Edexcel International A Level Maths · S1 · Statistics 1
Representing data
28 free, high-quality past-paper-style questions on Representing data for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 234 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
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A sports scientist wants to plan training groups for a running club. She models the 100 m times of the club's adult sprinters as being distributed symmetrically about 13.0 seconds. She then times 40 of the sprinters. For these 40 times the lower quartile is 12.6 s, the median is 12.9 s and the upper quartile is 13.7 s.
(a)
Give one advantage of using a model in this situation.
(1)
(b)
Timing the 40 sprinters is one stage of the modelling process. State the stage that comes next.
(1)
(c)
Use the quartiles to describe the skewness of the 40 times.
(2)
(d)
Comment on the model in the light of your answer to part (c), and suggest how it could be refined.
(1)
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Answer
Mark
Notes
Part (a)
any one, in context: quick/cheap way to predict the sprinters' times without timing everyone / helps plan the training groups / easy to adjust if the club changes
B1
Part (b)
compare the observed data (the 40 times) with the predictions of the model
B1
Part (c)
Q3 − Q2 = 0.8 and Q2 − Q1 = 0.3
M1
0.8 > 0.3, so the times are positively skewed
A1
Part (d)
the data are not symmetrical, so the model is not supported; refine it to a positively skewed model (e.g. with a median nearer 12.9 s) and test it again with new data
B1
Exam tip: When asked to comment on a model, link the statistic you found to the assumption the model made: here “symmetrical” is contradicted by the unequal halves of the box.
Common error: Dividing the frequency by the class boundary or the midpoint instead of by the class width. Write the width of each class in a separate column first.
The numbers of emails received in one day by 22 employees of a small firm were 24, 8, 36, 17, 29, 45, 15, 33, 21, 67, 26, 12, 38, 24, 31, 19, 42, 28, 17, 35, 51, 23
(a)
Draw an ordered stem-and-leaf diagram to show these data, including a key.
(3)
(b)
Find the median.
(1)
(c)
Find the lower quartile and the upper quartile.
(2)
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Answer
Mark
Notes
Part (a)
stems 0 to 6 with the leaves attached to the correct stems
22/2 = 11, so the mean of the 11th and 12th values: (26 + 28)/2 = 27
B1
Part (c)
22/4 = 5.5 → 6th value: Q1 = 19
B1
3 × 22/4 = 16.5 → 17th value: Q3 = 36
B1
Exam tip: Write the position before the value — “22/4 = 5.5, so the 6th value” — then count along the ordered leaves. A slip in counting still earns the method if the position is right.
The distances, in km, cycled in one week by the members of a cycling club are summarised as follows: smallest distance 12, lower quartile 38, median 47, upper quartile 59, largest distance 104. The three largest distances are 81, 92 and 104 km. For these data, an outlier is a value that is greater than Q3 + 1.5 × (Q3 − Q1) or less than Q1 − 1.5 × (Q3 − Q1).
(a)
Find the interquartile range.
(1)
(b)
Find the limits for outliers and identify any outliers.
(3)
(c)
State the value at which the upper whisker of a box plot of these data should end.
(1)
(d)
Describe the skewness of the distances, giving a reason.
(1)
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Answer
Mark
Notes
Part (a)
59 − 38 = 21 km
B1
Part (b)
59 + 1.5 × 21 or 38 − 1.5 × 21
M1
upper limit 90.5, lower limit 6.5
A1
92 and 104 are outliers; 12 > 6.5 so there are none at the lower end
A1
Part (c)
81 km (the largest value that is not an outlier)
B1
Part (d)
Q3 − Q2 = 12 > Q2 − Q1 = 9, so positive skew
B1
Model sketch
Figure: box plot of the distances, outliers marked ×
Common error: Ending the whisker at the fence (90.5) or at the largest outlier (104). The whisker stops at the last data value inside the fences.
The histogram in Figure 1 shows the heights, h cm, of the sunflowers grown in a community garden. No sunflower was shorter than 100 cm or taller than 200 cm. The top of every bar lies on a gridline.
(a)
Explain why the vertical axis shows frequency density rather than frequency.
(1)
(b)
Find the number of sunflowers in each class and the total number of sunflowers.
(3)
(c)
Estimate the number of sunflowers with heights between 125 cm and 150 cm.
(2)
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Answer
Mark
Notes
Part (a)
the classes have different widths, so the area of each bar, not its height, must represent the frequency
B1
Part (b)
frequency = frequency density × class width, e.g. 0.6 × 20
M1
12, 24, 38, 32, 16
A1
total = 122
A1
Part (c)
(130 − 125) × 2.4 + 38 + (150 − 140) × 1.6
M1
= 12 + 38 + 16 = 66
A1
Common error: Reading the bar heights as frequencies. On a histogram with unequal classes, multiply each height by its class width; the tall narrow bar is not necessarily the class with the most sunflowers.
Figure 1 is an ordered stem-and-leaf diagram, with split stems, of the resting pulse rates, in beats per minute, of the 21 members of a rowing club. For these data, an outlier is any value that is more than 1.5 × (Q3 − Q1) above the upper quartile or below the lower quartile.
(a)
Find the median pulse rate.
(1)
(b)
Find the lower quartile and the upper quartile.
(2)
(c)
Determine whether there are any outliers.
(2)
(d)
Describe the skewness of the pulse rates, giving a reason.
(1)
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Answer
Mark
Notes
Part (a)
21/2 = 10.5 → 11th value: 57
B1
Part (b)
21/4 = 5.25 → 6th value: Q1 = 52
B1
3 × 21/4 = 15.75 → 16th value: Q3 = 63
B1
(allow Q1 = 52, Q3 = 64 as the medians of the lower and upper halves, with ft in the later parts)
Exam tip: With split stems each stem appears twice: the first row holds leaves 0–4 and the second leaves 5–9. Count along every row, including any empty one, when you locate the 6th, 11th and 16th values.
A rail company recorded how late each of 80 late-running trains arrived at a station. The box plot in Figure 1 summarises the delays, in minutes. The ends of the box and whiskers, the median and the cross all lie on gridlines. An outlier is a value more than 1.5 times the interquartile range above the upper quartile or below the lower quartile.
(a)
Write down the median delay and find the interquartile range.
(2)
(b)
Show that the delay marked with a cross is an outlier, and explain why there can be no outliers at the lower end.
(2)
(c)
Describe the skewness of the delays, giving a reason.
(1)
(d)
Estimate how many of the 80 trains were delayed by more than 15 minutes.
(1)
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Answer
Mark
Notes
Part (a)
median = 9 minutes
B1
IQR = 15 − 6 = 9 minutes
B1
Part (b)
upper limit 15 + 1.5 × 9 = 28.5, and 34 > 28.5
B1
lower limit 6 − 13.5 = −7.5; every delay is positive, so none can be below it
B1
Part (c)
Q3 − Q2 = 6 > Q2 − Q1 = 3, so positive skew
B1
Part (d)
15 minutes is Q3, so about a quarter: 80 ÷ 4 = 20 trains
B1
Exam tip: Each whisker and each half of the box holds about a quarter of the data, however long it is drawn. A long section means the values there are spread out, not that there are more of them.
A bakery manager models the masses of the bakery's large white loaves as being distributed symmetrically about 800 g. The top part of Figure 1 shows the stages of the modelling process she follows; two stages are labelled P and Q. The lower part of Figure 1 is a box plot of the masses of a sample of 50 loaves; every value shown lies on a gridline.
(a)
State what should be written in the boxes labelled P and Q.
(2)
(b)
Use the box plot in Figure 1 to comment on whether the sample supports the manager's model.
(2)
(c)
State what the manager should do next.
(1)
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Answer
Mark
Notes
Part (a)
P: use the model to make predictions (e.g. of the median and the shape of the distribution of masses)
B1
Q: compare the sample data with the predictions of the model
B1
Part (b)
Q3 − Q2 = 806 − 803 = 3 and Q2 − Q1 = 803 − 796 = 7, so the masses are negatively skewed
M1
the sample is not symmetrical (and its median is 803 g, not 800 g), so it does not support the model
A1
Part (c)
refine the model (e.g. a negatively skewed model centred nearer 803 g) and test it again against new data
B1
Exam tip: When you judge a model, quote the numbers that test its assumption (here the two halves of the box) and then say which stage of the cycle comes next.
The lengths of 100 fish caught in a lake were measured to the nearest centimetre and grouped as shown.
Length (cm)
10–14
15–19
20–29
30–44
Frequency
24
30
p
q
On a histogram of these data, the bar for the class 10–14 is 1.5 cm wide and 8 cm tall.
(a)
Write down the class boundaries of the class 20–29.
(1)
(b)
Find the width and the height of the bar for the class 15–19.
(2)
(c)
The bar for the class 30–44 is 2 cm tall. Find its width and the value of q.
(3)
(d)
Find p and the height of the bar for the class 20–29.
(2)
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Answer
Mark
Notes
Part (a)
19.5 cm and 29.5 cm
B1
Part (b)
same class width (5 cm), so width 1.5 cm
B1
height = 8 × 30/24 = 10 cm
B1
Part (c)
class width 15 cm, and 5 cm ↔ 1.5 cm, so width = 4.5 cm
B1
area scale: 1.5 × 8 = 12 cm2 for 24 fish, so 0.5 cm2 per fish; area = 4.5 × 2 = 9 cm2
M1
q = 9 ÷ 0.5 = 18
A1
Part (d)
p = 100 − 24 − 30 − 18 = 28
B1
width 3 cm, area 28 × 0.5 = 14 cm2, height = 14 ÷ 3 = 4.67 cm
B1
Exam tip: Fix two scales from the bar you know: horizontal (cm per cm of length) and area (cm2 per fish). Every other bar then follows from width = class width × scale and height = area ÷ width.
The waiting times, t minutes, of patients at a clinic one week are summarised in the table. Two frequencies are missing.
Waiting time t (min)
Frequency
Frequency density
0 ≤ t < 5
14
5 ≤ t < 10
6.4
10 ≤ t < 20
46
20 ≤ t < 30
3.0
30 ≤ t < 60
18
(a)
Find the two missing frequencies.
(2)
(b)
Find the total number of patients and the frequency density of the class 30 ≤ t < 60.
(2)
(c)
Estimate the number of patients who waited more than 25 minutes.
(2)
(d)
Estimate the percentage of patients who waited less than 8 minutes.
(2)
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Answer
Mark
Notes
Part (a)
frequency = frequency density × class width: 6.4 × 5 = 32
B1
3.0 × 10 = 30
B1
Part (b)
14 + 32 + 46 + 30 + 18 = 140
B1
18 ÷ 30 = 0.6
B1
Part (c)
(30 − 25) × 3.0 + 18
M1
= 33
A1
Part (d)
14 + (8 − 5) × 6.4 = 33.2
M1
33.2 ÷ 140 × 100 = 23.7%
A1
Model sketch
Figure: histogram of the waiting times
Common error: Taking a fraction of the class's frequency using the wrong width, e.g. 3/10 of 32. The part 5 ≤ t < 8 is 3/5 of its class, or simply 3 × frequency density.
A fitness instructor recorded the number of press-ups completed in one minute by 15 members aged under 30 and 15 members aged 30 or over. The results are shown in the back-to-back stem-and-leaf diagram.
Under 30
Stem
30 or over
0
9
8
1
4 6 8 9
9 7 5 2
2
1 2 3 4 6 7 9
9 8 6 4 3 1 1
3
3 5
4 1
4
8
2
5
Key: 2 | 3 | 1 means 32 press-ups for the under-30 group and 31 for the 30-or-over group
For each group, an outlier is a value that is greater than Q3 + 1.5 × (Q3 − Q1) or less than Q1 − 1.5 × (Q3 − Q1).
(a)
Find the median and the quartiles for each group.
(4)
(b)
Determine whether there are any outliers in either group.
(3)
(c)
Compare the numbers of press-ups completed by the two groups.
under 30: IQR 12, limits 27 − 18 = 9 and 39 + 18 = 57
M1
all values lie between 9 and 57, so no outliers
A1
30 or over: IQR 11, limits 1.5 and 45.5, so 48 is an outlier
A1
Part (c)
the under-30 median (33) is higher than the 30-or-over median (23), so the younger members generally completed more press-ups
B1
the IQRs (12 and 11) are similar, so the spread of the middle half is about the same for both groups
B1
Exam tip: Read the left-hand leaves from the stem outwards: in “9 7 5 2 | 2 |” the values are 22, 25, 27 and 29. The key tells you which side is which group.
The times, x minutes, that 60 visitors spent at a museum exhibit have mean 23.4, median 26, mode 29 and standard deviation 6.5. The lower quartile is 19 and the upper quartile is 30. The shortest time was 4 minutes and the longest was 35 minutes.
(a)
Show, in two different ways, that the times are negatively skewed.
(2)
(b)
One measure of skewness is 3(mean − median) ÷ standard deviation. Evaluate this measure for these data.
(2)
(c)
An outlier is a value more than 2 standard deviations from the mean. Show that the shortest time is an outlier and that the longest is not.
(2)
(d)
The time of 4 minutes is found to be a recording error and is removed. Find the mean of the remaining 59 times.
(2)
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Answer
Mark
Notes
Part (a)
Q3 − Q2 = 4 < Q2 − Q1 = 7, so negative skew
B1
mean 23.4 < median 26 < mode 29, so negative skew
B1
Part (b)
3(23.4 − 26) ÷ 6.5
M1
= −1.2
A1
Part (c)
limits 23.4 ± 2 × 6.5, i.e. 10.4 and 36.4
M1
4 < 10.4 so 4 is an outlier; 35 < 36.4 so 35 is not
A1
Part (d)
Σx = 60 × 23.4 = 1404, so the remaining total is 1400
M1
mean = 1400 ÷ 59 = 23.7 min
A1
Exam tip: A coefficient given in the question is evaluated exactly as written; its sign tells you the direction of skew, so state that link in a few words.
The battery lives, in hours, of 20 phones of one model, each tested under the same conditions, were 8.3, 9.6, 7.4, 10.4, 5.2, 8.8, 7.9, 9.1, 12.9, 8.0, 7.1, 9.3, 8.6, 6.8, 9.8, 8.2, 10.1, 7.5, 9.0, 8.5 For these data, an outlier is a value that is greater than Q3 + 1.5 × (Q3 − Q1) or less than Q1 − 1.5 × (Q3 − Q1).
(a)
Find the median and the quartiles.
(3)
(b)
Show that exactly one of the values is an outlier.
(3)
(c)
Describe how the whiskers and any outliers should be drawn on a box plot of these data.
(2)
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Answer
Mark
Notes
Part (a)
ordered list, and positions 20/4 = 5, 20/2 = 10, 15 are whole numbers so average with the next value
12.9 > 12.075 is an outlier; the smallest value 5.2 > 5.075 is not, so exactly one
A1
Part (c)
lower whisker to 5.2; upper whisker to 10.4 (the largest value that is not an outlier)
B1
12.9 plotted separately as a cross
B1
Model sketch
Figure: box plot of the battery lives, outlier marked ×
Common error: Assuming the most extreme value at each end must be an outlier. Here 5.2 looks low but lies just inside the lower limit, so the lower whisker ends at 5.2.
The ages, in completed years, of the 132 members of a hill-walking club are summarised in the table.
Age (years)
10–14
15–19
20–29
30–49
50–79
Frequency
18
27
32
40
15
A histogram is to be drawn.
(a)
Explain why the class 15–19 has a class width of 5 years.
(1)
(b)
Calculate the frequency density of each class.
(2)
(c)
State the modal class.
(1)
(d)
Estimate the number of members aged from 25 to 34 years inclusive.
(2)
(e)
Explain why your answer to part (d) is only an estimate.
(1)
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Answer
Mark
Notes
Part (a)
ages are given in completed years, so the class contains ages from 15 up to (but not including) 20; boundaries 15 and 20 give width 5
B1
Part (b)
widths 5, 5, 10, 20, 30 used
M1
3.6, 5.4, 3.2, 2, 0.5
A1
Part (c)
15–19 (largest frequency density)
B1
Part (d)
25 ≤ age < 35: (30 − 25) × 3.2 + (35 − 30) × 2
M1
= 16 + 10 = 26
A1
Part (e)
it assumes the members' ages are spread evenly within each class, which may not be true
B1
Model sketch
Figure: histogram of the members' ages
Exam tip: For ages, “15–19” runs from the 15th birthday to just before the 20th, so the upper boundary is 20, not 19.5. Rounded measurements and completed years give different boundaries.
A roadside camera recorded the speeds, v km h−1, of cars passing a school. The histogram in Figure 1 represents the data, but the numbers on the frequency density axis have been removed. There were 36 cars in the class 40 ≤ v < 45. The top of every bar lies on a gridline.
(a)
Find the frequency density represented by one vertical grid unit.
(2)
(b)
Find the frequencies of the other four classes and the total number of cars.
(3)
(c)
Estimate the number of cars with speeds greater than 55 km h−1.
(2)
(d)
Estimate the percentage of these cars with speeds of at least 50 km h−1.
(1)
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Answer
Mark
Notes
Part (a)
frequency density of 40 ≤ v < 45 is 36 ÷ 5 = 7.2
M1
the bar is 9 units tall, so one unit represents 7.2 ÷ 9 = 0.8
A1
Part (b)
heights 3, 12, 5 and 1 units give frequency densities 2.4, 9.6, 4 and 0.8
M1
frequencies 24, 48, 40, 16
A1
total = 164 cars
A1
Part (c)
(60 − 55) × 4 + 16
M1
= 36
A1
Part (d)
(40 + 16) ÷ 164 × 100 = 34.1%
B1
Exam tip: When the scale is missing, fix it from the one class whose frequency you know: frequency density per grid unit. Then convert every other bar height before multiplying by its class width.
Figure 1 shows box plots of the distances, in metres, of the best throws in one season by the members of two javelin clubs, North and South. Every value shown lies on a gridline; the gridlines are 2 m apart. For each club, an outlier is a value that is greater than Q3 + 1.5(Q3 − Q1) or less than Q1 − 1.5(Q3 − Q1).
(a)
Find the interquartile range for each club.
(2)
(b)
Show that the throw marked with a cross for South is an outlier, and that North has no outliers.
(2)
(c)
Describe the skewness of each distribution, giving a reason.
(2)
(d)
Compare the distances thrown by the members of the two clubs.
(2)
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Answer
Mark
Notes
Part (a)
North: 46 − 36 = 10 m
B1
South: 52 − 40 = 12 m
B1
Part (b)
South: upper limit 52 + 18 = 70, and 74 > 70
B1
North: limits 36 − 15 = 21 and 46 + 15 = 61; the shortest (28 m) and longest (58 m) throws lie inside, so no outliers
South's median (44 m) is greater than North's (42 m), so South's members generally threw further
B1
South's IQR (12 m) is greater than North's (10 m), so South's distances are more spread out
B1
Common error: Misreading the scale. Find the gap between gridlines first (here 2 m) and count from the nearest labelled value; a quartile read one gridline out changes the IQR, the limits and the skewness.
Sixteen office workers took a typing course. Their typing speeds, in words per minute, were measured before and after the course and are shown in the back-to-back stem-and-leaf diagram at the top of Figure 1. Below it, Figure 1 also shows a box plot of the speeds before the course. An outlier is a value more than 1.5 × (Q3 − Q1) below Q1 or above Q3.
(a)
Find the median and the quartiles of the speeds after the course.
(3)
(b)
Show that exactly one of the speeds after the course is an outlier.
(3)
(c)
State the value at which the upper whisker of a box plot of the speeds after the course would end.
(1)
(d)
Use the box plot in Figure 1 and your answers to compare the typing speeds before and after the course.
(2)
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Answer
Mark
Notes
Part (a)
16 values: positions 4, 8 and 12 are whole numbers, so average each with the next value
M1
Q2 = (38 + 39)/2 = 38.5
A1
Q1 = (33 + 34)/2 = 33.5, Q3 = (43 + 45)/2 = 44
A1
Part (b)
IQR = 10.5, and 1.5 × 10.5 = 15.75
M1
limits 17.75 and 59.75
A1
only 66 > 59.75, and the smallest value 22 > 17.75, so exactly one outlier
A1
Part (c)
52 words per minute (the largest value that is not an outlier)
B1
Part (d)
the median rose from 28.5 to 38.5, so speeds were generally higher after the course
B1
the IQR is 10.5 in both cases, so the spread of the middle half of the speeds is unchanged
B1
Exam tip: On the left-hand side of a back-to-back diagram the leaves are read outwards from the stem: “8 2 | 1 |” means 12 and 18. The key tells you which side is which.
The masses, m kg, of the bags checked in for one flight were recorded. Figure 1 shows an incomplete histogram of the data: the bar for the class 20 ≤ m < 25 has not been drawn. There were 60 bags in this class.
(a)
Find the number of bags in each of the four classes shown in Figure 1.
(2)
(b)
Find the height of the missing bar and the total number of bags.
(2)
(c)
Estimate the number of bags with masses between 18 kg and 28 kg.
(2)
(d)
Bags heavier than 23 kg are charged an extra fee. Estimate the percentage of bags that were charged.
(1)
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Answer
Mark
Notes
Part (a)
frequency = frequency density × class width, e.g. 2 × 10
M1
20, 30, 50, 40
A1
Part (b)
60 ÷ 5 = 12
B1
20 + 30 + 50 + 60 + 40 = 200 bags
B1
Part (c)
(20 − 18) × 10 + 60 + (28 − 25) × 4
M1
= 20 + 60 + 12 = 92
A1
Part (d)
(25 − 23) × 12 + 40 = 64, and 64 ÷ 200 = 32%
B1
Common error: Drawing the missing bar with height 60. The height is the frequency density, 60 ÷ 5 = 12, so that the bar's area, not its height, represents the 60 bags.
Figure 1 shows the number of attempts, x, that each of 100 learner drivers needed to pass a practical driving test. For these data Σx = 261 and the standard deviation is 1.58 (to 3 s.f.).
(a)
Write down the mode and find the median.
(2)
(b)
Find the lower quartile and the upper quartile.
(2)
(c)
Find the mean number of attempts.
(1)
(d)
Give two different reasons why the data are positively skewed.
(2)
(e)
Evaluate 3(mean − median) ÷ standard deviation, and state what its sign shows.
(1)
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Answer
Mark
Notes
Part (a)
mode = 1
B1
the 50th and 51st values: cumulative frequencies 32 then 58, so both are 2 and the median = 2
B1
Part (b)
the 25th and 26th values are both 1, so Q1 = 1
B1
cumulative frequency 72 after x = 3 and 84 after x = 4, so the 75th and 76th values are 4: Q3 = 4
Exam tip: For a discrete distribution, build the cumulative frequencies first. The quartiles are then the values of x at the required positions, not class midpoints.
The numbers of hours, to the nearest hour, that 15 music students spent practising in one week were 6, 12, 3, 8, 21, 5, 9, 2, 7, 4, 10, 6, 3, 8, 5 A student drew the box plot in Figure 1 to represent these data. The student's box plot is not correct: it contains two errors. An outlier is a value greater than Q3 + 1.5 × (Q3 − Q1) or less than Q1 − 1.5 × (Q3 − Q1).
(a)
Find the median and the quartiles of the data.
(3)
(b)
Show that 21 is the only outlier.
(2)
(c)
Identify the two errors in the student's box plot in Figure 1 and state how each should be corrected.
21 > 16.5, and every other value lies between the limits (the next largest is 12), so 21 is the only outlier
A1
Part (c)
the median is drawn at 7; it should be at 6
B1
the upper whisker runs to 21; it should end at 12, with 21 plotted separately as a cross
B1
Common error: Running a whisker to the most extreme value without testing for outliers. Find the limits first; the whisker stops at the last data value inside them.
A courier recorded the distances, d km, of its deliveries from the depot on one day. The results are summarised in the table, where two frequencies are missing.
Distance d (km)
0 ≤ d < 2
2 ≤ d < 5
5 ≤ d < 8
8 ≤ d < 12
12 ≤ d < 20
Frequency
a
48
b
36
16
On a histogram of these data, the bar for 2 ≤ d < 5 is 8 cm tall, the bar for 0 ≤ d < 2 is 6 cm tall and the bar for 5 ≤ d < 8 is 7 cm tall.
(a)
Find the values of a and b.
(3)
(b)
Find the height of the bar for 12 ≤ d < 20.
(2)
(c)
Estimate the number of deliveries that were between 4 km and 10 km from the depot.
(2)
(d)
Estimates of the quartiles of these distances are Q1 = 3.09, Q2 = 5.79 and Q3 = 9.17. Describe the skewness of the distribution, giving a reason.
(2)
(e)
The longest delivery was 19.6 km. An outlier is a value greater than Q3 + 1.5 × (Q3 − Q1). Determine whether this delivery is an outlier.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
frequency density of 2 ≤ d < 5 is 48 ÷ 3 = 16, shown as 8 cm, so 1 cm ↔ frequency density 2
M1
0 ≤ d < 2: fd 12, a = 12 × 2 = 24
A1
5 ≤ d < 8: fd 14, b = 14 × 3 = 42
A1
Part (b)
fd = 16 ÷ 8 = 2
M1
height = 2 ÷ 2 = 1 cm
A1
Part (c)
(5 − 4) × 16 + 42 + (10 − 8) × 9
M1
= 16 + 42 + 18 = 76
A1
Part (d)
Q3 − Q2 = 3.38 and Q2 − Q1 = 2.70
M1
3.38 > 2.70, so positive skew
A1
Part (e)
IQR = 9.17 − 3.09 = 6.08
B1
limit = 9.17 + 1.5 × 6.08 = 18.29
M1
19.6 > 18.29, so it is an outlier
A1
Model sketch
Figure: histogram of the delivery distances
Exam tip: When bar heights are given in centimetres, find “1 cm ↔ ? units of frequency density” from the bar whose frequency you know, then convert every other height back to a frequency density before multiplying by the width.
A shop compares light bulbs from two suppliers. The lifetimes, in hundreds of hours, of 12 bulbs from supplier P were 41, 36, 47, 31, 44, 50, 38, 66, 43, 40, 48, 45 For a large sample of bulbs from supplier R: shortest 35, lower quartile 41, median 46, upper quartile 49, longest 52 (hundreds of hours), with no outliers. For these data, an outlier is a value that is greater than Q3 + 1.5 × (Q3 − Q1) or less than Q1 − 1.5 × (Q3 − Q1).
(a)
Find the median and the quartiles of the lifetimes of the bulbs from supplier P.
(3)
(b)
Show that supplier P's data contain exactly one outlier.
(2)
(c)
Describe the box plot for supplier P, stating the values at the ends of the whiskers.
(2)
(d)
Describe the skewness of each distribution, giving a reason in each case.
(2)
(e)
The shop wants bulbs that reliably last a long time. Recommend a supplier, giving two reasons based on the statistics, and state whether the bulb that lasted 6600 hours should affect the decision.
(3)
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Answer
Mark
Notes
Part (a)
ordered: 31, 36, 38, 40, 41, 43, 44, 45, 47, 48, 50, 66; positions 3, 6, 9 are whole numbers so average with the next
box from 39 to 47.5 with the median at 43.5; whiskers to 31 and 50
B1
66 plotted as a cross
B1
Part (d)
P: Q3 − Q2 = 4 < Q2 − Q1 = 4.5, so slight negative skew (accept “approximately symmetrical”, as 4 and 4.5 are close)
B1
R: Q3 − Q2 = 3 < Q2 − Q1 = 5, so negative skew
B1
Part (e)
supplier R: higher median (46 against 43.5), so R's bulbs generally last longer
B1
similar or smaller spread (IQR 8 against 8.5) and a longer shortest lifetime (35 against 31), so R is at least as reliable
B1
one unusually long-lived bulb is an outlier and should not drive the decision — the median and IQR are hardly affected by it
B1
Model sketch
Figure: box plots of the bulb lifetimes, outlier marked ×
Exam tip: A recommendation needs reasons from both location and spread, with the numbers quoted. Mention outliers only to say how far they should, or should not, influence the decision.
A web designer models the time that visitors spend on a page as being distributed symmetrically. To test the model, she records the times, t seconds, of 200 visits.
Time t (s)
0 ≤ t < 30
30 ≤ t < 60
60 ≤ t < 90
90 ≤ t < 150
150 ≤ t < 300
Frequency
12
45
54
54
35
For these data, estimates of the mean, median and standard deviation are 103.05 s, 83.9 s and 64.2 s.
(a)
Calculate the frequency density of each class.
(2)
(b)
State the modal class.
(1)
(c)
Estimate the number of visits that lasted more than 2 minutes.
(2)
(d)
One measure of skewness is 3(mean − median) ÷ standard deviation. Evaluate this measure and interpret its sign.
(2)
(e)
Comment on the designer's model. Give one advantage of using a model in this situation and suggest how the modelling process should continue.
(4)
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Answer
Mark
Notes
Part (a)
frequency ÷ class width
M1
0.4, 1.5, 1.8, 0.9, 0.233
A1
Part (b)
60 ≤ t < 90
B1
Part (c)
2 minutes = 120 s: (150 − 120) × 0.9 + 35
M1
= 27 + 35 = 62
A1
Part (d)
3(103.05 − 83.9) ÷ 64.2 = 0.895
B1
positive, so the times are positively skewed (a long tail of long visits)
B1
Part (e)
the data are positively skewed, so the symmetrical model is not supported
B1
also mean > median, consistent with this
B1
advantage, e.g. a model gives quick predictions for planning page content without recording every visit
B1
refine the model to a positively skewed one, then collect new data and compare again
B1
Model sketch
Figure: histogram of the visit times
Exam tip: In modelling questions, quote the statistic that contradicts the model's assumption (here the positive coefficient) and then describe the next stage of the cycle: refine, then test against new data.
The times, in seconds, taken by the 14 runners in each of two athletics squads, X and Y, to run 400 m are shown in the back-to-back stem-and-leaf diagram. One leaf in squad Y is shown as y, where 3 ≤ y ≤ 6.
Squad X
Stem
Squad Y
9 8 7 5 2
5
4 6 8
9 7 6 4 3 2 1
6
1 2 3 y 6 7 8
8 1
7
0 2 4
8
9
3
Key: 2 | 6 | 1 means 62 s for squad X and 61 s for squad Y
The median time for squad Y is 65 s. For each squad, an outlier is a value that is greater than Q3 + 1.5 × (Q3 − Q1) or less than Q1 − 1.5 × (Q3 − Q1).
(a)
Find the value of y.
(2)
(b)
Find the median and quartiles for squad X, and the quartiles for squad Y.
(4)
(c)
Show that squad X has no outliers and that squad Y has exactly one.
(3)
(d)
Describe the box plot for squad Y, stating the values at the ends of its whiskers.
(2)
(e)
Compare the performances of the two squads.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
14 values, so the median is the mean of the 7th and 8th values: (60 + y + 66)/2 = 65
box from 61 to 70 with the median at 65; whiskers to 54 and 74
B1
93 plotted as a cross
B1
Part (e)
median X (62.5 s) < median Y (65 s), so squad X generally ran faster
B1
both IQRs are 9 s, so the middle halves of the squads are equally consistent (squad Y also has one much slower runner)
B1
Model sketch
Figure: box plots of the 400 m times, outlier marked ×
Common error: Saying the squad with the higher median is “better”. In a race a lower time is better, so the context decides which way round the comparison goes.
A running app recorded the distances, d km, run in one week by 120 of its users. The histogram in Figure 1 represents the data. Below it, on the same scale, is a box plot of the same distances: the quartiles shown are estimates found from the grouped data, and the ends of the whiskers and the cross are actual distances. An outlier is a value more than 1.5 × (Q3 − Q1) above Q3 or below Q1.
(a)
Find the number of users in each class, and confirm that the total is 120.
(3)
(b)
Estimate the number of users who ran between 15 km and 32 km.
(3)
(c)
Use the box plot to describe the skewness of the distances, giving a reason.
(2)
(d)
Show that the distance of 57.4 km is an outlier but the distance of 51.8 km is not.
(3)
(e)
Give one feature of the distances that the histogram shows but the box plot does not.
57.4 > 53.5, so it is an outlier; 51.8 < 53.5, so it is not
A1
Part (e)
e.g. the modal class (20 ≤ d < 25) / how many users are in each class / the shape of the distribution inside the box
B1
Exam tip: A histogram and a box plot of the same data should tell the same story: here the long right-hand tail of the histogram matches the longer right half of the box and the long upper whisker.
A gardener grew three groups of 40 seedlings, each group with a different fertiliser, A, B or C, and measured their heights in mm after three weeks. The top row of Figure 1 shows a histogram for each group; the three histograms are drawn to the same scales. The bottom of Figure 1 shows the box plots of the three groups, numbered 1, 2 and 3 in a different order. For these data, use this outlier rule: an outlier is a value more than 1 × (Q3 − Q1) below Q1 or above Q3 (one interquartile range, not 1.5 interquartile ranges).
(a)
Describe the skewness of each of the distributions A, B and C.
(2)
(b)
Match each box plot to its histogram, giving a reason in each case.
(3)
(c)
For fertiliser A, estimate the number of seedlings taller than 25 mm.
(2)
(d)
For fertiliser C, the estimated quartiles are 26.7 mm, 38.0 mm and 45.7 mm. The shortest seedling was 3 mm and the next shortest was 9 mm. Show that 3 mm is an outlier but 9 mm is not.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
A: positive skew; B: symmetrical
B1
C: negative skew
B1
Part (b)
box plot 3 is A: the median is nearer the lower quartile and the upper whisker is long (positive skew)
B1
box plot 1 is B: the median is in the middle of the box and the whiskers are equal in length (symmetrical)
B1
box plot 2 is C: the median is nearer the upper quartile and there is a long lower tail (negative skew)
B1
Part (c)
(30 − 25) × 1.0 = 5, and the classes above 30 mm hold 0.6 × 10 + 0.4 × 10 + 0.2 × 10 = 12
M1
5 + 12 = 17
A1
Part (d)
IQR = 45.7 − 26.7 = 19.0
B1
lower limit 26.7 − 1 × 19.0 = 7.7
M1
3 < 7.7, so 3 mm is an outlier; 9 > 7.7, so 9 mm is not, and the lower whisker ends at 9
A1
Exam tip: To match a box plot to a histogram, look at where the median sits inside the box and which whisker is longer: both point the same way as the histogram's long tail.
The lengths of time, t hours, that cars stayed in a town-centre car park on one day are represented by the histogram in Figure 1, drawn on square grid paper. The key shows that one grid square represents 4 cars. No car stayed for 10 hours or more.
(a)
Find the number of cars in each class, and show that 360 cars used the car park.
(3)
(b)
Find the frequency density of the class 2 ≤ t < 4, and hence state the frequency density represented by one vertical grid unit.
(2)
(c)
Estimate the number of cars that stayed between 1.5 hours and 5 hours.
(3)
(d)
Parking is free for the first 3 hours. Estimate the percentage of these cars whose drivers had to pay.
(2)
(e)
The car park manager models the parking times as symmetrically distributed. Estimates from the data are: mean 2.81 hours, median 2.21 hours. Comment on the manager's model.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
count the squares in each bar: 18, 24, 28, 12, 8
M1
× 4: 72, 96, 112, 48, 32
A1
72 + 96 + 112 + 48 + 32 = 360
A1
Part (b)
112 ÷ 2 = 56 cars per hour
B1
the bar is 7 grid units tall, so one unit represents 8 cars per hour
B1
Part (c)
(2 − 1.5) × 96 = 48 and (5 − 4) × 24 = 24
M1
48 + 112 + 24
M1
= 184 cars
A1
Part (d)
(4 − 3) × 56 + 48 + 32 = 136
M1
136 ÷ 360 × 100 = 37.8%
A1
Part (e)
mean > median (2.81 > 2.21), so the times are positively skewed, as the long right-hand tail of the histogram also shows
B1
so the symmetrical model is not supported; it should be refined (e.g. to a positively skewed model) and tested again
B1
Exam tip: With a key square, area is the frequency: count whole squares in each bar and multiply by the key. Frequency densities follow afterwards from frequency ÷ class width.
The top part of Figure 1 is a stem-and-leaf diagram of the points scored by a basketball team in the 19 games of its season. The leaves in each row are in order, and two of them are shown as a and b. Below it, Figure 1 also shows a box plot of the same 19 scores. An outlier is a value more than 1.5 × (Q3 − Q1) above Q3 or below Q1.
(a)
Use the box plot in Figure 1 to find the values of a and b, explaining your method.
(3)
(b)
Show that the upper quartile is 84 and that 118 is the only outlier.
(3)
(c)
Describe the skewness of the scores, giving a reason.
(2)
(d)
In an extra play-off game the team scored 40 points. Find the median and quartiles of all 20 scores, and determine whether 40 is an outlier.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
19 values: Q1 is the 5th value (19/4 = 4.75) and the median is the 10th value (19/2 = 9.5)
118 > 114; the next largest score, 91, and the smallest, 52, lie inside the limits, so 118 is the only outlier
A1
Part (c)
Q3 − Q2 = 12 and Q2 − Q1 = 8
M1
12 > 8, so positive skew
A1
Part (d)
positions 5, 10, 15 are whole numbers, so average with the next value: Q1 = (63 + 64)/2 = 63.5
M1
Q2 = (70 + 72)/2 = 71, Q3 = (79 + 84)/2 = 81.5
A1
IQR = 18, lower limit 63.5 − 27 = 36.5; 40 > 36.5, so 40 is not an outlier
A1
Common error: Using the same quartile positions after a value is added. With 20 values the positions 5, 10 and 15 are whole numbers, so each quartile is the mean of two neighbouring values.
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