓ 1
Common error: A mean multiplied by the number of values gives the total. Use the total, not the means, when combining data.
97IGMC3-9-08r25-04Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium9 marks
No calculator
The table shows the age, x years, and the yearly maintenance cost, y thousand euros, of six aircraft in a small fleet. Age x: 2, 4, 6, 8, 10, 12 Cost y: 10, 14, 19, 21, 26, 30
Scatter diagram of the data
(a)
Describe the correlation between age and maintenance cost.
(1)
(b)
Find the mean age and the mean cost, giving the coordinates of the mean point.
(2)
(c)
A line of best fit passes through the mean point and the point (11, 28). Find the equation of the line in the form y = mx + c.
(3)
(d)
Use the line to estimate the yearly cost for an aircraft aged 9 years.
(2)
(e)
Explain why the line should not be used to estimate the cost for an aircraft aged 30 years.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 positive correlation
✓ 1
Part (b)
M1 42 ÷ 6 and 120 ÷ 6
✓ 1
A1 (7, 20)
✓ 1
Part (c)
M1 gradient = (28 − 20) ÷ (11 − 7)
✓ 1
A1 m = 2
✓ 1
A1 y = 2x + 6
✓ 1
Part (d)
M1 2 × 9 + 6
✓ 1
A1 24 thousand euros (FT from (c))
✓ 1
Part (e)
B1 30 years is outside the range of the data (2 to 12 years), so this is extrapolation and may be unreliable
✓ 1
Exam tip: A line of best fit should pass through the mean point. Only use it to estimate within the range of the data.
98IGMC3-9-08r25-05Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium7 marks
Calculator
A farmer records the number of eggs laid in one week by each of her 40 hens. Number of eggs: 3, 4, 5, 6, 7 Number of hens: 5, 8, k, 10, 5
(a)
Find the value of k.
(2)
(b)
Calculate the mean number of eggs laid per hen.
(3)
(c)
Find the percentage of the hens that laid at least 6 eggs.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 40 − (5 + 8 + 10 + 5)
✓ 1
A1 12
✓ 1
Part (b)
M1 at least three correct products of eggs × hens
✓ 1
A1 202 (FT from (a))
✓ 1
A1 5.05
✓ 1
Part (c)
M1 (10 + 5) ÷ 40 × 100
✓ 1
A1 37.5%
✓ 1
Exam tip: For a frequency table, the mean is the sum of (value × frequency) divided by the sum of the frequencies.
99IGMC3-9-08r25-06Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium7 marks
Calculator
Five students record the time, x hours, they spent practising their instrument in one week and their mark, y, in a music test. The maximum mark is 100. Hours x: 2, 4, 5, 7, 7 Mark y: 41, 52, 58, 66, 73
Scatter diagram with the line y = 5x + 33
(a)
Find the mean number of hours and the mean mark.
(2)
(b)
Show that the line y = 5x + 33 passes through the mean point.
(1)
(c)
Explain what the gradient of the line means in this context.
(1)
(d)
Explain why the line should not be used to predict the mark of a student who practises for 20 hours.
(1)
(e)
Use the line to estimate the mark of a student who practises for 6 hours.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 25 ÷ 5 and 290 ÷ 5
✓ 1
A1 5 hours and 58 marks
✓ 1
Part (b)
B1 5 × 5 + 33 = 58, which equals the mean mark
✓ 1
Part (c)
B1 each extra hour of weekly practice is associated with about 5 more marks
✓ 1
Part (d)
B1 the line gives 133, which is above the maximum mark of 100, and 20 is outside the range of the data
✓ 1
Part (e)
M1 5 × 6 + 33
✓ 1
A1 63
✓ 1
Common error: Correlation does not prove that practising causes higher marks. Describe the gradient as an association.
100IGMC3-9-08r25-09Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium6 marks
No calculator
120 pupils are asked which musical instrument they most like to play. 42 chose guitar, 30 chose piano and the same number of pupils chose drums as chose violin. The results are shown in a pie chart.
(a)
Find how many pupils chose drums.
(2)
(b)
Find the angle of the guitar sector in the pie chart.
(2)
(c)
Find the percentage of pupils who chose drums.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (120 − 42 − 30) ÷ 2
✓ 1
A1 24
✓ 1
Part (b)
M1 42 ÷ 120 × 360
✓ 1
A1 126°
✓ 1
Part (c)
M1 24 ÷ 120 × 100 (FT from (a))
✓ 1
A1 20%
✓ 1
Exam tip: In a pie chart, each person is worth 360° divided by the total, here 3° per pupil.
101IGMC3-9-08r25-14Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium8 marks
Calculator
A farm advisor records the rainfall, x mm, during the growing season and the crop yield, y tonnes per hectare, on eight farms. Rainfall x: 20, 35, 50, 65, 80, 95, 110, 75 Yield y: 3.1, 3.9, 4.6, 5.2, 5.8, 6.3, 6.6, 1.2
(a)
One farm does not follow the trend. Write down its rainfall and yield and suggest one possible reason.
(2)
(b)
Ignoring that farm, describe the correlation between rainfall and yield.
(1)
(c)
Calculate the mean yield for the seven farms that follow the trend. Give your answer correct to 3 significant figures.
(2)
(d)
Calculate the mean yield for all eight farms. Give your answer correct to 3 significant figures.
(2)
(e)
Comment on the effect of the unusual farm on the mean.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 (75, 1.2)
✓ 1
B1 a sensible reason, for example disease, pests or flooding on that farm
✓ 1
Part (b)
B1 positive correlation
✓ 1
Part (c)
M1 35.5 ÷ 7
✓ 1
A1 5.07 tonnes per hectare
✓ 1
Part (d)
M1 (35.5 + 1.2) ÷ 8
✓ 1
A1 4.59 tonnes per hectare
✓ 1
Part (e)
B1 it lowers the mean because its yield is much smaller than the others
✓ 1
Exam tip: An outlier can distort the mean. Always give a context-based reason when asked why a point does not fit.
102IGMC3-9-12r29-04Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium8 marks
Calculator
A shop owner recorded the number of ice creams sold, s, on six days with different midday temperatures, T °C. T: 12, 15, 18, 21, 24, 27 s: 41, 50, 58, 20, 75, 83 On one of the days the shop closed early.
(a)
Write down the coordinates of the point that is probably affected by the shop closing early and give a reason for your choice.
(2)
(b)
Calculate the mean number of ice creams sold, leaving out this day.
(2)
(c)
The line of best fit for the other five points passes through (12, 41) and (27, 83). Find its equation in the form s = mT + c.
(3)
(d)
Use the line to estimate the number of ice creams that would have been sold at 21 °C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 (21, 20)
✓ 1
B1 it does not follow the pattern of the other points (sales are much lower than the trend)
✓ 1
Part (b)
M1 (41 + 50 + 58 + 75 + 83)/5 = 307/5
✓ 1
A1 61.4
✓ 1
Part (c)
M1 gradient = (83 − 41)/(27 − 12)
✓ 1
A1 2.8
✓ 1
A1 s = 2.8T + 7.4
✓ 1
Part (d)
A1 66.2 (accept 66, ECF from (c))
✓ 1
Exam tip: An unusual point with a known explanation can be left out when drawing the line of best fit, but say so.
103IGMC3-9-12r29-05Averages, range and frequency tablesPaper CMedium6 marks
Calculator
The recovery times, t days, of 50 patients are summarised below. 0 < t ≤ 5: 8 patients 5 < t ≤ 10: 22 patients 10 < t ≤ 15: 14 patients 15 < t ≤ 20: 6 patients
(a)
Write down the modal class.
(1)
(b)
Write down the class that contains the median.
(1)
(c)
Calculate an estimate of the mean recovery time.
(3)
(d)
Explain why your answer to (c) is only an estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 5 < t ≤ 10
✓ 1
Part (b)
B1 5 < t ≤ 10 (the 25th and 26th values lie in this class, since 8 + 22 = 30)
✓ 1
Part (c)
M1 use of mid-values 2.5, 7.5, 12.5, 17.5
✓ 1
M1 sum of frequency × mid-value = 465
✓ 1
A1 9.3 days
✓ 1
Part (d)
B1 the exact recovery times are not known; each value is assumed to be at the middle of its class
✓ 1
Common error: Use the middle of each class interval, not its upper boundary, when estimating a mean from grouped data.
104IGMC3-9-12r29-06Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium8 marks
Calculator
A solar panel's daily energy output, y kWh, was recorded for six days with x hours of sunshine. x: 3, 4, 5, 6, 7, 8 y: 2.2, 3.0, 3.6, 4.4, 5.2, 5.6
Scatter diagram of the data
(a)
Describe the correlation between x and y.
(1)
(b)
Find the mean of x and the mean of y.
(2)
(c)
A line of best fit passes through the mean point and through (8, 5.5). Calculate its gradient.
(2)
(d)
Find the equation of the line of best fit.
(2)
(e)
Estimate the energy output on a day with 6.5 hours of sunshine.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 positive correlation
✓ 1
Part (b)
B1 mean of x = 5.5
✓ 1
B1 mean of y = 4.0
✓ 1
Part (c)
M1 (5.5 − 4.0)/(8 − 5.5)
✓ 1
A1 0.6
✓ 1
Part (d)
M1 y − 4 = 0.6(x − 5.5)
✓ 1
A1 y = 0.6x + 0.7
✓ 1
Part (e)
A1 4.6 kWh (ECF from (d))
✓ 1
Exam tip: A line of best fit should pass through the mean point, so use it as one of the two points when finding the equation.
105IGMC3-9-12r29-08Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium7 marks
Calculator
A shoe shop recorded the sizes of 20 + x pairs of boots sold in one week. Size: 5, 6, 7, 8, 9 Number of pairs: 3, x, 9, 6, 2 The mean size is 6.8.
(a)
Show that x = 10.
(4)
(b)
Write down the modal size.
(1)
(c)
Find the median size.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total of sizes = 15 + 6x + 63 + 48 + 18 = 144 + 6x
✓ 1
M1 total number of pairs = 20 + x
✓ 1
M1 (144 + 6x)/(20 + x) = 6.8
✓ 1
A1 144 + 6x = 136 + 6.8x, so 0.8x = 8 and x = 10
✓ 1
Part (b)
B1 6
✓ 1
Part (c)
M1 30 pairs, so the median is the mean of the 15th and 16th values (cumulative totals 3, 13, 22)
✓ 1
A1 7
✓ 1
Common error: Include the unknown frequency in the total frequency as well as in the sum of the products.
106IGMC3-9-11r31-03Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium8 marks
Calculator
A water company records the average daily temperature, x °C, and the volume of water used by a town, y megalitres (ML), on eight days. Temperature x: 14, 16, 18, 20, 22, 24, 26, 28 Water used y: 31, 34, 38, 40, 43, 47, 49, 54
Scatter diagram with the line y = 1.6x + 8.4
(a)
Find the mean temperature and the mean water usage.
(2)
(b)
Describe the correlation between temperature and water used.
(1)
(c)
A line of best fit for the data has equation \(y = 1.6x + 8.4\). Show that this line passes through the point (mean temperature, mean usage).
(2)
(d)
Use the line of best fit to estimate the water used when the temperature is 30 °C.
(2)
(e)
Explain why an estimate for a day with temperature 45 °C would be unreliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 mean temperature 21
✓ 1
B1 mean usage 42
✓ 1
Part (b)
B1 positive (accept strong positive)
✓ 1
Part (c)
M1 substitute \(x = 21\): \(1.6\times21 + 8.4\)
✓ 1
A1 \(=42\), which equals the mean usage (must see 42 and a conclusion)
✓ 1
Part (d)
M1 \(1.6\times30 + 8.4\)
✓ 1
A1 56.4 ML
✓ 1
Part (e)
B1 45 °C is outside the range of the data (extrapolation) so the trend may not continue
✓ 1
Common error: A line of best fit should only be used inside the range of the data; saying an estimate is unreliable needs a reason, such as extrapolation.
107IGMC3-9-11r31-04Averages, range and frequency tablesPaper CMedium6 marks
Calculator
In a game study, 20 players in group A play for a mean of 45 minutes per session. 30 players in group B play for a mean of 55 minutes per session.
(a)
Calculate the mean session length for all 50 players.
(3)
(b)
A 51st player joins the study. The mean session length for all 51 players is now 52 minutes. Find this player's session length. Use 51 minutes for your answer to the first part.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total for A: \(20\times45 = 900\) and total for B: \(30\times55 = 1650\)
✓ 1
M1 \((900 + 1650)\div50\)
✓ 1
A1 51 minutes
✓ 1
Part (b)
M1 new total \(= 51\times52 = 2652\)
✓ 1
M1 \(2652 - 2550\) (ECF from the first part)
✓ 1
A1 102 minutes
✓ 1
Common error: Do not average the two means (45 and 55) directly when the groups have different sizes; use the totals.
108IGMC3-9-11r31-08Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium6 marks
Calculator
In a pharmacology test, the dose of a drug, x mg, and the fall in blood pressure, y units, are recorded for seven patients. Dose x: 1, 2, 3, 4, 6, 8, 10 Fall y: 3, 5, 8, 9, 20, 16, 21
(a)
Describe the correlation between dose and fall in blood pressure.
(1)
(b)
One patient's result does not fit the pattern. Write down the dose of this patient.
(1)
(c)
A line of best fit passes through (2, 5) and (10, 21). Find its equation in the form \(y = mx + c\).
(3)
(d)
Use this line to estimate the fall in blood pressure for a dose of 6 mg.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 positive
✓ 1
Part (b)
B1 6 mg
✓ 1
Part (c)
M1 gradient \(\frac{21-5}{10-2}\)
✓ 1
A1 \(m = 2\)
✓ 1
A1 \(y = 2x + 1\)
✓ 1
Part (d)
B1 13 (ECF from the equation)
✓ 1
Common error: Spot an outlier by comparing each point with the trend, and do not use it to decide the line of best fit.
109IGMC3-9-11r31-09Averages, range and frequency tablesPaper CMedium8 marks
Calculator
The monthly rainfall, in mm, at a reservoir over 12 months is 38, 45, 52, 64, 45, 30, 22, 18, 27, 45, 56, 41.
(a)
Work out the range.
(1)
(b)
Find the median.
(2)
(c)
Calculate the mean.
(2)
(d)
The mean rainfall over 13 months is 41 mm. Find the rainfall in the 13th month.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 46 mm
✓ 1
Part (b)
M1 order the data (18, 22, 27, 30, 38, 41, 45, 45, 45, 52, 56, 64) and use the 6th and 7th values
✓ 1
A1 43 mm
✓ 1
Part (c)
M1 total 483
✓ 1
A1 40.25 mm
✓ 1
Part (d)
M1 total for 13 months \(= 13\times41 = 533\)
✓ 1
M1 \(533 - 483\)
✓ 1
A1 50 mm
✓ 1
Exam tip: For an even number of values, the median is the mean of the two middle values once the data are in order.
110IGMC3-9-12r34-05Averages, range and frequency tablesPaper CMedium5 marks
Calculator
The table shows the time, t minutes, that 50 people spent on one visit to a library.
Time (t minutes)
0 < t ≤ 10
10 < t ≤ 20
20 < t ≤ 30
30 < t ≤ 40
40 < t ≤ 60
Frequency
6
14
18
8
4
(a)
Write down the modal class.
(1)
(b)
Calculate an estimate of the mean time.
(3)
(c)
Explain why your answer to (b) is only an estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 20 < t ≤ 30
✓ 1
Part (b)
M1 use of midpoints 5, 15, 25, 35, 50
✓ 1
M1 (5×6 + 15×14 + 25×18 + 35×8 + 50×4) = 1170
✓ 1
A1 1170 ÷ 50 = 23.4 minutes
✓ 1
Part (c)
B1 the exact times are not known; midpoints are used for each class
✓ 1
Common error: Use the midpoint of each class, not the lower or upper boundary, and divide by the total frequency of 50.
111IGMC3-9-12r34-07Averages, range and frequency tablesPaper CMedium6 marks
No calculator
The monthly energy output (in GWh) of two wind farms over seven months is Farm A: 42, 55, 48, 61, 50, 47, 53 Farm B: 45, 49, 51, 52, 50, 53, 54
(a)
Find the median output of farm A.
(2)
(b)
Find the range of farm B.
(2)
(c)
Farm A has median 50 and range 19. Compare the two farms using the median and the range.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 ordered list 42, 47, 48, 50, 53, 55, 61
✓ 1
A1 50 GWh
✓ 1
Part (b)
M1 54 − 45
✓ 1
A1 9 GWh
✓ 1
Part (c)
B1 farm B has a higher median (51 compared with 50), so a higher typical output
✓ 1
B1 farm B has a smaller range, so its output is more consistent
✓ 1
Exam tip: When comparing, give one comment about the average and one about spread, each written in context.
112IGMC3-9-12r34-08Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium8 marks
Calculator
Six swimmers record the hours of practice per week, x, and their 100 m time, y seconds.
x
2
4
5
7
8
10
y
88
80
77
70
65
58
Scatter diagram of the data
(a)
Describe the type of correlation.
(1)
(b)
Find the mean point (mean of x, mean of y).
(2)
(c)
A line of best fit passes through the mean point and the point (10, 57). Find its equation in the form y = mx + c.
(3)
(d)
Use your equation to estimate the time for a swimmer who practises for 9 hours per week.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 negative correlation
✓ 1
Part (b)
M1 36 ÷ 6 and 438 ÷ 6
✓ 1
A1 (6, 73)
✓ 1
Part (c)
M1 gradient = (57 − 73) ÷ (10 − 6)
✓ 1
A1 m = −4
✓ 1
A1 y = −4x + 97
✓ 1
Part (d)
M1 substitute x = 9 (ECF from (c))
✓ 1
A1 61 seconds
✓ 1
Common error: A line of best fit must pass through the mean point; check that your equation gives y = 73 when x = 6.
113IGMC3-9-12r34-09Averages, range and frequency tablesPaper CMedium6 marks
No calculator
The number of new members joining five library branches in one week is 12, 15, x, 20 and 2x. The mean of these five numbers is 19.
(a)
Find the value of x.
(3)
(b)
Find the median of the five numbers.
(2)
(c)
Find the range.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (12 + 15 + x + 20 + 2x) ÷ 5 = 19
✓ 1
M1 47 + 3x = 95
✓ 1
A1 x = 16
✓ 1
Part (b)
M1 ordered list 12, 15, 16, 20, 32 (ECF from (a))
✓ 1
A1 16
✓ 1
Part (c)
A1 20 (ECF from (a))
✓ 1
Exam tip: Substitute your value back to check that the five numbers really have mean 19.
114IGMC3-9-10r36-05Averages, range and frequency tablesPaper CMedium6 marks
No calculator
In a cycling stage race, five riders in a team had these times, in hours, for one stage: 3.8, 4.5, 4.1, 5.2, 3.9. A sixth rider then finishes in t hours.
(a)
Find the mean time of the first five riders.
(2)
(b)
The mean time of all six riders is 4.4 hours. Find the value of t.
(2)
(c)
Find the median time of all six riders.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 21.5 ÷ 5
✓ 1
A1 4.3
✓ 1
Part (b)
M1 6 × 4.4 − 21.5, or 26.4 − 21.5
✓ 1
A1 4.9
✓ 1
Part (c)
M1 puts the six times in order and identifies the 3rd and 4th values, 4.1 and 4.5
✓ 1
A1 4.3
✓ 1
Exam tip: When a mean is given, multiply it by the number of values to get the total; this turns a mean problem into a simple sum.
115IGMC3-9-10r36-06Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium7 marks
Calculator
The table shows the number of server outages per day recorded for a number of days at a data centre. Outages per day: 0, 1, 2, 3, 4 Number of days: 4, 9, 8, k, 5 The mean number of outages per day is 2.25.
M1 total days = 26 + k, and total outages ÷ total days = 2.25
✓ 1
Part (b)
M1 45 + 3k = 2.25(26 + k)
✓ 1
M1 0.75k = 13.5
✓ 1
A1 k = 18
✓ 1
Part (c)
M1 there are 44 days, so the median is the mean of the 22nd and 23rd values
✓ 1
A1 3
✓ 1
Common error: With a frequency table the total frequency is the sum of the frequencies, including the unknown k, so remember to add k to the number of days.
116IGMC3-9-10r36-07Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium6 marks
Calculator
Archaeologists dig down through a site and record the depth, d m, and the estimated age, a years, of six objects. Depth d (m): 0.5, 1.0, 1.5, 2.0, 2.5, 3.0 Age a (years): 300, 500, 800, 1000, 1300, 1500
Scatter diagram of the data
(a)
Describe the correlation between depth and age.
(1)
(b)
Find the mean depth and the mean age.
(2)
(c)
A line of best fit passes through the mean point and through the point (3.0, 1500). Calculate the gradient of this line.
(2)
(d)
Explain what the gradient represents in this context.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 strong positive correlation
✓ 1
Part (b)
B1 mean depth = 1.75 m
✓ 1
B1 mean age = 900 years
✓ 1
Part (c)
M1 (1500 − 900) ÷ (3.0 − 1.75)
✓ 1
A1 480
✓ 1
Part (d)
B1 for each extra metre of depth, the age increases by about 480 years
✓ 1
Exam tip: When interpreting a gradient, say how much the quantity on the vertical axis changes for each one unit increase of the horizontal quantity, using the units.
117IGMC3-9-10r36-09Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium6 marks
Calculator
A museum stores pottery found at an archaeological site. The first group of 12 pots has a mean mass of 340 g. A second group of 8 pots has a mean mass of 415 g.
(a)
Find the total mass of the first group of pots.
(1)
(b)
Calculate the mean mass of all 20 pots.
(3)
(c)
A 21st pot is added. The mean mass of all 21 pots is 372 g. Find the mass of the 21st pot.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 12 × 340 = 4080 g
✓ 1
Part (b)
M1 8 × 415 = 3320
✓ 1
M1 (4080 + 3320) ÷ 20
✓ 1
A1 370
✓ 1
Part (c)
M1 21 × 372 − 7400, or 7812 − 7400
✓ 1
A1 412
✓ 1
Exam tip: Do not average two means directly; find each total mass first, add the totals and divide by the total number of items.
118IGMC3-9-11r38-05Averages, range and frequency tablesPaper CMedium6 marks
Calculator
The lengths, L cm, of 50 fish landed at a harbour are summarised below. 20 < L ≤ 30: frequency 6 30 < L ≤ 40: frequency 14 40 < L ≤ 50: frequency 18 50 < L ≤ 60: frequency 9 60 < L ≤ 70: frequency 3
(a)
Write down the modal class.
(1)
(b)
Write down the class that contains the median.
(1)
(c)
Calculate an estimate of the mean length.
(3)
(d)
Explain why your answer to (c) is only an estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 40 < L ≤ 50
✓ 1
Part (b)
B1 40 < L ≤ 50 (the 25th and 26th values lie in this class; cumulative frequencies 20 then 38)
✓ 1
Part (c)
M1 mid-values 25, 35, 45, 55, 65 used
✓ 1
M1 Σ(frequency × mid-value) = 2140
✓ 1
A1 2140 ÷ 50 = 42.8 cm
✓ 1
Part (d)
B1 the exact lengths are not known; mid-values are assumed for each class
✓ 1
Exam tip: Write the mid-values in a column next to the frequencies so the products are easy to check.
119IGMC3-9-11r38-06Averages, range and frequency tablesPaper CMedium6 marks
No calculator
A fashion designer prices six scarves, in £: 14, 9, x, 12, 9, 16. The mean price is £11.
(a)
Find the value of x.
(3)
(b)
Write down the mode.
(1)
(c)
Find the median price.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total price = 6 × 11 = 66
✓ 1
M1 66 − (14 + 9 + 12 + 9 + 16) = 66 − 60
✓ 1
A1 x = 6
✓ 1
Part (b)
B1 9
✓ 1
Part (c)
M1 prices in order 6, 9, 9, 12, 14, 16 (ECF from (a))
✓ 1
A1 (9 + 12) ÷ 2 = 10.5
✓ 1
Common error: Writing the middle value of an unordered list as the median.
120IGMC3-9-11r38-07Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium6 marks
Calculator
A robotics team tests how the payload carried, p kg, affects the running time, t minutes, of a warehouse robot on one battery charge. Payload p: 2, 4, 6, 8, 10 Running time t: 55, 49, 44, 38, 33
Scatter diagram of the data
(a)
Describe the correlation between p and t.
(1)
(b)
Find the mean of p and the mean of t. A line of best fit passes through this mean point.
(2)
(c)
The line of best fit has gradient −2.7. Use the line to estimate the running time when p = 9.
(2)
(d)
Explain what the gradient −2.7 means in this context.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 negative correlation
✓ 1
Part (b)
B1 mean of p = 6
✓ 1
B1 mean of t = 219 ÷ 5 = 43.8
✓ 1
Part (c)
M1 43.8 − 2.7 × (9 − 6)
✓ 1
A1 35.7 minutes
✓ 1
Part (d)
B1 each extra kilogram of payload reduces the running time by about 2.7 minutes
✓ 1
Exam tip: A line of best fit passes through the mean point, so use it as a fixed point when you apply a given gradient.
121IGMC3-9-11r38-10Averages, range and frequency tablesPaper CMedium7 marks
Calculator
A harbour office records the number of fishing boats, x, that leave port each day. Number of boats x: 1, 2, 3, 4 Number of days: 4, 7, k, 5 The mean number of boats per day is 2.5.
(a)
Find the value of k.
(3)
(b)
Find the median number of boats per day.
(2)
(c)
On one more day, 5 boats leave port. Calculate the new mean number of boats per day. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total boats = 4 + 14 + 3k + 20 = 38 + 3k, and total days = 16 + k
M1 20 days, so the median is between the 10th and 11th values (cumulative frequencies 4, 11)
✓ 1
A1 2
✓ 1
Part (c)
M1 (50 + 5) ÷ 21
✓ 1
A1 2.62
✓ 1
Exam tip: Form an equation from Σfx ÷ Σf = mean, then multiply out the fraction.
122IGMC3-9-11r43-04Averages, range and frequency tablesPaper CMedium5 marks
Calculator
A pharmacist times how long 40 tablets take to dissolve in water. The table shows the results. Time t (minutes): 0 < t ≤ 2, 2 < t ≤ 4, 4 < t ≤ 6, 6 < t ≤ 8, 8 < t ≤ 10 Frequency: 4, 10, 14, 8, 4
(a)
Write down the modal class.
(1)
(b)
Calculate an estimate of the mean time.
(3)
(c)
Explain why your answer to part (b) is only an estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 4 < t ≤ 6
✓ 1
Part (b)
M1 mid-values 1, 3, 5, 7, 9 multiplied by the frequencies and added: 4+30+70+56+36 = 196
✓ 1
M1 196 ÷ 40
✓ 1
A1 4.9
✓ 1
Part (c)
B1 the exact times are not known; each tablet is assumed to have the mid-value of its class
✓ 1
Exam tip: Use the mid-value of each class, not the class width or the upper boundary, when estimating a mean from grouped data.
123IGMC3-9-11r43-05Averages, range and frequency tablesPaper CMedium8 marks
Calculator
A player scores 42, 55, x, 61 and 38 points in five games. The mean score for the five games is 50 points.
(a)
Find the value of x.
(3)
(b)
The player scores y points in a sixth game. The mean score for all six games is 51 points. Find y.
(3)
(c)
Find the median of the six scores. Use your values from parts (a) and (b), or ECF from them.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total score = 5 × 50 = 250
✓ 1
M1 250 − (42 + 55 + 61 + 38)
✓ 1
A1 54
✓ 1
Part (b)
M1 total for six games = 6 × 51 = 306
✓ 1
M1 306 − 250
✓ 1
A1 56
✓ 1
Part (c)
M1 scores in order 38, 42, 54, 55, 56, 61
✓ 1
A1 54.5
✓ 1
Common error: To find a missing value from a mean, work with the total (mean × number of values), not with the mean itself.
124IGMC3-9-11r43-06Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium7 marks
Calculator
A water company monitors household use. 12 households in street A use a mean of 150 litres per day. 8 households in street B use a mean of 200 litres per day.
(a)
Work out the total volume of water used per day by the 12 households in street A.
(1)
(b)
Calculate the mean volume per day for all 20 households.
(3)
(c)
A 21st household joins the study. The mean for all 21 households is now 171 litres per day. Find the daily use of the 21st household.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 1800 (litres)
✓ 1
Part (b)
M1 8 × 200 = 1600
✓ 1
M1 (1800 + 1600) ÷ 20
✓ 1
A1 170 (litres)
✓ 1
Part (c)
M1 21 × 171 = 3591
✓ 1
M1 3591 − 3400
✓ 1
A1 191 (litres)
✓ 1
Exam tip: Never average two means directly unless the groups are the same size; combine the totals instead.
125IGMC3-9-11r43-07Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium8 marks
Calculator
A doctor records the dose of a drug, x mg, and the fall in blood pressure, y mmHg, for five patients. Dose x: 10, 20, 30, 40, 50 Fall y: 4, 9, 13, 18, 21
Scatter diagram with the line of best fit through (30, 13) and (50, 22)
(a)
Find the mean of x and the mean of y.
(2)
(b)
Describe the correlation and say what it shows about dose and blood pressure.
(2)
(c)
A line of best fit passes through (30, 13) and (50, 22). Use it to estimate the fall in blood pressure for a dose of 36 mg.
(3)
(d)
Explain why the line is unreliable for a dose of 120 mg.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 30
✓ 1
B1 13
✓ 1
Part (b)
B1 positive correlation
✓ 1
B1 a larger dose is linked to a larger fall in blood pressure
✓ 1
Part (c)
M1 gradient (22 − 13) ÷ (50 − 30) = 0.45
✓ 1
M1 13 + 0.45 × (36 − 30)
✓ 1
A1 15.7 (mmHg)
✓ 1
Part (d)
B1 120 mg is far outside the range of the data (10 mg to 50 mg), so this is extrapolation
✓ 1
Exam tip: Describe correlation using its type (positive, negative or none) and then link it to the two quantities in the context.
126IGMC3-9-11r43-13Averages, range and frequency tablesPaper CMedium9 marks
Calculator
A water company records the volume pumped, in megalitres (ML), on seven days. 18 22 19 25 22 31 21
(a)
Find the median volume.
(2)
(b)
Calculate the mean volume. Give your answer correct to 3 significant figures.
(2)
(c)
The 31 ML was recorded on the day of a burst pipe. It is removed. Calculate the mean of the other six values. Give your answer correct to 3 significant figures.
(3)
(d)
Find the median of the six values. Explain why the median is less affected than the mean by removing the 31 ML.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 values in order 18, 19, 21, 22, 22, 25, 31
✓ 1
A1 22
✓ 1
Part (b)
M1 158 ÷ 7
✓ 1
A1 22.6
✓ 1
Part (c)
M1 158 − 31 = 127
✓ 1
M1 127 ÷ 6
✓ 1
A1 21.2
✓ 1
Part (d)
B1 21.5
✓ 1
B1 the median moved from 22 to 21.5 (change 0.5) but the mean moved by about 1.4, because the mean uses the size of every value including the extreme one
✓ 1
Common error: An extreme value pulls the mean towards it but changes the median very little.
127IGMC3-9-11r45-03Averages, range and frequency tablesPaper CMedium7 marks
Calculator
A hospital records how long 50 patients wait in an emergency department.
Waiting time, t minutes
0 < t ≤ 10
10 < t ≤ 20
20 < t ≤ 30
30 < t ≤ 60
Frequency
8
15
17
10
(a)
Write down the modal class.
(1)
(b)
Find the class that contains the median.
(1)
(c)
Calculate an estimate of the mean waiting time.
(4)
(d)
Explain why your answer to part (c) is only an estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
20 < t ≤ 30
✓ 1
Part (b)
The 25th and 26th values lie in the third class (cumulative frequencies 8, 23, 40): 20 < t ≤ 30
✓ 1
Part (c)
M1 uses midpoints 5, 15, 25, 45
✓ 1
M1 Σfx = 5×8 + 15×15 + 25×17 + 45×10 = 1140
✓ 1
M1 divides by 50
✓ 1
A1 22.8 minutes
✓ 1
Part (d)
The exact waiting times are not known; the midpoint is assumed to represent each class
✓ 1
Exam tip: For grouped data use the midpoint of each class; the class 30 < t ≤ 60 has midpoint 45, not 30.
128IGMC3-9-11r45-05Averages, range and frequency tablesPaper CMedium6 marks
Calculator
A Formula 1 driver completes 6 consecutive laps. The mean lap time is 92.5 s. Five of the lap times, in seconds, are 91.8, 93.2, 92.0, 94.1, 90.9.
Exam tip: If you are given a mean, multiply by the number of values to get the total before looking for a missing value.
129IGMC3-9-11r45-06Averages, range and frequency tablesPaper CMedium7 marks
Calculator
A hospital ward has 12 nurses with a mean age of 34 years and 8 doctors with a mean age of 45 years.
(a)
Calculate the total of the ages of the 12 nurses.
(1)
(b)
Calculate the mean age of all 20 nurses and doctors.
(3)
(c)
A 21st person joins the group and the mean age of all 21 people becomes exactly 39 years. Find the age of the new person.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
12 × 34 = 408 years
✓ 1
Part (b)
M1 total of doctors' ages = 8 × 45 = 360
✓ 1
M1 (408 + 360) ÷ 20
✓ 1
A1 38.4 years
✓ 1
Part (c)
M1 21 × 39 = 819
✓ 1
M1 subtracts the previous total 768
✓ 1
A1 51 years
✓ 1
Common error: Do not find the mean of two means by adding and halving unless the groups are the same size; combine the totals instead.
130IGMC3-9-11r45-11Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium8 marks
Calculator
A Formula 1 team tests 8 drivers on a simulator. The table shows the hours of simulator practice, x, and the best lap time, y seconds.
x
2
4
6
7
9
10
12
14
y
98
96
95
93
92
90
87
85
Scatter diagram with the line of best fit through (8, 92) and (20, 80)
(a)
Find the mean of x and the mean of y.
(2)
(b)
Describe the correlation.
(1)
(c)
A line of best fit passes through the mean point (8, 92) and through the point (20, 80). Find the equation of the line in the form y = mx + c.
(3)
(d)
Use your equation to estimate the best lap time after 11 hours of practice.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 mean of x = 8
✓ 1
B1 mean of y = 92
✓ 1
Part (b)
Negative correlation (more practice, lower lap time)
✓ 1
Part (c)
M1 gradient = (80 − 92) ÷ (20 − 8)
✓ 1
A1 m = −1
✓ 1
A1 y = −x + 100
✓ 1
Part (d)
M1 substitutes x = 11 (ECF from (c))
✓ 1
A1 89 s
✓ 1
Common error: A line of best fit should pass through the mean point; check that your equation gives 92 when x = 8.
131IGMC3-9-10r46-04Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium7 marks
Calculator
A beekeeper records the maximum daily temperature, x °C, and the number of bees, y, leaving a hive in one minute on eight days. The scatter diagram shows points rising from lower left to upper right, close to a straight line. A line of best fit is drawn through the points (14, 20) and (26, 68).
(a)
Describe the correlation between temperature and the number of bees leaving the hive.
(1)
(b)
Calculate the gradient of the line of best fit.
(2)
(c)
Find the equation of the line of best fit in the form y = mx + c.
(2)
(d)
Use your equation to estimate the number of bees leaving the hive in one minute when the temperature is 21 °C.
(1)
(e)
Explain why the equation should not be used to estimate the number of bees when the temperature is 40 °C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 positive correlation
✓ 1
Part (b)
M1 (68 − 20) ÷ (26 − 14)
✓ 1
A1 4
✓ 1
Part (c)
M1 substitutes a point, e.g. 20 = 4 × 14 + c
✓ 1
A1 y = 4x − 36
✓ 1
Part (d)
B1 4 × 21 − 36 = 48 (ECF from (c))
✓ 1
Part (e)
B1 40 °C is outside the range of the data, so the estimate is an extrapolation and unreliable
✓ 1
Exam tip: For an estimate from a line of best fit, substitute into the equation of the line rather than reading values from the nearest data point.
132IGMC3-9-10r46-10Averages, range and frequency tablesPaper CMedium6 marks
Calculator
The table shows the number of days that 25 patients stayed in a hospital ward after a minor operation. The frequency for 3 days is p. Days | 1 | 2 | 3 | 4 | 5 Frequency | 4 | 9 | p | 6 | 3
(a)
Find the value of p.
(1)
(b)
Find the median number of days.
(2)
(c)
Calculate the mean number of days.
(2)
(d)
Work out the total number of days that the 25 patients stayed in the ward.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 25 − (4 + 9 + 6 + 3) = 3
✓ 1
Part (b)
M1 the 13th value
✓ 1
A1 2
✓ 1
Part (c)
M1 (4 + 18 + 9 + 24 + 15) ÷ 25
✓ 1
A1 2.8
✓ 1
Part (d)
B1 25 × 2.8 = 70 (ECF from (c))
✓ 1
Common error: In a frequency table the median position comes from the cumulative frequency, not from the middle of the list of values.
133IGMC3-9-10r46-11Scatter diagrams, correlation, cumulative frequency and histogramsPaper CMedium6 marks
Calculator
A Formula 1 team studies how tyre wear affects lap time. For six laps they record the age of the tyres, x laps, and the lap time, y seconds: (4, 82.1), (8, 82.6), (12, 83.2), (16, 83.5), (20, 84.1), (24, 84.6).
Scatter diagram of the data
(a)
Describe the correlation between tyre age and lap time.
(1)
(b)
Show that the mean point is (14, 83.35).
(2)
(c)
A line of best fit passes through the mean point and has gradient 0.12. Estimate the lap time when the tyres are 18 laps old.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 positive correlation
✓ 1
Part (b)
M1 84 ÷ 6 = 14 and 500.1 ÷ 6
✓ 1
A1 x̄ = 14 and ȳ = 83.35
✓ 1
Part (c)
M1 uses y − 83.35 = 0.12(x − 14) or equivalent
✓ 1
M1 substitutes x = 18
✓ 1
A1 83.83 s
✓ 1
Exam tip: A line of best fit must pass through the mean point; use it as a fixed point when you build the equation.
134IGMC-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.
135IGMC-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.
136IGMC3-9-06d-13Averages, range and frequency tablesPaper CHard9 marks
Calculator
The table shows the number of goals scored by a team in each of its matches. Goals: 0, 1, 2, 3, 4 Number of matches: 4, x, 9, 5, 2 The mean number of goals per match is 1.75.
(a)
Form an equation in x and solve it.
(4)
(b)
Find the median number of goals.
(2)
(c)
In the next 12 matches the team scores a mean of 2.5 goals per match. Find the mean for all 40 matches.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total goals = x + 18 + 15 + 8 = x + 41
✓ 1
M1 total matches = x + 20
✓ 1
M1 (x + 41) = 1.75(x + 20)
✓ 1
A1 x = 8
✓ 1
Part (b)
M1 28 matches so the 14th and 15th values (cumulative totals 4, 12, 21)
✓ 1
A1 2
✓ 1
Part (c)
M1 12 × 2.5 = 30 goals
✓ 1
M1 (49 + 30) ÷ 40
✓ 1
A1 1.975
✓ 1
Exam tip: Set up the mean as total goals ÷ total matches, with both written in terms of the unknown.
137IGMC3-9-08r7-13Scatter diagrams, correlation, cumulative frequency and histogramsPaper CHard6 marks
Calculator
In a drug trial the mean reaction time of 24 participants in group A is 62 ms. The mean reaction time of the 16 participants in group B is m ms. The mean of all 40 participants is 65.8 ms.
(a)
Find the value of m.
(3)
(b)
A participant in group A with a reaction time of 80 ms leaves the trial. Calculate the mean reaction time of the remaining 39 participants. Give your answer to 3 significant figures.
(2)
(c)
Explain why the median of all 40 participants cannot be found from the two group means.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total for all 40 = 40 × 65.8 = 2632
✓ 1
M1 total for A = 24 × 62 = 1488
✓ 1
A1 m = (2632 − 1488) ÷ 16 = 71.5
✓ 1
Part (b)
M1 (2632 − 80) ÷ 39
✓ 1
A1 65.4 ms
✓ 1
Part (c)
R1 the median depends on the individual values, which the means do not give
✓ 1
Common error: Do not average two means directly unless the groups are the same size; combine the totals.
138IGMC3-9-10r14-12Averages, range and frequency tablesPaper CHard5 marks
Calculator
An inspector records the number of turbines on each wind farm in a region.
Number of turbines
2
3
4
5
6
Number of farms
5
8
x
6
3
The mean number of turbines per farm is 3.75.
(a)
Write an expression, in terms of x, for the mean number of turbines per farm.
Exam tip: With a frequency table, mean = Σ(value × frequency) ÷ Σ(frequency); an unknown frequency appears in both the top and the bottom.
139IGMC3-9-10r14-13Scatter diagrams, correlation, cumulative frequency and histogramsPaper CHard4 marks
Calculator
A swimming club has two groups. Group A has 15 swimmers with a mean 50 m time of 48.4 s. Group B has 25 swimmers. The mean time of all 40 swimmers is 52.2 s.
(a)
Calculate the total of the times of the swimmers in group A.
(1)
(b)
Calculate the total of the times of all 40 swimmers.
(1)
(c)
Calculate the mean time of group B, correct to 3 significant figures. Use unrounded values.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 15 × 48.4 = 726 s
✓ 1
Part (b)
B1 40 × 52.2 = 2088 s
✓ 1
Part (c)
M1 (2088 − 726) ÷ 25
✓ 1
A1 54.5 s
✓ 1
Common error: You cannot average two means directly; each mean must be weighted by the size of its group, so work with totals.
140IGMC3-9-09r16-13Averages, range and frequency tablesPaper CHard4 marks
No calculator
The number of pods in each of some roasting batches is recorded. The frequency of batches with 2 pods is f.
Pods
0
1
2
3
4
Batches
5
9
f
7
3
The mean number of pods is 1.8.
(a)
How many batches have more than 2 pods?
(1)
(b)
Find the value of f.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
7 + 3 = 10
✓ 1
Part (b)
M1 sum of pods = 0 + 9 + 2f + 21 + 12 = 42 + 2f and total batches = 24 + f
Common error: The mean uses the total of (value × frequency) divided by the total frequency, and the total frequency includes the unknown f.
141IGMC3-9-09r16-14Averages, range and frequency tablesPaper CHard6 marks
Calculator
A coffee importer has two deliveries of sacks. Delivery A has 20 sacks with mean mass 24.5 kg. Delivery B has 30 sacks with mean mass 21.0 kg.
(a)
Find the total mass of the sacks in Delivery A.
(1)
(b)
Calculate the mean mass of all 50 sacks.
(2)
(c)
A third delivery of 10 sacks arrives. The mean mass of all 60 sacks is 23.0 kg. Find the mean mass of the sacks in the third delivery.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
20 × 24.5 = 490 kg
✓ 1
Part (b)
M1 (490 + 30 × 21.0) ÷ 50 = 1120 ÷ 50
✓ 1
A1 22.4 kg
✓ 1
Part (c)
M1 total mass of 60 sacks = 60 × 23.0 = 1380
✓ 1
M1 mass of third delivery = 1380 − 1120 = 260
✓ 1
A1 260 ÷ 10 = 26.0 kg
✓ 1
Exam tip: To combine means, convert each mean to a total first; never average two means unless the groups are the same size.
142IGMC3-9-08r19-14Averages, range and frequency tablesPaper CHard5 marks
No calculator
Five whole numbers, written in order from smallest to largest, are the numbers of coins found in five graves. The mean is 8, the median is 7, the mode is 6 and the range is 7. Find the five numbers.
(a)
Find the five numbers.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sum of the five numbers = 5 × 8 = 40
✓ 1
B1 middle number = 7
✓ 1
B1 smallest two numbers are both 6 (the mode must be 6 and no number below the median can exceed 7)
✓ 1
M1 largest = 6 + 7 = 13, so fourth number = 40 − 6 − 6 − 7 − 13
✓ 1
A1 6, 6, 7, 8, 13
✓ 1
Exam tip: Use each statistic in turn: the mean gives the total, the median gives the middle value, and the mode and range fix the ends.
143IGMC3-9-08r25-08Averages, range and frequency tablesPaper CHard6 marks
Calculator
The numbers of passengers who missed a connecting flight on five days are five whole numbers. Written in order of size, the five numbers have mode 3, median 4, mean 6 and range 9.
(a)
Show that the sum of the five numbers is 30.
(1)
(b)
Explain why the two smallest numbers must both be 3.
(2)
(c)
Find the largest number.
(1)
(d)
Find the fourth number, in order of size.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5 × 6 = 30
✓ 1
Part (b)
B1 the mode is 3, so 3 appears at least twice
✓ 1
B1 the median is 4, so only the two smallest values can be below it and no larger value can equal 3
✓ 1
Part (c)
A1 3 + 9 = 12
✓ 1
Part (d)
M1 30 − (3 + 3 + 4 + 12)
✓ 1
A1 8
✓ 1
Exam tip: Use each statistic one at a time: the mean gives the total, the median gives the middle value and the range links the smallest and largest.
144IGMC3-9-12r29-13Averages, range and frequency tablesPaper CHard9 marks
Calculator
A pharmacy recorded the number of customers on five days. The numbers are positive integers with mode 6, median 7, mean 8 and range 7.
(a)
Show that the total number of customers over the five days is 40.
(1)
(b)
Find the five numbers, in order.
(4)
(c)
On a sixth day there were 10 customers. Find the new mean, to 3 significant figures.
(2)
(d)
Find the new median.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5 × 8 = 40
✓ 1
Part (b)
M1 median is the third value, so it is 7
✓ 1
M1 mode is 6, so 6 occurs at least twice and the two smallest values are both 6; the range is 7, so the largest value is 13
✓ 1
M1 6 + 6 + 7 + d + 13 = 40, so d = 8 (fourth value)
✓ 1
A1 6, 6, 7, 8, 13
✓ 1
Part (c)
M1 (40 + 10)/6
✓ 1
A1 8.33
✓ 1
Part (d)
M1 ordered list 6, 6, 7, 8, 10, 13
✓ 1
A1 7.5 (the mean of 7 and 8)
✓ 1
Common error: Do not stop once the median and mode are found; use every condition, including the range and the total.
145IGMC3-9-12r34-06Scatter diagrams, correlation, cumulative frequency and histogramsPaper CHard6 marks
Calculator
A swimming coach records the number of lengths swum by a group of children in a session.
Lengths
20
25
30
35
40
Number of children
4
7
x
6
3
The mean number of lengths is 29.5.
(a)
Show that the total number of lengths swum by the children is 585 + 30x.
(2)
(b)
Find the value of x.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 4×20 + 7×25 + 6×35 + 3×40 (= 585)
✓ 1
A1 adding 30 × x to give 585 + 30x
✓ 1
Part (b)
B1 total number of children is 20 + x
✓ 1
M1 (585 + 30x) ÷ (20 + x) = 29.5
✓ 1
M1 585 + 30x = 590 + 29.5x
✓ 1
A1 x = 10
✓ 1
Exam tip: Set up the mean as total ÷ number of children, multiply both sides by (20 + x), then collect the x terms.
146IGMC3-9-10r36-14Averages, range and frequency tablesPaper CHard7 marks
Calculator
Five riders in a cycling challenge each complete a whole number of laps. The mode is 7, the median is 8, the mean is 9 and the range is 7.
(a)
Show that the total number of laps completed by the five riders is 45.
(1)
(b)
Find the number of laps completed by each of the five riders.
(4)
(c)
A sixth rider completes 8 laps. Calculate the new mean number of laps. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 5 × 9 = 45
✓ 1
Part (b)
M1 the median is the 3rd value, so it is 8, and the mode of 7 means two riders completed 7 laps, which are the two smallest values
✓ 1
M1 smallest 7 and largest 7 + 7 = 14
✓ 1
M1 4th value = 45 − 7 − 7 − 8 − 14
✓ 1
A1 7, 7, 8, 9, 14
✓ 1
Part (c)
M1 (45 + 8) ÷ 6
✓ 1
A1 8.83
✓ 1
Exam tip: Put the unknown values in order from smallest to largest and use each given statistic (mode, median, mean, range) to fix one value at a time.
147IGMC3-9-11r38-14Scatter diagrams, correlation, cumulative frequency and histogramsPaper CHard8 marks
Calculator
A robotics lab times 15 test runs of a sorting robot. The mean time is 42.4 s. Two times were recorded wrongly: 63 s should have been 36 s, and 72 s should have been 27 s.
(a)
Work out the total of the 15 recorded times.
(2)
(b)
Find the correct mean time for the 15 runs.
(3)
(c)
A 16th run is timed. The mean of all 16 correct times is exactly 38 s. Find the time of the 16th run.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 15 × 42.4
✓ 1
A1 636 s
✓ 1
Part (b)
M1 corrected total = 636 − 63 + 36 − 72 + 27
✓ 1
A1 564
✓ 1
A1 564 ÷ 15 = 37.6 s
✓ 1
Part (c)
M1 16 × 38 = 608
✓ 1
M1 608 − 564
✓ 1
A1 44 s
✓ 1
Common error: Correcting the mean by the average of the two errors instead of correcting the total and dividing again.
148IGMC3-9-11r43-08Averages, range and frequency tablesPaper CHard9 marks
Calculator
In a game, each player collects some rare items. The table shows the number of items collected. The mean number of items collected is 2.5. Items collected: 1, 2, 3, 4 Number of players: 4, x, 7, 5
(a)
Find the value of x.
(4)
(b)
Find the median number of items collected.
(2)
(c)
Four more players join and each collects 5 items. Calculate the new mean. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 total items = 4 + 2x + 21 + 20 = 45 + 2x
✓ 1
M1 number of players = 16 + x
✓ 1
M1 (45 + 2x) ÷ (16 + x) = 2.5
✓ 1
A1 x = 10
✓ 1
Part (b)
M1 26 players, so 13th and 14th values (cumulative frequencies 4, 14)
✓ 1
A1 2 (ECF from (a))
✓ 1
Part (c)
M1 total items now 65 + 4 × 5 = 85
✓ 1
M1 85 ÷ 30
✓ 1
A1 2.83
✓ 1
Common error: When a frequency is unknown, write the total number of items and the total number of players as expressions in the unknown before forming the equation.
149IGMC3-9-11r45-07Averages, range and frequency tablesPaper CHard5 marks
Calculator
A survey asks beekeepers how many times they were stung in one season.
Number of stings
0
1
2
3
Frequency
5
8
x
4
The mean number of stings is 1.6.
(a)
Write down, in terms of x, the total number of beekeepers and the total number of stings.
(2)
(b)
Form an equation using the mean and solve it to find x.
Exam tip: Mean = total ÷ frequency, so with an unknown frequency write both the total and the number of items as expressions first.
150IGMC3-9-10r46-05Averages, range and frequency tablesPaper CHard6 marks
No calculator
Five whole numbers, written in order of size, represent the number of operations carried out in a hospital theatre on five days. The numbers have mode 6, median 7, mean 9 and range 9.
(a)
Show that the five numbers add up to 45.
(1)
(b)
Explain why the two smallest numbers must both be 6.
(2)
(c)
Find the largest number.
(1)
(d)
Find the fourth number.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 5 × 9 = 45
✓ 1
Part (b)
B1 the median is 7, so the two smallest numbers are at most 7 and the mode needs 6 to appear at least twice
✓ 1
B1 the numbers above the median are at least 7, so the two 6s must be the two smallest
✓ 1
Part (c)
B1 6 + 9 = 15
✓ 1
Part (d)
M1 45 − (6 + 6 + 7 + 15)
✓ 1
A1 11
✓ 1
Exam tip: Use the median, mode and range one at a time to fix individual values, then use the mean to find the last one.
151IGMC3-9-10r46-06Averages, range and frequency tablesPaper CHard7 marks
Calculator
A beekeeper has two apiaries. Apiary A has 12 hives with a mean honey yield of 21.5 kg per hive. Apiary B has 8 hives with a mean yield of 26.0 kg per hive.
(a)
Calculate the total yield of the hives in apiary A.
(1)
(b)
Calculate the mean yield per hive for all 20 hives in apiaries A and B together.
(3)
(c)
A third apiary, C, has 5 hives. The mean yield for all 25 hives in A, B and C is 24.0 kg. Calculate the mean yield per hive in apiary C.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 12 × 21.5 = 258 kg
✓ 1
Part (b)
M1 total for B = 8 × 26.0 = 208
✓ 1
M1 (258 + 208) ÷ 20
✓ 1
A1 23.3 kg
✓ 1
Part (c)
M1 total for 25 hives = 24.0 × 25 = 600
✓ 1
M1 (600 − 258 − 208) ÷ 5
✓ 1
A1 26.8 kg
✓ 1
Common error: Do not average two means directly unless the groups are the same size; combine the totals instead.
152IGMC-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.
153IGMC-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.
154IGMC3-9-11r43-12Scatter diagrams, correlation, cumulative frequency and histogramsPaper EEasy7 marks
No calculator
In a game, 160 players complete level A. The shortest time is 6 minutes and the longest is 52 minutes. The lower quartile is 14 minutes, the median is 20 minutes and the upper quartile is 31 minutes. On level B, a different group has a median time of 24 minutes and an interquartile range of 7 minutes.
(a)
Find the range of the times on level A.
(1)
(b)
Find the interquartile range of the times on level A.
(2)
(c)
Find the number of players on level A whose time is between the median and the upper quartile.
(2)
(d)
Compare the times on the two levels.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 46 (minutes)
✓ 1
Part (b)
M1 31 − 14
✓ 1
A1 17 (minutes)
✓ 1
Part (c)
M1 one quarter of 160 (or 25% of 160)
✓ 1
A1 40
✓ 1
Part (d)
B1 level B has the higher median, so times are typically longer on level B
✓ 1
B1 level B has the smaller interquartile range, so times are more consistent on level B
✓ 1
Exam tip: A comparison needs one statement about an average (median) and one about spread (interquartile range), both in context.
155IGMC-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.
156IGMC-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.
157IGMC-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.
158IGMC-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.
159IGMC-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
Histogram of the visitors' ages
(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.
160IGMC3-9-06d-09Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium8 marks
Calculator
The lifetimes, t hours, of 80 e-scooter batteries are summarised by cumulative frequencies. t ≤ 10: 6; t ≤ 20: 26; t ≤ 30: 46; t ≤ 40: 66; t ≤ 50: 76; t ≤ 60: 80. Assume that values are spread evenly within each class.
(a)
Find the number of batteries with a lifetime greater than 40 hours and at most 50 hours.
(1)
(b)
Estimate the median lifetime.
(3)
(c)
Estimate the interquartile range.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
76 − 66 = 10
✓ 1
Part (b)
M1 40th value lies in 20 < t ≤ 30
✓ 1
M1 20 + (40 − 26)/20 × 10
✓ 1
A1 27
✓ 1
Part (c)
M1 20th and 60th values
✓ 1
A1 Q1 = 10 + (20 − 6)/20 × 10 = 17
✓ 1
A1 Q3 = 30 + (60 − 46)/20 × 10 = 37
✓ 1
A1 IQR = 20
✓ 1
Exam tip: On a cumulative frequency curve, quartiles are read at 1/4 and 3/4 of the total frequency (here the 20th and 60th values).
161IGMC3-9-06d-10Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium6 marks
Calculator
A histogram shows the waiting times, t minutes, of 120 clinic patients. The bar for 0 < t ≤ 10 represents 15 patients. The bar for 30 < t ≤ 50 represents 18 patients. The bar for 10 < t ≤ 25 has height 3.2 (frequency density). The only other bar is for 25 < t ≤ 30.
(a)
Find the number of patients represented by the bar for 10 < t ≤ 25.
(2)
(b)
Find the number of patients represented by the bar for 25 < t ≤ 30.
(2)
(c)
Find the height of the bar for 25 < t ≤ 30.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 3.2 × 15
✓ 1
A1 48
✓ 1
Part (b)
M1 120 − (15 + 48 + 18)
✓ 1
A1 39 (ECF from (a))
✓ 1
Part (c)
M1 39 ÷ 5
✓ 1
A1 7.8 (ECF from (b))
✓ 1
Common error: In a histogram the area, not the height, represents frequency; frequency density = frequency ÷ class width.
162IGMC3-9-06d-11Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The speeds, v km/h, of 200 cars passing a sensor are summarised by cumulative frequencies. v ≤ 30: 12; v ≤ 40: 50; v ≤ 50: 120; v ≤ 60: 176; v ≤ 70: 196; v ≤ 80: 200. Assume that values are spread evenly within each class.
(a)
The speed limit is 60 km/h. Find the percentage of cars travelling faster than the limit.
(2)
(b)
Estimate the number of cars with a speed of 45 km/h or less.
(2)
(c)
The fastest 10% of cars are fined. Estimate the lowest speed at which a car is fined.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (200 − 176)/200
✓ 1
A1 12%
✓ 1
Part (b)
M1 50 + (45 − 40)/10 × 70
✓ 1
A1 85
✓ 1
Part (c)
M1 fastest 10% starts at the 180th car
✓ 1
M1 60 + (180 − 176)/20 × 10
✓ 1
A1 62
✓ 1
Exam tip: "Fastest 10%" means the 90th percentile, which is the 180th of 200 values.
163IGMC3-9-08r7-07Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The times taken by 80 players to finish a game level are summarised.
Time t (minutes)
0 < t ≤ 10
10 < t ≤ 20
20 < t ≤ 30
30 < t ≤ 40
40 < t ≤ 60
Frequency
6
14
30
20
10
Assume that the times in each class are evenly spread.
(a)
Write down the cumulative frequencies for the five classes.
(2)
(b)
State the class that contains the median.
(1)
(c)
Estimate the median time by linear interpolation. Give your answer correct to 3 significant figures.
(2)
(d)
Estimate how many players took more than 35 minutes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 6, 20, 50
✓ 1
B1 70, 80
✓ 1
Part (b)
B1 20 < t ≤ 30 (the 40th value lies there)
✓ 1
Part (c)
M1 20 + (40 − 20) ÷ 30 × 10
✓ 1
A1 26.7 minutes
✓ 1
Part (d)
M1 cumulative frequency at 35 = 50 + 0.5 × 20 = 60
✓ 1
A1 80 − 60 = 20
✓ 1
Common error: The median is the (n ÷ 2)th value of the cumulative frequency, not half of the largest time.
164IGMC3-9-08r7-08Averages, range and frequency tablesPaper EMedium7 marks
Calculator
A histogram shows the time taken for a drug to take effect in 118 patients. The heights of the bars are frequency densities in patients per minute: the bar for 0 to 10 minutes has height 1.2, for 10 to 20 minutes height 2.8, for 20 to 30 minutes height 3.6, for 30 to 50 minutes height 1.5 and for 50 to 80 minutes height 0.4.
Histogram of the times for the drug to take effect
(a)
Calculate the number of patients in the class from 20 to 30 minutes.
(1)
(b)
Calculate the number of patients in the class from 30 to 50 minutes.
(1)
(c)
Show that the total number of patients is 118.
(2)
(d)
Estimate the mean time, using the midpoint of each class. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 3.6 × 10 = 36
✓ 1
Part (b)
B1 1.5 × 20 = 30
✓ 1
Part (c)
M1 frequencies 12, 28, 36, 30, 12 (width × height for each bar)
✓ 1
A1 12 + 28 + 36 + 30 + 12 = 118
✓ 1
Part (d)
M1 midpoints 5, 15, 25, 40, 65
✓ 1
M1 5×12 + 15×28 + 25×36 + 40×30 + 65×12 = 3360
✓ 1
A1 3360 ÷ 118 = 28.5 minutes
✓ 1
Common error: In a histogram, frequency = frequency density × class width; the height is not the frequency.
165IGMC3-9-08r7-12Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
A histogram shows the length of gaming sessions of players. The bar for 0 to 20 minutes has height 0.9. The bar for 20 to 30 minutes represents 36 players. The bar for 30 to 60 minutes has height 1.5. The bar for 60 to 120 minutes has height 0.25.
(a)
Calculate the number of players in the class from 0 to 20 minutes.
(1)
(b)
Calculate the height of the bar for the class from 20 to 30 minutes.
(1)
(c)
Calculate the number of players in the classes from 30 to 60 and from 60 to 120 minutes.
(2)
(d)
Estimate the proportion of all players whose sessions last longer than 45 minutes. Assume the players in each class are evenly spread. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.9 × 20 = 18
✓ 1
Part (b)
B1 36 ÷ 10 = 3.6
✓ 1
Part (c)
B1 1.5 × 30 = 45
✓ 1
B1 0.25 × 60 = 15
✓ 1
Part (d)
M1 players from 45 to 60: 15 ÷ 30 × 45 = 22.5
✓ 1
M1 (22.5 + 15) ÷ (18 + 36 + 45 + 15) = 37.5 ÷ 114
✓ 1
A1 0.329
✓ 1
Common error: Area, not height, represents frequency, so a half-width bar section holds half of that bar's players.
166IGMC3-9-08r13-06Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The time t minutes for 80 patients to reach their lowest blood pressure after a dose is summarised in the table. 0 < t ≤ 20: 6 20 < t ≤ 40: 18 40 < t ≤ 60: 20 60 < t ≤ 80: 24 80 < t ≤ 120: 12 In parts (b) and (c) assume that the times are evenly spread within each class.
(a)
Complete the cumulative frequency for each class.
(2)
(b)
Estimate the number of patients whose time was more than 100 minutes.
(2)
(c)
Estimate the median time.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 6, 24, 44
✓ 1
B1 68, 80
✓ 1
Part (b)
M1 cumulative frequency at 100 = 68 + (20 ÷ 40) × 12
✓ 1
A1 74, so 80 − 74 = 6 patients
✓ 1
Part (c)
B1 the 40th value, lying in 40 < t ≤ 60
✓ 1
M1 40 + (40 − 24) ÷ 20 × 20
✓ 1
A1 56 minutes
✓ 1
Exam tip: The median of 80 values is at the 40th value on the cumulative frequency scale. Find the class containing it, then interpolate.
167IGMC3-9-08r13-10Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium5 marks
Calculator
The times taken by 150 players to complete Level 1 of a game are shown on a cumulative frequency curve. Reading from the curve: lower quartile 31 minutes, median 40 minutes, upper quartile 52 minutes. For Level 2, the median is 36 minutes and the interquartile range is 14 minutes.
(a)
Write down the cumulative frequency value used to read the upper quartile for Level 1.
(1)
(b)
Find the interquartile range for Level 1.
(1)
(c)
Compare the times for the two levels.
(2)
(d)
The slowest 10% of players are offered extra help. Write down the cumulative frequency value at which the cut-off time is read.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.75 × 150 = 112.5
✓ 1
Part (b)
B1 52 − 31 = 21 minutes
✓ 1
Part (c)
R1 Level 2 has the lower median, so players are generally quicker
✓ 1
R1 Level 2 has the smaller interquartile range, so times are more consistent
✓ 1
Part (d)
B1 0.9 × 150 = 135
✓ 1
Exam tip: When comparing distributions, comment on an average (median) and on a spread (interquartile range) and refer to the context.
168IGMC3-9-10r14-09Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium6 marks
Calculator
80 swimmers each swim 200 m. The cumulative frequency table shows their times, t seconds.
Time
t ≤ 120
t ≤ 130
t ≤ 140
t ≤ 150
t ≤ 160
t ≤ 170
Cumulative frequency
5
15
35
65
75
80
Assume the times are spread evenly within each class (linear interpolation).
(a)
Find the number of swimmers whose time is greater than 140 s but not greater than 150 s.
(1)
(b)
Estimate the median time, correct to 3 significant figures.
(2)
(c)
Estimate the interquartile range, correct to 3 significant figures. Use unrounded quartiles.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 65 − 35 = 30
✓ 1
Part (b)
M1 40th value lies in 140 < t ≤ 150: 140 + (5 ÷ 30) × 10
Common error: Quartile positions are ¼ and ¾ of the total cumulative frequency (here 20 and 60), not of the highest time.
169IGMC3-9-10r14-10Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
A histogram shows the ages, a years, of members of a library. The bar heights (frequency density) are:
Age
0 < a ≤ 10
10 < a ≤ 20
20 < a ≤ 40
40 < a ≤ 70
Frequency density
1.2
3.5
2.0
0.6
Histogram of the members' ages
(a)
Calculate the number of members aged over 20 and up to 40.
(2)
(b)
Show that the total number of members is 105.
(2)
(c)
Calculate an estimate of the mean age, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequency = frequency density × class width = 2.0 × 20
✓ 1
A1 40
✓ 1
Part (b)
M1 other frequencies 12, 35 and 18
✓ 1
A1 12 + 35 + 40 + 18 = 105
✓ 1
Part (c)
M1 midpoints 5, 15, 30, 55 used with frequencies: 60 + 525 + 1200 + 990 = 2775
✓ 1
M1 2775 ÷ 105
✓ 1
A1 26.4 years
✓ 1
Common error: On a histogram with unequal class widths, the height is frequency density; the frequency is the area of the bar, not its height.
170IGMC3-9-10r14-14Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium4 marks
Calculator
The cumulative frequency table shows the daily energy output, x MWh, of a wind farm over 180 days.
Output
x ≤ 20
x ≤ 40
x ≤ 60
x ≤ 80
x ≤ 100
Cumulative frequency
12
52
121
165
180
Assume the values are spread evenly within each class.
(a)
Find the number of days with output greater than 60 MWh and not more than 80 MWh.
(1)
(b)
Estimate the 90th percentile of the daily output, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 165 − 121 = 44
✓ 1
Part (b)
B1 90th percentile is the 162nd value (0.9 × 180)
✓ 1
M1 60 + ((162 − 121) ÷ 44) × 20
✓ 1
A1 78.6 MWh
✓ 1
Exam tip: For a percentile, find its position as a fraction of the total cumulative frequency and then interpolate inside the class where that position falls.
171IGMC3-9-09r16-06Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium6 marks
Calculator
The speeds of 50 vehicles crossing a bridge are recorded.
Speed v (km/h)
40 < v ≤ 50
50 < v ≤ 60
60 < v ≤ 70
70 < v ≤ 80
80 < v ≤ 90
Frequency
6
14
20
8
2
(a)
Write down the modal class.
(1)
(b)
Find the percentage of vehicles with speed greater than 70 km/h.
(2)
(c)
Calculate an estimate of the mean speed.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
60 < v ≤ 70
✓ 1
Part (b)
8 + 2 = 10 vehicles
✓ 1
10 ÷ 50 × 100 = 20%
✓ 1
Part (c)
M1 using mid-values 45, 55, 65, 75, 85
✓ 1
M1 sum of (frequency × mid-value) = 3110
✓ 1
A1 3110 ÷ 50 = 62.2 km/h
✓ 1
Exam tip: For grouped data use the mid-value of each class, and divide by the total frequency, not the number of classes.
172IGMC3-9-09r16-07Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium6 marks
Calculator
The lifetimes, t hours, of 80 probe batteries are summarised.
t
0 < t ≤ 20
20 < t ≤ 40
40 < t ≤ 60
60 < t ≤ 80
80 < t ≤ 100
Frequency
4
16
30
22
8
(a)
Complete the cumulative frequency for t ≤ 20, t ≤ 40, t ≤ 60, t ≤ 80 and t ≤ 100.
(2)
(b)
How many batteries lasted more than 60 hours?
(1)
(c)
A cumulative frequency graph is drawn by joining the points (20, 4), (40, 20), (60, 50), ... with straight lines. Use straight-line interpolation between (40, 20) and (60, 50) to estimate the median lifetime. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
4, 20, 50
✓ 1
72, 80
✓ 1
Part (b)
80 − 50 = 30
✓ 1
Part (c)
M1 median is the 40th value
✓ 1
M1 40 + (40 − 20) ÷ (50 − 20) × 20
✓ 1
A1 53.3 hours
✓ 1
Exam tip: The median of a cumulative frequency graph is read at half of the total frequency, plotted against the upper class boundaries.
173IGMC3-9-09r16-08Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The roasting times, t minutes, of 72 batches of coffee are shown in the table. A histogram is to be drawn.
t
0 < t ≤ 8
8 < t ≤ 12
12 < t ≤ 14
14 < t ≤ 20
Frequency
12
24
18
18
(a)
Calculate the frequency density for each class.
(2)
(b)
Estimate the number of batches with roasting time greater than 10 and at most 12 minutes.
(2)
(c)
Estimate the number of batches with roasting time greater than 6 and at most 9 minutes.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
1.5 and 6
✓ 1
9 and 3
✓ 1
Part (b)
M1 2 × 6 (or 24 × 2/4)
✓ 1
A1 12
✓ 1
Part (c)
2 × 1.5 = 3 batches in 6 < t ≤ 8
✓ 1
1 × 6 = 6 batches in 8 < t ≤ 9
✓ 1
9
✓ 1
Common error: The height of a histogram bar is frequency density (frequency ÷ class width), not frequency, when the class widths are unequal.
174IGMC3-9-09r16-10Averages, range and frequency tablesPaper EMedium5 marks
No calculator
Twelve strain-gauge readings on a bridge cable, in microstrain, are: 23 31 18 27 35 22 29 40 25 33 20 28
Exam tip: With 12 values, the quartiles are the medians of the lower six and upper six ordered values.
175IGMC3-9-09r16-11Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium6 marks
Calculator
The breaking loads of 120 test beams are summarised by a cumulative frequency graph, drawn by joining these points with straight lines:
Load (kN)
30
40
50
60
70
80
Cumulative frequency
8
24
48
84
108
120
Cumulative frequency graph of the breaking loads
(a)
How many beams failed at a load greater than 60 kN?
(1)
(b)
Use straight-line interpolation to estimate the lower quartile.
(2)
(c)
Use straight-line interpolation to estimate the upper quartile.
(2)
(d)
Work out the interquartile range.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
120 − 84 = 36
✓ 1
Part (b)
M1 the 30th value lies between (40, 24) and (50, 48)
✓ 1
A1 40 + (30 − 24) ÷ (48 − 24) × 10 = 42.5 kN
✓ 1
Part (c)
M1 the 90th value lies between (60, 84) and (70, 108)
✓ 1
A1 60 + (90 − 84) ÷ (108 − 84) × 10 = 62.5 kN
✓ 1
Part (d)
62.5 − 42.5 = 20 kN (ECF)
✓ 1
Exam tip: Lower and upper quartiles are read at 1/4 and 3/4 of the total cumulative frequency.
176IGMC3-9-08r19-09Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium8 marks
Calculator
The cumulative frequency table shows the time, t minutes, taken for 72 rides on a cycle route.
Time t
t ≤ 20
t ≤ 40
t ≤ 60
t ≤ 80
t ≤ 100
Cumulative frequency
6
21
47
66
72
In parts (b) and (c) estimate each value using linear interpolation within the class, as if reading from a smooth cumulative frequency curve.
(a)
Find the number of rides with 40 < t ≤ 60.
(1)
(b)
Estimate the median time, correct to 3 significant figures.
(3)
(c)
Estimate the interquartile range, correct to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 47 − 21 = 26
✓ 1
Part (b)
B1 36th value
✓ 1
M1 40 + (36 − 21)/26 × 20
✓ 1
A1 51.5 minutes
✓ 1
Part (c)
B1 18th and 54th values used
✓ 1
A1 lower quartile 36
✓ 1
A1 upper quartile 67.4 (accept 67.37)
✓ 1
A1 interquartile range 31.4 minutes
✓ 1
Common error: The median is at n/2 on the cumulative frequency axis, not at half of the highest time.
177IGMC3-9-08r19-10Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium4 marks
No calculator
A histogram shows the distances, d km, cycled each week by commuters. The vertical axis is frequency density. The bar heights are: 0 < d ≤ 10: 2.4; 10 < d ≤ 25: 3.2; 25 < d ≤ 35: 5.0; 35 < d ≤ 60: 0.8.
Histogram of the weekly distances
(a)
Work out the number of commuters in the class 10 < d ≤ 25.
(2)
(b)
Work out the total number of commuters.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 3.2 × 15
✓ 1
A1 48
✓ 1
Part (b)
M1 frequencies of the other classes 24, 50 and 20 (any two correct)
✓ 1
A1 142
✓ 1
Common error: On a histogram the frequency is the area of the bar (density × width), not its height.
178IGMC3-9-08r19-11Averages, range and frequency tablesPaper EMedium6 marks
Calculator
The table shows the lengths, l cm, of 50 bones found at a dig.
Length l
0 < l ≤ 10
10 < l ≤ 20
20 < l ≤ 30
30 < l ≤ 40
Frequency
8
14
18
10
(a)
Write down the modal class.
(1)
(b)
Write down the class that contains the median.
(1)
(c)
Calculate an estimate of the mean length.
(3)
(d)
Explain why your answer to (c) is only an estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 20 < l ≤ 30
✓ 1
Part (b)
B1 20 < l ≤ 30 (cumulative frequencies 8, 22, 40; median between the 25th and 26th values)
✓ 1
Part (c)
B1 mid-values 5, 15, 25, 35
✓ 1
M1 (5×8 + 15×14 + 25×18 + 35×10) / 50
✓ 1
A1 21.0 cm
✓ 1
Part (d)
B1 the exact lengths are not known; mid-values are assumed
✓ 1
Exam tip: For grouped data use mid-values, and divide by the total frequency, not the number of classes.
179IGMC3-9-08r25-10Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium11 marks
Calculator
The delay, t minutes, of 100 flights at an airport is recorded. 0 < t ≤ 10: 12 flights 10 < t ≤ 20: 28 flights 20 < t ≤ 30: 36 flights 30 < t ≤ 40: 16 flights 40 < t ≤ 60: 8 flights In this question, assume that the cumulative frequency increases at a constant rate within each class (straight-line interpolation).
(a)
Write down the cumulative frequency for the end of each class.
(2)
(b)
Estimate the median delay. Give your answer correct to 3 significant figures.
(3)
(c)
Estimate the number of flights with a delay of more than 35 minutes.
(2)
(d)
Estimate the interquartile range. Take the lower quartile as the value at cumulative frequency 25 and the upper quartile as the value at 75. Give your answer correct to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 12, 40, 76
✓ 1
B1 92, 100
✓ 1
Part (b)
M1 the 50th value lies in the class 20 < t ≤ 30
✓ 1
M1 20 + (10 ÷ 36) × 10
✓ 1
A1 22.8 minutes
✓ 1
Part (c)
M1 cumulative frequency at 35 is 76 + 0.5 × 16 = 84
✓ 1
A1 16
✓ 1
Part (d)
M1 lower quartile = 10 + (13 ÷ 28) × 10
✓ 1
M1 upper quartile = 20 + (35 ÷ 36) × 10
✓ 1
A1 14.6 and 29.7
✓ 1
A1 15.1 minutes
✓ 1
Exam tip: Use the position (such as 25 or 75 out of 100) to find which class the quartile lies in, then interpolate within that class.
180IGMC3-9-08r25-13Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The delays, in minutes, of eleven flights are: 14, 3, 30, 9, 6, 22, 12, 18, 5, 15, 8
(a)
Find the median delay.
(2)
(b)
Find the lower quartile and the upper quartile.
(2)
(c)
Find the interquartile range.
(1)
(d)
Flights with another airline have a median delay of 12 minutes and an interquartile range of 5 minutes. Compare the delays of the two airlines.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 data in order: 3, 5, 6, 8, 9, 12, 14, 15, 18, 22, 30
✓ 1
A1 12 minutes
✓ 1
Part (b)
B1 lower quartile 6
✓ 1
B1 upper quartile 18
✓ 1
Part (c)
A1 12 minutes (FT from (b))
✓ 1
Part (d)
B1 the typical (median) delay is the same for both airlines
✓ 1
B1 the other airline has the smaller interquartile range, so its delays are more consistent
✓ 1
Common error: The interquartile range is the upper quartile minus the lower quartile, not the range of all the data.
181IGMC3-9-12r29-07Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The cumulative frequency table shows the recovery times, t days, of 80 patients. t ≤ 2: 6 t ≤ 4: 22 t ≤ 6: 44 t ≤ 8: 66 t ≤ 10: 76 t ≤ 12: 80 Use linear interpolation between the values in the table where an estimate is needed.
(a)
How many patients took more than 6 days to recover?
(1)
(b)
Estimate the median recovery time, to 3 significant figures.
(3)
(c)
Estimate the interquartile range, to 3 significant figures.
Exam tip: The quartiles are found at 25% and 75% of the total frequency, which here means the 20th and 60th values.
182IGMC3-9-12r29-09Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The amounts spent, £s, by 160 customers in a shop are summarised below. 0 < s ≤ 10: 18 10 < s ≤ 30: 56 30 < s ≤ 40: 36 40 < s ≤ 60: 30 60 < s ≤ 100: 20 A histogram is drawn with frequency density on the vertical axis.
(a)
Calculate the frequency density for the classes 0 < s ≤ 10, 30 < s ≤ 40 and 60 < s ≤ 100.
(3)
(b)
The bar for 10 < s ≤ 30 is 5.6 cm tall. Find the height of the bar for 60 < s ≤ 100.
(2)
(c)
Estimate the number of customers who spent more than £50, assuming customers are spread evenly within each class.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 1.8
✓ 1
B1 3.6
✓ 1
B1 0.5
✓ 1
Part (b)
M1 scale = 5.6 ÷ 2.8 = 2 cm per unit of frequency density
✓ 1
A1 1.0 cm
✓ 1
Part (c)
M1 half of the class 40 < s ≤ 60 plus the class 60 < s ≤ 100: 15 + 20
✓ 1
A1 35
✓ 1
Common error: In a histogram it is the area of a bar, not its height, that is proportional to frequency; divide frequency by class width.
183IGMC3-9-12r29-11Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The daily energy output, E MWh, of a wind farm was recorded over 120 days. E ≤ 20: 8 E ≤ 40: 36 E ≤ 60: 84 E ≤ 80: 112 E ≤ 100: 120 (cumulative frequencies). Use linear interpolation between the table values where an estimate is needed.
(a)
How many days had an output greater than 40 MWh and at most 60 MWh?
(1)
(b)
Estimate the 90th percentile of the output, to 3 significant figures.
(3)
(c)
Estimate the number of days on which the output was greater than 70 MWh.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 48
✓ 1
Part (b)
M1 90% of 120 = 108th value, between E = 60 and E = 80
✓ 1
M1 60 + (108 − 84)/(112 − 84) × 20
✓ 1
A1 77.1 MWh
✓ 1
Part (c)
M1 days with E ≤ 70: 84 + (70 − 60)/(80 − 60) × (112 − 84)
✓ 1
A1 98
✓ 1
A1 120 − 98 = 22 days
✓ 1
Exam tip: A percentile is found from a fraction of the total frequency, then located in the table by interpolation.
184IGMC3-9-11r31-05Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The daily water use of 80 households is recorded. Volume used, v litres: 0 < v ≤ 100, 100 < v ≤ 200, 200 < v ≤ 300, 300 < v ≤ 400, 400 < v ≤ 500 Number of households: 8, 22, 30, 14, 6 A cumulative frequency curve is drawn using these data. When estimating, assume the curve is a straight line between the plotted points (linear interpolation).
(a)
Write down the cumulative frequencies for the five upper class boundaries.
(2)
(b)
Estimate the median volume. Give your answer to 3 significant figures.
(3)
(c)
Estimate the number of households that use more than 350 litres.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 at least three correct
✓ 1
B1 8, 30, 60, 74, 80
✓ 1
Part (b)
M1 median is the 40th value, which lies in the class 200 < v ≤ 300
✓ 1
M1 \(200 + 100\times\frac{40-30}{60-30}\)
✓ 1
A1 233 litres
✓ 1
Part (c)
M1 cumulative frequency at 350 is \(60 + \frac{1}{2}\times14\) (= 67) or \(\frac{1}{2}\times14 + 6\)
✓ 1
A1 13
✓ 1
Exam tip: On a cumulative frequency curve the values are plotted at the upper class boundaries, and the median is the (n ÷ 2)th value.
185IGMC3-9-11r31-06Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
A drug trial records the time, t minutes, taken for a medicine to start working in each patient. The results are shown in a histogram with these bars. 0 < t ≤ 10: frequency density 1.2 10 < t ≤ 20: frequency density 2.8 20 < t ≤ 40: frequency density 1.5 40 < t ≤ 80: frequency density 0.3
Histogram of the times for the medicine to start working
(a)
Calculate the number of patients in the class 20 < t ≤ 40.
(2)
(b)
Show that 82 patients were in the trial.
(2)
(c)
Estimate the mean time taken. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequency = frequency density × class width = \(1.5\times20\)
✓ 1
A1 30
✓ 1
Part (b)
M1 frequencies 12, 28, 30 and 12 found (at least three correct)
Common error: In a histogram the frequency is the area of the bar (frequency density × class width), not the height.
186IGMC3-9-11r31-13Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The repair times, t hours, of 120 burst water pipes are recorded. 0 < t ≤ 2: 18 pipes 2 < t ≤ 3: 21 pipes 3 < t ≤ 6: 54 pipes 6 < t ≤ 12: 27 pipes A histogram is to be drawn.
(a)
Calculate the frequency density for each class.
(2)
(b)
Write down the class that has the tallest bar.
(1)
(c)
Estimate the number of pipes that took more than 8 hours to repair.
(2)
(d)
Estimate the percentage of pipes repaired in less than 4 hours.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 at least two correct
✓ 1
B1 9, 21, 18, 4.5
✓ 1
Part (b)
B1 2 < t ≤ 3
✓ 1
Part (c)
M1 \((12-8)\times4.5\)
✓ 1
A1 18
✓ 1
Part (d)
M1 \(18 + 21 + 1\times18 = 57\)
✓ 1
A1 47.5 %
✓ 1
Common error: The tallest bar is the class with the greatest frequency density, which is not always the class with the greatest frequency.
187IGMC3-9-12r34-10Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The cumulative frequency table shows the ages, a years, of 150 members of a library.
Age
a ≤ 10
a ≤ 20
a ≤ 30
a ≤ 40
a ≤ 50
a ≤ 60
a ≤ 80
Cumulative frequency
0
12
40
82
120
140
150
Assume that members are spread evenly within each class.
(a)
Write down the number of members aged over 30 and up to 40.
(1)
(b)
Estimate the median age, correct to 3 significant figures.
(3)
(c)
Estimate the number of members older than 45.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 82 − 40 = 42
✓ 1
Part (b)
B1 median is the 75th value, in the class 30 < a ≤ 40
✓ 1
M1 30 + (75 − 40) ÷ 42 × 10
✓ 1
A1 38.3 years
✓ 1
Part (c)
B1 cumulative frequency at 45 is 82 + 0.5 × 38 = 101
✓ 1
M1 150 − 101
✓ 1
A1 49
✓ 1
Common error: The median of 150 values is at the 75th value on the cumulative frequency axis, not at the 150th.
188IGMC3-9-12r34-11Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium5 marks
Calculator
The table shows how many days 135 library loans lasted.
Loan length (d days)
0 < d ≤ 5
5 < d ≤ 10
10 < d ≤ 20
20 < d ≤ 40
Frequency
30
45
36
24
A histogram is to be drawn for this data.
(a)
Calculate the frequency density for each class.
(3)
(b)
Estimate the number of loans that lasted more than 15 days. Assume loans are spread evenly within each class.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 frequency density = frequency ÷ class width
✓ 1
B1 6 and 9 for the first two classes
✓ 1
B1 3.6 and 1.2 for the last two classes
✓ 1
Part (b)
M1 half of the 10 < d ≤ 20 class: 18 (ECF from (a) if used)
✓ 1
A1 18 + 24 = 42
✓ 1
Exam tip: On a histogram the area of a bar, not its height, represents frequency; use frequency density when class widths differ.
189IGMC3-9-12r34-14Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium6 marks
No calculator
Over a period of 120 days, the daily energy output of two wind farms was summarised.
Lower quartile
Median
Upper quartile
Farm P (MWh)
51
62
70
Farm Q (MWh)
54
58
63
(a)
Work out the interquartile range for each farm.
(2)
(b)
Compare the daily output of the two farms.
(2)
(c)
Estimate the number of days on which the output of farm Q was 54 MWh or less.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 farm P: 70 − 51 = 19
✓ 1
B1 farm Q: 63 − 54 = 9
✓ 1
Part (b)
B1 farm P has the higher median, so a higher typical output
✓ 1
B1 farm Q has the smaller interquartile range, so more consistent output (ECF from (a))
✓ 1
Part (c)
M1 25% of 120
✓ 1
A1 30
✓ 1
Common error: The interquartile range is the difference between the quartiles, not the range of all the data.
190IGMC3-9-10r36-08Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium6 marks
Calculator
The ride distances, in km, of two cycling clubs are summarised below. Club P: smallest 24, lower quartile 38, median 52, upper quartile 64, largest 90 Club Q: smallest 30, lower quartile 44, median 50, upper quartile 58, largest 70
Box plots of the ride distances of the two clubs
(a)
Find the interquartile range for each club.
(2)
(b)
Compare the distances ridden by the two clubs. Give one comparison of the typical distance and one comparison of the spread.
(2)
(c)
There are 120 riders in Club P. Estimate the number who rode between 38 km and 64 km.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 Club P: 64 − 38 = 26
✓ 1
B1 Club Q: 58 − 44 = 14
✓ 1
Part (b)
B1 the median distances are similar (52 km for P and 50 km for Q), with P slightly higher
✓ 1
B1 Club P's distances are more spread out (IQR 26 km compared with 14 km), so Club Q's riders are more consistent
✓ 1
Part (c)
M1 half of the riders lie between the lower and upper quartiles
✓ 1
A1 60
✓ 1
Common error: The interquartile range is the upper quartile minus the lower quartile, not the largest value minus the smallest value.
191IGMC3-9-10r36-10Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium6 marks
Calculator
The cumulative frequency table shows the response times, t milliseconds, of 80 requests to a data centre server. Assume that a cumulative frequency graph is drawn by joining the plotted points with straight lines. Response time t ≤ 10: 8 t ≤ 20: 26 t ≤ 30: 52 t ≤ 40: 70 t ≤ 50: 80
Cumulative frequency graph of the response times
(a)
Write down the frequency for each of the classes 0 < t ≤ 10, 10 < t ≤ 20 and 20 < t ≤ 30.
(2)
(b)
Use the graph to estimate the median response time. Give your answer correct to 3 significant figures.
(2)
(c)
Estimate the number of requests that took more than 35 ms.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differences of cumulative frequencies
✓ 1
A1 8, 18, 26
✓ 1
Part (b)
M1 median is the 40th value, between (20, 26) and (30, 52)
✓ 1
A1 25.4 (accept 25 to 26 from a graph)
✓ 1
Part (c)
M1 cumulative frequency at 35 ms is 61 (by interpolation between (30, 52) and (40, 70))
✓ 1
A1 80 − 61 = 19 (accept 18 to 20 from a graph)
✓ 1
Common error: A cumulative frequency graph is plotted against the upper class boundaries, and the median is read at n/2 on the vertical axis.
192IGMC3-9-10r36-11Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
The table shows the widths, w mm, of 100 pottery shards found at a dig. 0 < w ≤ 10: frequency 15 10 < w ≤ 20: frequency 29 20 < w ≤ 40: frequency 35 40 < w ≤ 60: frequency 16 60 < w ≤ 100: frequency 5 A histogram is drawn using frequency density on the vertical axis.
(a)
Calculate the frequency density for the class 20 < w ≤ 40 and for the class 60 < w ≤ 100.
(2)
(b)
Estimate the number of shards with a width greater than 50 mm. Assume the shards in each class are evenly spread.
(2)
(c)
Calculate an estimate of the mean width. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 35 ÷ 20 = 1.75
✓ 1
B1 5 ÷ 40 = 0.125
✓ 1
Part (b)
M1 half of the 40 < w ≤ 60 class, 8, added to the 5 in the top class
Common error: In a histogram the area of a bar, not its height, represents the frequency, so divide each frequency by the class width.
193IGMC3-9-11r38-08Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium8 marks
Calculator
The masses of 80 fish in a net are summarised in a cumulative frequency table. Use linear interpolation within a class, as if reading a cumulative frequency graph. Mass m (g): ≤ 200, ≤ 400, ≤ 600, ≤ 800, ≤ 1000, ≤ 1200 Cumulative frequency: 3, 10, 30, 50, 70, 80
(a)
Write down the number of fish with mass greater than 800 g.
(1)
(b)
Estimate the median mass.
(2)
(c)
Estimate the interquartile range.
(3)
(d)
Fish heavier than 1000 g are sold as premium fish. Find the percentage of the fish that are premium.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 80 − 50 = 30
✓ 1
Part (b)
M1 40th value lies in the class 600 to 800: 600 + (10 ÷ 20) × 200
✓ 1
A1 700 g
✓ 1
Part (c)
M1 lower quartile at the 20th value: 400 + (10 ÷ 20) × 200 = 500
✓ 1
M1 upper quartile at the 60th value: 800 + (10 ÷ 20) × 200 = 900
✓ 1
A1 interquartile range = 400 g
✓ 1
Part (d)
M1 (80 − 70) ÷ 80
✓ 1
A1 12.5%
✓ 1
Exam tip: The quartiles are at ¼ and ¾ of the total frequency (here the 20th and 60th values), not at the quarter points of the mass scale.
194IGMC3-9-11r38-09Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium8 marks
Calculator
A clothing retailer records the time, t minutes, that customers spend in a changing room. The results are drawn as a histogram with these frequency densities. 0 < t ≤ 5: frequency density 2.4 5 < t ≤ 10: frequency density 4.0 10 < t ≤ 20: frequency density 3.0 20 < t ≤ 40: 10 customers (frequency density not given)
(a)
Find the number of customers in the class 0 < t ≤ 5.
(2)
(b)
Find the frequency density for the class 20 < t ≤ 40.
(2)
(c)
Estimate the percentage of customers who spend more than 15 minutes in the changing room.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequency density × class width = 2.4 × 5
✓ 1
A1 12
✓ 1
Part (b)
M1 10 ÷ 20
✓ 1
A1 0.5
✓ 1
Part (c)
B1 total number of customers = 12 + 20 + 30 + 10 = 72
✓ 1
M1 customers in 15 < t ≤ 20 = 3.0 × 5 = 15
✓ 1
M1 (15 + 10) ÷ 72 × 100
✓ 1
A1 34.7%
✓ 1
Common error: Reading the height of a histogram bar as the frequency. Frequency = frequency density × class width.
195IGMC3-9-11r38-11Averages, range and frequency tablesPaper EMedium6 marks
No calculator
The lengths, in cm, of 11 fish caught in a net are shown in order. 14, 16, 17, 19, 21, 22, 24, 25, 28, 30, 33
(a)
Write down the median length.
(1)
(b)
Find the lower quartile and the upper quartile.
(2)
(c)
Find the interquartile range.
(1)
(d)
A twelfth fish of length 38 cm is added. Find the new median.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 22 cm
✓ 1
Part (b)
B1 lower quartile = 17 cm
✓ 1
B1 upper quartile = 28 cm
✓ 1
Part (c)
B1 28 − 17 = 11 cm
✓ 1
Part (d)
M1 6th and 7th values are 22 and 24
✓ 1
A1 (22 + 24) ÷ 2 = 23 cm
✓ 1
Exam tip: With an even number of values the median is the mean of the two middle values.
196IGMC3-9-11r43-09Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium8 marks
Calculator
A water company repairs 80 pipes. The table shows the cumulative frequency of the repair time t hours. Assume that repair times within each class are evenly spread. Time: t ≤ 2, t ≤ 4, t ≤ 6, t ≤ 8, t ≤ 10 Cumulative frequency: 6, 20, 50, 74, 80
(a)
Find the number of repairs that took more than 4 hours but not more than 6 hours.
(1)
(b)
Estimate the median repair time. Give your answer correct to 3 significant figures.
(3)
(c)
Estimate the interquartile range. Give your answer correct to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 30
✓ 1
Part (b)
M1 40th value lies in the class 4 < t ≤ 6
✓ 1
M1 4 + (40 − 20) ÷ 30 × 2
✓ 1
A1 5.33 (hours)
✓ 1
Part (c)
B1 lower quartile (20th value) = 4
✓ 1
M1 upper quartile is the 60th value, 6 + (60 − 50) ÷ 24 × 2
✓ 1
A1 6.83
✓ 1
A1 2.83 (hours)
✓ 1
Exam tip: Quartiles are found at one quarter and three quarters of the total cumulative frequency, then located in their class.
197IGMC3-9-11r43-10Averages, range and frequency tablesPaper EMedium7 marks
No calculator
Players of an online game wait in a lobby before a match. A histogram shows the waiting times t minutes. The vertical axis is frequency density. The bars are: 0 < t ≤ 10 with height 3 10 < t ≤ 15 with height 7 15 < t ≤ 30 with height 3 30 < t ≤ 60 with height 1
Histogram of the waiting times
(a)
Find the number of players who waited more than 10 minutes but not more than 15 minutes.
(2)
(b)
Find the total number of players.
(2)
(c)
Estimate the number of players who waited more than 20 minutes.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 7 × 5
✓ 1
A1 35
✓ 1
Part (b)
M1 adding the frequencies 30, 35, 45 and 30
✓ 1
A1 140
✓ 1
Part (c)
M1 (30 − 20) × 3 = 30
✓ 1
M1 adding 30 × 1 = 30 for the last bar
✓ 1
A1 60
✓ 1
Common error: In a histogram the frequency is the area of the bar (frequency density × class width), not its height.
198IGMC3-9-11r45-08Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium7 marks
Calculator
Recovery times, in days, are recorded for 80 hospital patients after a minor operation. The cumulative frequency for each recovery time t is shown.
Time (days)
t ≤ 5
t ≤ 10
t ≤ 15
t ≤ 20
t ≤ 30
Cumulative frequency
8
24
52
72
80
A cumulative frequency graph joins consecutive points with straight lines. Use linear interpolation between points in your working.
(a)
Find the number of patients whose recovery time was more than 20 days.
(1)
(b)
Estimate the median recovery time.
(3)
(c)
Estimate the interquartile range.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
80 − 72 = 8
✓ 1
Part (b)
M1 the 40th value lies between t = 10 (24) and t = 15 (52)
✓ 1
M1 10 + (40 − 24) ÷ (52 − 24) × 5
✓ 1
A1 12.9 days
✓ 1
Part (c)
M1 uses the 20th and 60th values
✓ 1
A1 lower quartile 8.75 and upper quartile 17
✓ 1
A1 IQR = 8.25 days
✓ 1
Exam tip: Quartiles are read at one quarter and three quarters of the total cumulative frequency (20 and 60 here), not at 25 and 75.
199IGMC3-9-11r45-09Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium6 marks
Calculator
The durations of 120 tyre stints in a racing series are grouped. A histogram is to be drawn.
Duration, d minutes
0 < d ≤ 5
5 < d ≤ 15
15 < d ≤ 20
20 < d ≤ 40
Frequency
10
40
40
30
(a)
Calculate the frequency density for the classes 5 < d ≤ 15 and 15 < d ≤ 20.
(2)
(b)
State which class has the tallest bar in the histogram.
(1)
(c)
Estimate the number of stints with a duration between 10 and 18 minutes, assuming the stints are spread evenly within each class.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 40 ÷ 10 = 4
✓ 1
B1 40 ÷ 5 = 8
✓ 1
Part (b)
15 < d ≤ 20 (highest frequency density, 8)
✓ 1
Part (c)
M1 uses frequency density × width for each part: 5 × 4 and 3 × 8
✓ 1
A1 20 + 24
✓ 1
A1 44
✓ 1
Common error: The height of a histogram bar is frequency density (frequency ÷ class width), not frequency, when the classes have unequal widths.
200IGMC3-9-11r45-10Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium6 marks
Calculator
Team X records its 11 pit stop times, in seconds, in order: 2.3, 2.4, 2.4, 2.5, 2.6, 2.6, 2.7, 2.9, 3.1, 3.4, 4.8. Team Y has a median pit stop time of 2.5 s and an interquartile range of 0.3 s.
(a)
Find the median time for Team X.
(1)
(b)
Find the lower quartile and the upper quartile for Team X.
(2)
(c)
Find the interquartile range for Team X.
(1)
(d)
Compare the pit stop times of the two teams, referring to an average and to the spread.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
2.6 s (6th value)
✓ 1
Part (b)
B1 lower quartile 2.4 s (3rd value)
✓ 1
B1 upper quartile 3.1 s (9th value)
✓ 1
Part (c)
3.1 − 2.4 = 0.7 s (ECF from (b))
✓ 1
Part (d)
B1 Team Y is quicker on average (median 2.5 s is less than 2.6 s)
✓ 1
B1 Team Y is more consistent (smaller IQR, 0.3 s compared with 0.7 s)
✓ 1
Exam tip: When comparing two sets of data, make one comment about a typical value and one about spread, and use the figures in your answer.
201IGMC3-9-10r46-08Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium9 marks
Calculator
A Formula 1 team records 54 pit stop times, in seconds. The results are shown in a histogram with frequency density on the vertical axis. The bar heights (frequency densities) are: 12 for 2 < t ≤ 2.5, 36 for 2.5 < t ≤ 3, 20 for 3 < t ≤ 4 and 5 for 4 < t ≤ 6.
Histogram of the pit stop times (the class 6 < t ≤ 10 is not drawn)
(a)
Calculate the number of pit stops with times in the class 3 < t ≤ 4.
(2)
(b)
Show that the numbers in the other three classes add up to 34.
(2)
(c)
A further class 6 < t ≤ 10 has 8 pit stops. Find the height of its bar.
(2)
(d)
Estimate the number of pit stops with times between 2.75 s and 3.5 s.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequency = frequency density × class width = 20 × 1
✓ 1
A1 20
✓ 1
Part (b)
M1 6 + 18 + 10
✓ 1
A1 34 (so the total is 20 + 34 = 54)
✓ 1
Part (c)
M1 frequency density = frequency ÷ class width = 8 ÷ 4
✓ 1
A1 2
✓ 1
Part (d)
M1 2.75 to 3 is half the class 2.5–3: 0.25 × 36 = 9
✓ 1
M1 3 to 3.5 is half the class 3–4: 0.5 × 20 = 10
✓ 1
A1 19
✓ 1
Exam tip: In a histogram the area of a bar, not its height, represents the frequency.
202IGMC3-9-10r46-09Scatter diagrams, correlation, cumulative frequency and histogramsPaper EMedium6 marks
Calculator
The number of overtakes in each of 11 Formula 1 races is recorded: 14, 17, 19, 21, 22, 25, 26, 28, 31, 33, 40.
(a)
Write down the median.
(1)
(b)
Find the lower quartile, the upper quartile and the interquartile range.
(3)
(c)
For a second set of 11 races the median is 24 and the interquartile range is 7. Compare the two sets of data.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 25
✓ 1
Part (b)
B1 lower quartile 19
✓ 1
B1 upper quartile 31
✓ 1
B1 31 − 19 = 12
✓ 1
Part (c)
B1 first set has a slightly higher median / typically more overtakes
✓ 1
B1 first set has a larger interquartile range, so it is more spread out / less consistent
✓ 1
Exam tip: When comparing two data sets, always make one comment about an average and one about spread, each in the context of the question.
203IGMC-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.
204IGMC-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.
205IGMC-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)
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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.
206IGMC-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
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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.
207IGMC3-9-06d-12Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard10 marks
Calculator
The masses, m kg, of 60 parcels are shown in a histogram. The heights of the bars (frequency density) are: 0 < m ≤ 2: 6; 2 < m ≤ 5: 8; 5 < m ≤ 10: 3.2; 10 < m ≤ 20: 0.8.
Histogram of the parcel masses
(a)
Find the frequency of each class.
(4)
(b)
Show that the total number of parcels is 60.
(1)
(c)
State the class that contains the median.
(1)
(d)
Calculate an estimate of the mean mass of a parcel.
(4)
Show mark scheme
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Notes
Part (a)
12
✓ 1
24
✓ 1
16
✓ 1
8
✓ 1
Part (b)
12 + 24 + 16 + 8 = 60
✓ 1
Part (c)
2 < m ≤ 5
✓ 1
Part (d)
M1 midpoints 1, 3.5, 7.5, 15
✓ 1
M1 12×1 + 24×3.5 + 16×7.5 + 8×15 = 336
✓ 1
M1 336 ÷ 60
✓ 1
A1 5.6
✓ 1
Common error: Multiply each frequency density by its class width to get the frequency; the heights themselves are not frequencies.
208IGMC3-9-06d-14Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard9 marks
Calculator
A hydrologist records monthly rainfall, x mm, and reservoir level, y m, for six months. (x, y): (20, 3.0), (30, 3.6), (40, 4.3), (50, 5.0), (60, 5.5), (70, 6.2) A line of best fit is drawn so that it passes through the mean point and through (20, 2.9).
Scatter diagram of the data
(a)
Describe the correlation.
(1)
(b)
Find the coordinates of the mean point.
(2)
(c)
Find the gradient of the line of best fit.
(2)
(d)
Find the equation of the line in the form y = mx + c.
(2)
(e)
Use your equation to estimate the reservoir level when the rainfall is 55 mm.
(2)
Show mark scheme
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Notes
Part (a)
Positive (correlation)
✓ 1
Part (b)
x̄ = 45
✓ 1
ȳ = 4.6
✓ 1
Part (c)
M1 (4.6 − 2.9)/(45 − 20)
✓ 1
A1 0.068
✓ 1
Part (d)
M1 4.6 = 0.068 × 45 + c
✓ 1
A1 y = 0.068x + 1.54
✓ 1
Part (e)
M1 0.068 × 55 + 1.54
✓ 1
A1 5.28 (ECF from (d))
✓ 1
Exam tip: A line of best fit passes through the mean point, which gives a second point for finding its equation.
209IGMC3-9-08r7-11Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard9 marks
Calculator
The daily water use of 120 households is summarised in a cumulative frequency table.
Use u (litres)
u ≤ 100
u ≤ 150
u ≤ 200
u ≤ 250
u ≤ 300
Cumulative frequency
12
36
66
96
120
Use linear interpolation within classes, assuming the values are evenly spread.
(a)
Estimate the median.
(2)
(b)
Estimate the lower quartile and the upper quartile.
(2)
(c)
Calculate the interquartile range.
(1)
(d)
The company charges extra to the 10% of households that use the most water. Estimate the smallest daily use of a household in this group.
(2)
(e)
Estimate the number of households that use more than 220 litres.
(2)
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Notes
Part (a)
M1 60th value lies in 150 to 200: 150 + (60 − 36) ÷ 30 × 50
M1 108th value lies in 250 to 300: 250 + (108 − 96) ÷ 24 × 50
✓ 1
A1 275 litres
✓ 1
Part (e)
M1 cumulative frequency at 220 = 66 + 20 ÷ 50 × 30 = 78
✓ 1
A1 120 − 78 = 42
✓ 1
Common error: Quartiles are found at one quarter and three quarters of the total frequency (30th and 90th values here).
210IGMC3-9-08r7-14Averages, range and frequency tablesPaper EHard7 marks
Calculator
The ping times of 50 players are recorded in a grouped table.
Ping p (ms)
0 < p ≤ 20
20 < p ≤ 40
40 < p ≤ 60
60 < p ≤ 100
Frequency
8
x
y
6
The estimated mean ping, using class midpoints, is 38.4 ms.
(a)
Write down an equation for x and y from the total number of players.
(1)
(b)
Show that 30x + 50y = 1360.
(2)
(c)
Find x and y.
(3)
(d)
Find the class with the highest frequency density.
(1)
Show mark scheme
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Notes
Part (a)
B1 x + y = 36
✓ 1
Part (b)
M1 8×10 + 30x + 50y + 6×80 = 38.4 × 50
✓ 1
A1 560 + 30x + 50y = 1920 so 30x + 50y = 1360
✓ 1
Part (c)
M1 substitute y = 36 − x: 30x + 50(36 − x) = 1360
✓ 1
A1 x = 22
✓ 1
A1 y = 14
✓ 1
Part (d)
B1 20 < p ≤ 40 (frequency density 22 ÷ 20 = 1.1 is the largest)
✓ 1
Exam tip: Write the mean equation using total frequency multiplied by the mean on one side.
211IGMC3-9-08r13-07Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard7 marks
Calculator
The times t minutes taken by players to complete a game level are shown in a histogram. The table gives the frequencies for four of the classes. 0 < t ≤ 10: 12 10 < t ≤ 20: 30 20 < t ≤ 30: not given 30 < t ≤ 50: 36 50 < t ≤ 90: 24 The bar for 20 < t ≤ 30 has height 4.2 (frequency density).
(a)
Calculate the frequency density for the class 30 < t ≤ 50.
(1)
(b)
Find the frequency for the class 20 < t ≤ 30.
(1)
(c)
Write down the modal class.
(1)
(d)
Calculate an estimate of the mean time.
(4)
Show mark scheme
Marking point
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Notes
Part (a)
B1 36 ÷ 20 = 1.8
✓ 1
Part (b)
B1 4.2 × 10 = 42
✓ 1
Part (c)
B1 20 < t ≤ 30 (it has the greatest frequency density, 4.2)
✓ 1
Part (d)
M1 total frequency 12 + 30 + 42 + 36 + 24 = 144
✓ 1
M1 midpoints 5, 15, 25, 40, 70 used
✓ 1
M1 5×12 + 15×30 + 25×42 + 40×36 + 70×24 = 4680
✓ 1
A1 4680 ÷ 144 = 32.5 minutes (ECF from (b))
✓ 1
Common error: In a histogram with unequal widths the modal class is the one with the greatest frequency density, not the tallest frequency.
212IGMC3-9-08r13-09Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard8 marks
Calculator
A pump at a water works runs for x hours each day. Over 100 days the times are shown in a histogram with these frequency densities (days per hour). 0 < x ≤ 2: 5 2 < x ≤ 5: 8 5 < x ≤ 6: 14 6 < x ≤ 10: k
(a)
Find the frequency for the class 2 < x ≤ 5.
(1)
(b)
Find the value of k.
(3)
(c)
A day is chosen at random. Find the probability that the pump ran for more than 5 hours.
(2)
(d)
Estimate the number of days on which the pump ran for between 4 and 6 hours. Assume that the days are evenly spread within each class.
(2)
Show mark scheme
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Notes
Part (a)
B1 8 × 3 = 24
✓ 1
Part (b)
M1 frequency of last class = 4k
✓ 1
M1 10 + 24 + 14 + 4k = 100
✓ 1
A1 k = 13
✓ 1
Part (c)
M1 (14 + 4 × 13) ÷ 100 = 66 ÷ 100
✓ 1
A1 0.66 (or 33/50) (ECF from (b))
✓ 1
Part (d)
M1 (5 − 4) × 8 + (6 − 5) × 14
✓ 1
A1 22
✓ 1
Common error: In a histogram the frequency is area (frequency density × class width), not the height of the bar.
213IGMC3-9-08r13-13Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard7 marks
Calculator
A water company records the time t minutes taken to locate 200 leaks. Assume that the times are evenly spread within each class. 0 < t ≤ 10: 12 10 < t ≤ 20: 18 20 < t ≤ 30: 40 30 < t ≤ 40: 40 40 < t ≤ 60: 50 60 < t ≤ 100: 40
(a)
Write down the cumulative frequency for each class.
(2)
(b)
Estimate the lower quartile.
(2)
(c)
Estimate the upper quartile.
(2)
(d)
Find the interquartile range.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 12, 30, 70
✓ 1
B1 110, 160, 200
✓ 1
Part (b)
M1 50th value lies in 20 < t ≤ 30; 20 + (50 − 30) ÷ 40 × 10
✓ 1
A1 25 minutes
✓ 1
Part (c)
M1 150th value lies in 40 < t ≤ 60; 40 + (150 − 110) ÷ 50 × 20
✓ 1
A1 56 minutes
✓ 1
Part (d)
A1 56 − 25 = 31 minutes (ECF from (b) and (c))
✓ 1
Common error: The lower quartile is at 1/4 of the total frequency (the 50th value here), not at 1/4 of the largest time.
214IGMC3-9-10r14-11Averages, range and frequency tablesPaper EHard6 marks
Calculator
A histogram shows the time t hours that members of a swimming club train in a week. The class 0 < t ≤ 2 has frequency 14 and its bar is 3.5 cm high.
(a)
The class 2 < t ≤ 5 has frequency 36. Calculate the height of its bar, in cm.
(3)
(b)
The bar for the class 6 < t ≤ 10 is 1.5 cm high. Calculate the frequency of this class.
(3)
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Marking point
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Notes
Part (a)
M1 frequency densities 14 ÷ 2 = 7 and 36 ÷ 3 = 12
✓ 1
M1 scale 3.5 ÷ 7 = 0.5 cm per unit of frequency density
✓ 1
A1 6 cm
✓ 1
Part (b)
M1 frequency density = 1.5 ÷ 0.5 = 3
✓ 1
M1 frequency = 3 × 4
✓ 1
A1 12
✓ 1
Exam tip: Find the scale first: bar height ÷ frequency density gives the same number of centimetres for every bar.
215IGMC3-9-08r19-12Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard8 marks
Calculator
A histogram shows the times, t minutes, of server maintenance tasks. One centimetre of height represents a frequency density of 0.8 tasks per minute. The bar heights are: 0 < t ≤ 10: 2 cm; 10 < t ≤ 25: 3 cm; 25 < t ≤ 30: 6 cm; 30 < t ≤ 60: 1.5 cm.
Histogram; bar heights are given in centimetres
(a)
Show that 24 tasks took between 25 and 30 minutes.
(2)
(b)
Find the total number of tasks.
(3)
(c)
Estimate the number of tasks with 20 < t ≤ 40.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequency density = 6 × 0.8 = 4.8
✓ 1
A1 4.8 × 5 = 24
✓ 1
Part (b)
M1 frequency = density × width for each bar
✓ 1
A1 16, 36 and 36 for the other three bars (any two correct)
✓ 1
A1 112
✓ 1
Part (c)
M1 uses 5 minutes of the 10–25 bar and 10 minutes of the 30–60 bar
✓ 1
M1 2.4 × 5 = 12 and 1.2 × 10 = 12
✓ 1
A1 12 + 24 + 12 = 48
✓ 1
Common error: Only part of a bar is used when the interval starts or ends inside a class; use density × the part of the width.
216IGMC3-9-08r19-13Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard9 marks
Calculator
The cumulative frequency table shows the power, p kW, used by 120 servers in room A of a data centre.
Power p
p ≤ 2
p ≤ 4
p ≤ 6
p ≤ 8
p ≤ 10
Cumulative frequency
10
34
76
112
120
Use linear interpolation within the class to estimate each value.
(a)
Estimate the median power, correct to 3 significant figures.
(3)
(b)
Estimate the interquartile range, correct to 3 significant figures.
(4)
(c)
Room B has median power 5.8 kW and interquartile range 2.4 kW. Compare the two rooms.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 60th value
✓ 1
M1 4 + (60 − 34)/42 × 2
✓ 1
A1 5.24 kW
✓ 1
Part (b)
B1 30th and 90th values
✓ 1
A1 lower quartile 3.67
✓ 1
A1 upper quartile 6.78
✓ 1
A1 interquartile range 3.11 kW
✓ 1
Part (c)
B1 room B has the higher median, so typically uses more power per server
✓ 1
B1 room B has the smaller interquartile range, so its power use is more consistent
✓ 1
Exam tip: Quote both the median and the interquartile range when comparing two distributions.
217IGMC3-9-08r25-11Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard7 marks
Calculator
A histogram shows the masses of apples from an orchard. The heights of the bars are frequency densities. 100 < m ≤ 120 g: height 0.4 120 < m ≤ 130 g: height 1.6 130 < m ≤ 150 g: height 2.0 150 < m ≤ 190 g: height 0.5
Histogram of the masses of the apples
(a)
Calculate the number of apples with a mass in the class 150 < m ≤ 190.
(2)
(b)
Calculate the total number of apples.
(2)
(c)
Calculate an estimate of the mean mass of the apples. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 class width 40 and frequency = 0.5 × 40
✓ 1
A1 20
✓ 1
Part (b)
M1 frequencies 8, 16 and 40 found
✓ 1
A1 84
✓ 1
Part (c)
M1 uses midpoints 110, 125, 140, 170 with the frequencies
Common error: Frequency = frequency density × class width. Do not read the bar height as the frequency when class widths are unequal.
218IGMC3-9-08r25-12Averages, range and frequency tablesPaper EHard7 marks
Calculator
The table shows the lengths, t minutes, of songs in a playlist. The frequency of the class 3 < t ≤ 4 is x. 2 < t ≤ 3: 6 songs 3 < t ≤ 4: x songs 4 < t ≤ 5: 19 songs 5 < t ≤ 7: 15 songs
(a)
Show that the sum of (midpoint × frequency) for all the songs is 190.5 + 3.5x.
(2)
(b)
The estimated mean length of a song is 4.51 minutes. Find the value of x.
(4)
(c)
Explain why the mean calculated from this table is only an estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 midpoints 2.5, 3.5, 4.5 and 6
✓ 1
A1 15 + 3.5x + 85.5 + 90 = 190.5 + 3.5x
✓ 1
Part (b)
M1 (190.5 + 3.5x) ÷ (40 + x) = 4.51
✓ 1
M1 190.5 + 3.5x = 180.4 + 4.51x
✓ 1
M1 10.1 = 1.01x
✓ 1
A1 x = 10
✓ 1
Part (c)
B1 the exact lengths are not known, so the midpoint is used to represent each class
✓ 1
Exam tip: With an unknown frequency, write the total frequency as an expression too, then form and solve an equation.
219IGMC3-9-12r29-12Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard10 marks
Calculator
The ages, a years, of 120 people who caught an infection are shown in a histogram. The frequency densities are: 0 < a ≤ 10: 1.5 10 < a ≤ 25: 3.2 25 < a ≤ 30: x 30 < a ≤ 50: 1.8 50 < a ≤ 60: 0.9
(a)
Calculate the number of people in the class 10 < a ≤ 25.
Common error: The width of each bar is the class width, so a bar from 25 to 30 has width 5, not 1.
220IGMC3-9-12r29-14Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard10 marks
Calculator
The first-fault times, t months, of 140 solar inverters are summarised in a cumulative frequency table. t ≤ 10: 12 t ≤ 20: p t ≤ 30: 96 t ≤ 40: 128 t ≤ 50: 140 Linear interpolation is used for all estimates. The estimated median is 25 months.
(a)
Find the value of p.
(4)
(b)
Estimate the interquartile range, to 3 significant figures.
(4)
(c)
Calculate the percentage of inverters whose first fault came after more than 40 months.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 the median is the 70th value, between t = 20 and t = 30
Exam tip: Write the interpolation as a fraction of the way across the class so an unknown value can be solved for as an ordinary equation.
221IGMC3-9-11r31-07Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard7 marks
Calculator
The time, t minutes, that 100 gamers spend on a game in one evening is shown in a histogram. 0 < t ≤ 20: 16 gamers, drawn as a bar 4 cm high 20 < t ≤ 50: 45 gamers 50 < t ≤ 60: 24 gamers 60 < t ≤ 120: 15 gamers The same vertical scale is used for all bars.
(a)
Calculate the height of the bar for 20 < t ≤ 50.
(3)
(b)
Calculate the height of the bar for 50 < t ≤ 60.
(2)
(c)
Estimate the number of gamers who spend more than 80 minutes on the game.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequency density of first bar \(= 16\div20 = 0.8\), so 1 cm represents 0.2 (or 0.8 per 4 cm)
✓ 1
M1 frequency density for 20 < t ≤ 50 is \(45\div30 = 1.5\)
✓ 1
A1 7.5 cm
✓ 1
Part (b)
M1 frequency density \(24\div10 = 2.4\) (or ratio \(2.4\div0.8 = 3\))
✓ 1
A1 12 cm
✓ 1
Part (c)
M1 \((120-80)\times\frac{15}{60}\) (assume equal spread in the class)
✓ 1
A1 10
✓ 1
Exam tip: With a fixed vertical scale, bar height is proportional to frequency density, so compare heights through the ratio of frequency densities.
222IGMC3-9-11r31-10Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard8 marks
Calculator
The concentration of a medicine, c units, in the blood of 120 patients is recorded. The cumulative frequencies are: c ≤ 20: 10 c ≤ 40: 34 c ≤ 60: 70 c ≤ 80: 106 c ≤ 100: 120 For each estimate, assume the cumulative frequency curve is a straight line between the plotted points (linear interpolation). Use unrounded values in later calculations.
(a)
Estimate the lower quartile. Give your answer to 3 significant figures.
(3)
(b)
Estimate the upper quartile. Give your answer to 3 significant figures.
(3)
(c)
Hence find the interquartile range, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 lower quartile is the 30th value, in the class 20 < c ≤ 40
✓ 1
M1 \(20 + 20\times\frac{30-10}{34-10}\)
✓ 1
A1 36.7
✓ 1
Part (b)
M1 upper quartile is the 90th value, in the class 60 < c ≤ 80
✓ 1
M1 \(60 + 20\times\frac{90-70}{106-70}\)
✓ 1
A1 71.1
✓ 1
Part (c)
M1 upper quartile − lower quartile (ECF)
✓ 1
A1 34.4
✓ 1
Common error: The quartiles are the (n ÷ 4)th and (3n ÷ 4)th values, here the 30th and 90th, and interpolation must be done inside the correct class.
223IGMC3-9-11r31-11Averages, range and frequency tablesPaper EHard5 marks
Calculator
The time, t minutes, taken by players to complete a game level is recorded. 0 < t ≤ 10: 6 players 10 < t ≤ 20: 15 players 20 < t ≤ 30: x players 30 < t ≤ 50: 8 players The estimated mean time, using midpoints, is 21.25 minutes.
(a)
Write down the midpoint of the class 30 < t ≤ 50.
(1)
(b)
Show that \(3.75x = 41.25\).
(3)
(c)
Find the value of x.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 40
✓ 1
Part (b)
B1 total number of players \(= 29 + x\)
✓ 1
M1 \(\sum fx = 6\times5+15\times15+25x+8\times40\) \((=575+25x)\)
✓ 1
A1 \(575 + 25x = 21.25(29 + x)\) leading to \(3.75x = 41.25\) with a correct step shown
✓ 1
Part (c)
A1 11
✓ 1
Exam tip: For a grouped mean with an unknown frequency, write the total of (midpoint × frequency) and the total frequency as expressions in the unknown, then form an equation.
224IGMC3-9-12r34-12Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard8 marks
Calculator
A histogram shows the wind speed, v m/s, recorded on 100 days at a wind farm. The frequency densities for the classes are
Wind speed (v)
0 < v ≤ 4
4 < v ≤ 6
6 < v ≤ 10
10 < v ≤ 20
Frequency density
5
12
9
not given
(a)
Work out the frequency for the class 4 < v ≤ 6.
(2)
(b)
Find the frequency density of the class 10 < v ≤ 20.
(3)
(c)
Calculate an estimate of the mean wind speed.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 frequency = frequency density × class width = 12 × 2
✓ 1
A1 24
✓ 1
Part (b)
B1 frequencies of the first and third classes are 20 and 36
✓ 1
M1 100 − 20 − 24 − 36 = 20 (ECF from (a))
✓ 1
A1 20 ÷ 10 = 2
✓ 1
Part (c)
M1 use of midpoints 2, 5, 8 and 15
✓ 1
M1 (2×20 + 5×24 + 8×36 + 15×20) ÷ 100
✓ 1
A1 7.48 m/s
✓ 1
Common error: Do not use the frequency density in place of the frequency when calculating the mean.
225IGMC3-9-12r34-13Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard9 marks
Calculator
The cumulative frequency table shows the times, t minutes, taken by 80 swimmers to swim 1 km.
Time
t ≤ 10
t ≤ 15
t ≤ 20
t ≤ 25
t ≤ 30
t ≤ 35
t ≤ 45
Cumulative frequency
0
3
12
28
52
72
80
Assume that times are spread evenly within each class.
(a)
Write down the number of swimmers who took more than 20 minutes and up to 25 minutes.
(1)
(b)
Estimate the median time.
(2)
(c)
Estimate the interquartile range.
(3)
(d)
The slowest 15% of swimmers took longer than T minutes. Estimate T.
(3)
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Part (a)
B1 28 − 12 = 16
✓ 1
Part (b)
M1 40th value in the class 25 < t ≤ 30: 25 + (40 − 28) ÷ 24 × 5
Exam tip: Read the quartiles at 25% and 75% of the total frequency, then subtract; the lower and upper quartiles give the spread of the middle half.
226IGMC3-9-10r36-12Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard8 marks
Calculator
The cumulative frequency table shows the finishing times, t minutes, of 120 riders in a cycling event. Assume that a cumulative frequency graph is drawn by joining the plotted points with straight lines. Finishing time t ≤ 120: 6 t ≤ 140: 24 t ≤ 160: 72 t ≤ 180: 108 t ≤ 200: 118 t ≤ 220: 120
Cumulative frequency graph of the finishing times
(a)
Estimate the median finishing time.
(2)
(b)
Estimate the interquartile range.
(3)
(c)
Estimate the number of riders who took longer than 190 minutes.
(2)
(d)
In the previous year's event the interquartile range was 41 minutes. Compare the spread of finishing times in the two years.
(1)
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Part (a)
M1 median is the 60th value, between (140, 24) and (160, 72), e.g. 140 + (36 ÷ 48) × 20
✓ 1
A1 155 (accept 153 to 157 from a graph)
✓ 1
Part (b)
M1 lower quartile is the 30th value and upper quartile is the 90th value
✓ 1
A1 lower quartile 142.5 and upper quartile 170 (accept 1 minute either way from a graph)
✓ 1
A1 27.5 (accept 26 to 29)
✓ 1
Part (c)
M1 cumulative frequency at 190 minutes is 113 (by interpolation between (180, 108) and (200, 118))
✓ 1
A1 120 − 113 = 7 (accept 6 to 8 from a graph)
✓ 1
Part (d)
B1 this year's finishing times were less spread out, because the interquartile range (27.5 minutes) is smaller than 41 minutes (ECF from (b))
✓ 1
Common error: For quartiles use n/4 and 3n/4 on the vertical axis (here the 30th and 90th values), then read across to the horizontal axis.
227IGMC3-9-10r36-13Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard9 marks
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The histogram shows the durations, t minutes, of 150 jobs run on a data centre server. Classes have different widths, and the vertical axis shows frequency density. 0 < t ≤ 5: frequency density 6 5 < t ≤ 10: frequency density 10 10 < t ≤ 20: frequency density k 20 < t ≤ 40: frequency density 1.5
(a)
Find the number of jobs in the class 0 < t ≤ 5 and in the class 5 < t ≤ 10.
(2)
(b)
Find the number of jobs in the class 20 < t ≤ 40.
(1)
(c)
Find the value of k.
(3)
(d)
Calculate an estimate of the median duration.
(3)
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Part (a)
M1 frequency = frequency density × class width, e.g. 6 × 5
✓ 1
A1 30 and 50
✓ 1
Part (b)
B1 1.5 × 20 = 30
✓ 1
Part (c)
M1 150 − 30 − 50 − 30
✓ 1
A1 40 jobs in the class 10 < t ≤ 20
✓ 1
A1 k = 40 ÷ 10 = 4
✓ 1
Part (d)
M1 median is the 75th value, which lies in the class 5 < t ≤ 10 (cumulative frequencies 30 and 80)
✓ 1
M1 5 + (75 − 30) ÷ 10
✓ 1
A1 9.5
✓ 1
Exam tip: Frequency = frequency density × class width. Use cumulative totals to find which class contains the median, then interpolate inside that class.
228IGMC3-9-11r38-12Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard9 marks
Calculator
The lengths, L cm, of the fish in a catch are shown in a histogram. 10 < L ≤ 20: frequency density 1.2 20 < L ≤ 30: frequency density 2.6 30 < L ≤ 50: 38 fish (frequency density not given) 50 < L ≤ 80: frequency density 0.4
(a)
Find the frequency density for the class 30 < L ≤ 50.
(2)
(b)
Calculate an estimate of the mean length. Give your answer correct to 3 significant figures.
(4)
(c)
Estimate the median length. Give your answer correct to 3 significant figures.
(3)
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Part (a)
M1 38 ÷ 20
✓ 1
A1 1.9
✓ 1
Part (b)
B1 frequencies 12, 26 and 12 found (density × width), total 88
B1 44th value lies in 30 < L ≤ 50 (cumulative frequencies 38 then 76)
✓ 1
M1 30 + (44 − 38) ÷ 38 × 20
✓ 1
A1 33.2 cm
✓ 1
Common error: Using the width-times-density product only for the first class and treating later classes as if all widths were equal.
229IGMC3-9-11r38-13Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard9 marks
Calculator
The assembly times of 120 robots are summarised in a cumulative frequency table. Use linear interpolation within a class. Time t (minutes): ≤ 10, ≤ 20, ≤ 30, ≤ 40, ≤ 50, ≤ 60 Cumulative frequency: 8, 30, 66, 100, 116, 120
(a)
Estimate the number of robots that took more than 45 minutes to assemble.
(2)
(b)
Estimate the median time. Give your answer correct to 3 significant figures.
(3)
(c)
Estimate the interquartile range. Give your answer correct to 3 significant figures.
(4)
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Part (a)
M1 number up to 45 minutes = 100 + (5 ÷ 10) × 16 = 108
Exam tip: Locate the cumulative frequency you need first, then work out which class it falls in before interpolating.
230IGMC3-9-11r43-11Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard9 marks
Calculator
A water company records the water used in one quarter by 200 households. The table shows the cumulative frequency of the volume u m3. Assume that values within each class are evenly spread. Volume: u ≤ 10, u ≤ 20, u ≤ 30, u ≤ 40, u ≤ 50, u ≤ 60 Cumulative frequency: 16, 56, 120, 172, 192, 200
(a)
Estimate the median volume. Give your answer correct to 3 significant figures.
(3)
(b)
Households that use more than 45 m3 pay a higher rate. Estimate the number of these households.
(3)
(c)
The 5% of households with the greatest use receive a warning letter. Estimate the smallest volume that receives a letter.
(3)
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Part (a)
M1 100th value lies in the class 20 < u ≤ 30
✓ 1
M1 20 + (100 − 56) ÷ 64 × 10
✓ 1
A1 26.9
✓ 1
Part (b)
M1 172 + (45 − 40) ÷ 10 × 20
✓ 1
A1 182
✓ 1
A1 18
✓ 1
Part (c)
M1 5% of 200 = 10 households, so the 190th value
✓ 1
M1 40 + (190 − 172) ÷ 20 × 10
✓ 1
A1 49 (m3)
✓ 1
Exam tip: For percentiles, convert the percentage to a position in the ordered data first, then interpolate inside the class that contains it.
231IGMC3-9-11r43-14Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard11 marks
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An online game studies hours played per week, x, and the score, y points. Five players play 4, 7, 9, 12 and 13 hours. A sixth player plays h hours. The mean number of hours for the six players is 10. The six scores are 52, 63, 66, 77, 78 and 84.
(a)
Find the value of h.
(3)
(b)
A line of best fit passes through (4, 52) and (16, 88). Find the equation of the line in the form y = mx + c.
(3)
(c)
Show that the line of best fit passes through the mean point of the data.
(2)
(d)
Explain what the gradient means in this context.
(1)
(e)
Use the line to estimate the score of a player who plays 11.5 hours per week.
(2)
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Part (a)
M1 total hours = 6 × 10 = 60
✓ 1
M1 60 − (4 + 7 + 9 + 12 + 13)
✓ 1
A1 15
✓ 1
Part (b)
M1 gradient (88 − 52) ÷ (16 − 4)
✓ 1
A1 m = 3
✓ 1
A1 y = 3x + 40
✓ 1
Part (c)
M1 mean score = 420 ÷ 6 = 70
✓ 1
A1 when x = 10, 3 × 10 + 40 = 70, so the line passes through (10, 70)
✓ 1
Part (d)
B1 each extra hour played per week is linked to 3 more points in the score
✓ 1
Part (e)
M1 3 × 11.5 + 40
✓ 1
A1 74.5 (points)
✓ 1
Exam tip: The line of best fit always passes through the mean point, which gives a quick check on any equation you find.
232IGMC3-9-11r45-12Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard7 marks
Calculator
The waiting times of 120 patients at a clinic are summarised by cumulative frequency.
Waiting time (min)
≤ 10
≤ 20
≤ 30
≤ 40
≤ 60
Cumulative frequency
9
33
69
100
120
Assume that consecutive points are joined by straight lines (linear interpolation).
(a)
Estimate the median waiting time.
(3)
(b)
Estimate the number of patients who waited more than 45 minutes.
(3)
(c)
Write your answer to part (b) as a percentage of all the patients.
(1)
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Notes
Part (a)
M1 the 60th value lies between 20 min (33) and 30 min (69)
✓ 1
M1 20 + (60 − 33) ÷ (69 − 33) × 10
✓ 1
A1 27.5 minutes
✓ 1
Part (b)
M1 cumulative frequency at 45 minutes = 100 + (45 − 40) ÷ 20 × 20
✓ 1
A1 105
✓ 1
A1 120 − 105 = 15
✓ 1
Part (c)
15 ÷ 120 × 100 = 12.5% (ECF from (b))
✓ 1
Exam tip: 'More than' means subtract the cumulative frequency from the total.
233IGMC3-9-11r45-14Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard9 marks
Calculator
A histogram shows the sizes of 94 bee colonies, in thousands of bees, x.
Size, x
0 < x ≤ 20
20 < x ≤ 30
30 < x ≤ 50
50 < x ≤ 80
Frequency density
0.9
2.4
k
0.4
(a)
Find the value of k.
(4)
(b)
Estimate the number of colonies with a size greater than 40 thousand bees.
(2)
(c)
Calculate an estimate of the mean size of a colony.
B1 40 < x ≤ 50 has 10 × 2 = 20 colonies (ECF from (a))
✓ 1
A1 20 + 12 = 32
✓ 1
Part (c)
M1 uses midpoints 10, 25, 40, 65
✓ 1
M1 Σfx = 180 + 600 + 1600 + 780 = 3160
✓ 1
A1 33.6 thousand bees
✓ 1
Exam tip: Convert frequency density to frequency by multiplying by the class width before adding or averaging.
234IGMC3-9-10r46-12Scatter diagrams, correlation, cumulative frequency and histogramsPaper EHard9 marks
Calculator
The masses of honey from 100 hives are summarised in the table. Assume the hives within each class are spread evenly, so cumulative frequency can be found by linear interpolation. Mass, m kg | 40 < m ≤ 50 | 50 < m ≤ 60 | 60 < m ≤ 70 | 70 < m ≤ 80 | 80 < m ≤ 90 | 90 < m ≤ 100 Frequency | 10 | 20 | 40 | 20 | 7 | 3
(a)
Complete the cumulative frequency for each upper class boundary.
(2)
(b)
Estimate the median mass.
(2)
(c)
Estimate the interquartile range.
(3)
(d)
Estimate the number of hives with a mass between 55 kg and 65 kg.
(2)
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Part (a)
B1 10, 30, 70
✓ 1
B1 90, 97, 100
✓ 1
Part (b)
M1 50th value lies in 60 < m ≤ 70, 20 hives into a class with 40 hives
✓ 1
A1 60 + (20 ÷ 40) × 10 = 65 kg
✓ 1
Part (c)
M1 lower quartile: 25th value lies in 50 < m ≤ 60, 50 + (15 ÷ 20) × 10 = 57.5
✓ 1
M1 upper quartile: 75th value in class 70–80, 70 + (5 ÷ 20) × 10 = 72.5
✓ 1
A1 72.5 − 57.5 = 15 kg
✓ 1
Part (d)
M1 cumulative frequency at 55 is 10 + 0.5 × 20 = 20 and at 65 is 30 + 0.5 × 40 = 50
✓ 1
A1 50 − 20 = 30
✓ 1
Exam tip: Locate the class containing the required position first, then interpolate within that class.
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