Edexcel International A Level Maths · S1 · Statistics 1
Measures of location and dispersion
28 free, high-quality past-paper-style questions on Measures of location and dispersion 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 · 231 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-S1-2-B-01S1 · ASEasy5 marks
The number of books, x, borrowed in one month by each of 50 members of a library is shown in the table.
Books, x
0
1
2
3
4
Frequency
9
17
12
8
4
Figure: bar chart of the numbers of books borrowed
(a)
Write down the mode.
(1)
(b)
Find the median.
(2)
(c)
Find the mean number of books borrowed.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
mode = 1 book
B1
Part (b)
50/2 = 25, so the median is the mean of the 25th and 26th values; cumulative frequencies 9, 26, …
M1
both are 1, so median = 1 book
A1
Part (c)
Σfx = 0 + 17 + 24 + 24 + 16 = 81
M1
mean = 81 ÷ 50 = 1.62 books
A1
Exam tip: Write the cumulative frequencies above the table. The position of the median (here the 25th and 26th values) can then be read off without listing all 50 values.
IQR = 33 − 23 = 10 mm and range = 40 − 18 = 22 mm, ft their quartiles
B1
Part (c)
new order 14, 18, 22, 23, 24, 26, 27, … ; the median is still the 6th value
M1
median = 26 mm
A1
Common error: Finding the quartiles from the data in the order given. Always rearrange the values in order first; the position rule only works on ordered data.
mean = 576 ÷ 18 = 32 cm; σ2 = 19 080 ÷ 18 − 322 = 36, so σ = 6 cm
A1
Exam tip: When values are added, update both sums and n, then use the formulae again. Here each new fish is exactly one standard deviation from the mean, so neither statistic changes.
The midday temperature, x °C, was recorded at a weather station on 10 days. The mean was 18.4 °C and the standard deviation was 3.5 °C. The temperatures are converted to degrees Fahrenheit, y °F, using y = 1.8x + 32.
(a)
Find the mean temperature in °F.
(2)
(b)
Find the standard deviation of the temperatures in °F.
(2)
(c)
Find the variance of the temperatures in °F.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
y = 1.8 × 18.4 + 32
M1
= 65.12 °F (65.1 °F)
A1
Part (b)
σy = 1.8 × 3.5, the + 32 does not affect the spread
M1
= 6.3 °F
A1
Part (c)
6.32 = 39.69 (°F)2 (39.7), ft their (b)
B1
Common error: Adding 32 to the standard deviation. Adding a constant moves every temperature by the same amount, so it changes the mean but not the spread.
Figure 1 shows the masses, in kg, of the 15 suitcases checked in for a small flight.
(a)
Find the median mass.
(1)
(b)
Find the lower quartile and the upper quartile.
(2)
(c)
Find the interquartile range and the range.
(1)
(d)
Given that Σx = 444, find the mean mass.
(1)
(e)
One more suitcase, of mass 26 kg, is then checked in. Find the median mass of the 16 suitcases.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
15/2 = 7.5, so the 8th value: median = 27 kg
B1
Part (b)
15/4 = 3.75, so Q1 is the 4th value = 20 kg
B1
3 × 15/4 = 11.25, so Q3 is the 12th value = 38 kg
B1
Part (c)
IQR = 38 − 20 = 18 kg and range = 50 − 15 = 35 kg, ft their quartiles
B1
Part (d)
444 ÷ 15 = 29.6 kg
B1
Part (e)
16/2 = 8: mean of the 8th and 9th values, (26 + 27) ÷ 2 = 26.5 kg
B1
Exam tip: A stem-and-leaf diagram already lists the data in order, so count leaves to reach the position you need. Read the key before writing any value.
Figure 1 shows how many minutes late each of 10 trains arrived at a station one morning. A train that arrived on time is shown at 0.
(a)
Find the median number of minutes late.
(1)
(b)
Find the mean number of minutes late.
(1)
(c)
Find the standard deviation of the number of minutes late.
(2)
(d)
The train that was 10 minutes late had been held by a signal failure. State, with a reason, whether the mean or the median better describes a typical delay that morning.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
mean of the 5th and 6th values: median = 2 minutes
The distances, d km, cycled in one week by each of the 50 members of a cycling club are summarised in Figure 1. The classes are 10 ≤ d < 20, 20 ≤ d < 30, and so on up to 60 ≤ d < 70.
(a)
Write down the modal class.
(1)
(b)
Estimate the mean distance cycled.
(3)
(c)
Use interpolation to estimate the median distance.
position 25; cumulative frequencies 4, 13, 28, so the median is in 30–40: 30 + (25 − 13) ÷ 15 × 10
M1
= 38 km
A1
Exam tip: For grouped continuous data the median is at position n/2, here the 25th value, and interpolation is done across the class boundaries 30 and 40.
The masses of 120 eggs, each recorded to the nearest gram, are summarised below.
Mass (g)
45–49
50–54
55–59
60–64
65–74
Frequency
12
31
40
26
11
(a)
Write down the class boundaries and the class width of the class 65–74.
(1)
(b)
Identify the modal class, giving a reason.
(1)
(c)
Use interpolation to estimate the median mass.
(2)
(d)
Estimate the interquartile range of the masses.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
boundaries 64.5 g and 74.5 g, width 10 g
B1
Part (b)
55–59: it has the highest frequency density, 40 ÷ 5 = 8 (the last class has density 1.1)
B1
Part (c)
position 60; cumulative frequencies 12, 43, 83, so the median is in 55–59: 54.5 + (60 − 43) ÷ 40 × 5
M1
= 56.6 g (56.625)
A1
Part (d)
Q1: position 30, in 50–54: 49.5 + (30 − 12) ÷ 31 × 5
M1
Q1 = 52.40…
A1
Q3: position 90, in 60–64: 59.5 + (90 − 83) ÷ 26 × 5 = 60.84…
A1
IQR ≈ 8.44 g
A1
Model sketch
Figure: histogram of the egg masses
Common error: Interpolating from the stated limits (55 and 59) instead of the class boundaries (54.5 and 59.5). Masses to the nearest gram from 54.5 up to 59.5 are all recorded as 55–59.
The times, t minutes, taken by 200 runners to complete a 10 km road race are summarised below.
Time, t (min)
35 ≤ t < 40
40 ≤ t < 45
45 ≤ t < 50
50 ≤ t < 55
55 ≤ t < 65
65 ≤ t < 80
Frequency
8
26
54
62
38
12
Figure: histogram of the race times
(a)
Use interpolation to estimate the median time.
(2)
(b)
Estimate the 10th and the 90th percentiles.
(3)
(c)
Hence find the 10th to 90th interpercentile range.
(1)
(d)
The fastest 5% of runners receive a medal. Estimate the time a runner needed to beat to receive a medal.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
position 100; cumulative frequencies 8, 34, 88, 150, so 50 + (100 − 88) ÷ 62 × 5
M1
= 51.0 minutes (50.97…)
A1
Part (b)
P10: position 20, 40 + (20 − 8) ÷ 26 × 5 or P90: position 180, 55 + (180 − 150) ÷ 38 × 10
M1
P10 = 42.3 minutes (42.30…)
A1
P90 = 62.9 minutes (62.89…)
A1
Part (c)
62.89… − 42.30… = 20.6 minutes, ft their (b)
B1
Part (d)
5% of 200 = 10th runner: 40 + (10 − 8) ÷ 26 × 5
M1
= 40.4 minutes (40.38…)
A1
Exam tip: The 5th, 10th and 90th percentiles use the same interpolation as the median: find the position r × n ÷ 100 first, then the class it falls in.
Common error: Decoding with the wrong operation order: the coding subtracts 400 and then divides by 25, so decoding multiplies by 25 and then adds 400. The standard deviation is only multiplied by 25.
Two groups of students sat the same test. Group P had 24 students, with mean mark 58 and standard deviation 9. Group Q had 16 students, with mean mark 66 and standard deviation 7.
(a)
Show that, for the marks x of group P, Σx2 = 82 680.
(2)
(b)
Find the mean mark of all 40 students.
(2)
(c)
Find the standard deviation of the marks of all 40 students.
(3)
(d)
Explain why the standard deviation of all 40 marks is larger than the standard deviation of either group.
the two group means are different (58 and 66), so combining the groups adds extra spread between them
B1
Common error: Averaging the two means, (58 + 66) ÷ 2 = 62. The groups are different sizes, so the larger group must count more: add the totals and divide by 40.
Two machines, A and B, fill bags of flour labelled “500 g”. For a sample of bags from machine A, the mean mass was 502.0 g and the standard deviation was 3.2 g. For a sample of 20 bags from machine B, the masses x grams gave Σx = 10 010 and Σx2 = 5 010 420.
(a)
Find the mean and the standard deviation of the masses of the bags from machine B.
(3)
(b)
Compare the masses of the bags filled by the two machines.
(2)
(c)
Suggest, with a reason, which machine is more likely to produce bags with a mass below 500 g.
machine A’s bags have a higher mean mass (502.0 g against 500.5 g), so they are heavier on average
B1
machine A has a smaller standard deviation (3.2 g against 4.56 g), so its masses are more consistent
B1
Part (c)
machine B: its mean is closer to 500 g and its masses are more spread out
B1
Exam tip: Each comparison needs the statistic, the direction and what it means in context. “Machine A has a smaller standard deviation, so its masses are more consistent” does all three.
σ2 = 3397 ÷ 80 − 6.22 = 42.4625 − 38.44 = 4.0225, so σ ≈ 2.01 cm
A1
Part (d)
position 40; cumulative frequencies 8, 22, 44, so the median is in 5–6: 5 + (40 − 22) ÷ 22 × 1
M1
= 5.82 cm (5.818…)
A1
Common error: Taking the bar heights as the frequencies. In a histogram the area of a bar gives the frequency, so multiply each frequency density by its class width.
The speeds, v km h−1, of 160 cars passing a speed camera were recorded. Figure 1 shows the cumulative frequency polygon, with the points plotted at the upper class boundaries.
(a)
Use Figure 1, or interpolation, to estimate the median speed.
(2)
(b)
Estimate the interquartile range of the speeds.
(3)
(c)
Drivers whose speed is in the fastest 10% are sent a warning letter. Estimate the lowest speed for which a letter is sent.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
position 80, which lies in 50–60: 50 + (80 − 40) ÷ 52 × 10
Exam tip: Reading across from a cumulative frequency and down to the speed axis is the same as linear interpolation in the class, because the polygon joins the plotted points with straight lines.
A consumer group measured the life, in hours, of a sample of batteries of each of two brands, P and Q, in the same toy. Figure 1 shows box plots of the results.
(a)
Write down the median and find the interquartile range for each brand.
(2)
(b)
Compare the battery lives of the two brands.
(2)
(c)
A customer wants the brand for which the greater proportion of batteries last at least 24 hours. Using Figure 1, state which brand this is, giving a reason.
(1)
(d)
The lives of the brand P batteries are converted into minutes. Find the median and the interquartile range in minutes.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
P: median 28 h, IQR = 32 − 24 = 8 h
B1
Q: median 30 h, IQR = 38 − 22 = 16 h
B1
Part (b)
brand Q has the higher median (30 h against 28 h), so its batteries tend to last slightly longer
B1
brand Q has the larger IQR (16 h against 8 h), so its battery lives are much less consistent
B1
Part (c)
brand P: its lower quartile is 24 h, so about 75% of P batteries last at least 24 h; for Q, 24 h lies above the lower quartile (22 h), so fewer than 75% of Q batteries do
B1
Part (d)
median = 28 × 60 = 1680 minutes
B1
IQR = 8 × 60 = 480 minutes, ft their (a)
B1
Exam tip: A comparison needs one statement about location and one about spread, each with the statistic, the direction and its meaning in context.
A bakery weighs 40 loaves, recording each mass, x grams, to the nearest 5 g. The masses are coded using y = x − 800⁄5. Figure 1 shows the coded masses.
(a)
Find the mean of y.
(2)
(b)
Find the standard deviation of y.
(2)
(c)
Hence find the mean and the standard deviation of the masses of the loaves.
(3)
(d)
Loaves with a recorded mass of less than 795 g are sold at a reduced price. Find how many of the 40 loaves are sold at a reduced price.
The resting pulse rates, in beats per minute (bpm), of 11 athletes and 13 non-athletes were measured. Figure 1 shows the results.
(a)
Find the median and the quartiles of the pulse rates of the athletes.
(3)
(b)
Find the median and the interquartile range of the pulse rates of the non-athletes.
(2)
(c)
Compare the pulse rates of the two groups.
(2)
(d)
One more athlete, with a pulse rate of 59 bpm, is measured. Find the median pulse rate of the 12 athletes.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
median = 6th value = 61 bpm
B1
11/4 = 2.75, so Q1 = 3rd value = 55 bpm
B1
8.25, so Q3 = 9th value = 67 bpm
B1
Part (b)
median = 7th value = 72 bpm
B1
Q1 = 4th value = 67, Q3 = 10th value = 81, so IQR = 14 bpm (allow Q1 = 65.5, Q3 = 82.5, IQR = 17 from the (n + 1)/4 rule)
B1
Part (c)
the athletes have the lower median (61 bpm against 72 bpm), so their resting pulse rates are generally lower
B1
the athletes have the smaller IQR (12 bpm against 14 bpm), so their rates are less spread out, ft their IQRs
B1
Part (d)
mean of the 6th and 7th values: (59 + 61) ÷ 2 = 60 bpm
B1
Common error: Reading the left-hand leaves in the wrong direction. On the athletes’ side the leaves get larger moving away from the stem, so 7 3 1 1 | 6 means 61, 61, 63 and 67.
A driving school recorded the number of attempts, x, that each of its 60 students needed to pass a test. The mean number of attempts was 2.3.
Attempts, x
1
2
3
4
5
Frequency
18
p
12
q
4
(a)
Show that p = 20 and q = 6.
(4)
(b)
Find the standard deviation of the number of attempts.
(3)
(c)
Find the median and the interquartile range.
(3)
(d)
The school charges a registration fee of $10 plus $15 for each attempt. Find the mean and the standard deviation of the amount paid by the 60 students.
The waiting times, x minutes, of 14 patients at a clinic one morning were 6 8 9 11 12 12 14 15 17 18 20 22 25 91 For these data Σx = 280 and Σx2 = 11 414.
(a)
Find the median and the interquartile range of the waiting times.
(3)
(b)
Find the mean and the standard deviation of the waiting times.
(3)
(c)
The patient who waited 91 minutes had arrived without an appointment, and their time is removed. Find the mean and the standard deviation of the remaining 13 times.
(4)
(d)
Find the median of the remaining 13 times.
(1)
(e)
Using your answers, explain which pair of statistics, median and interquartile range or mean and standard deviation, better summarises the original 14 waiting times.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
14/2 = 7, so the median is the mean of the 7th and 8th values = 14.5 minutes
B1
Q1: 14/4 = 3.5 → 4th value = 11; Q3: 10.5 → 11th value = 20
M1
IQR = 9 minutes
A1
Part (b)
mean = 280 ÷ 14 = 20 minutes
B1
σ2 = 11 414 ÷ 14 − 202 = 815.29 − 400 = 415.29
M1
σ = 20.4 minutes
A1
Part (c)
Σx = 280 − 91 = 189 and Σx2 = 11 414 − 912 = 3133
M1
mean = 189 ÷ 13 = 14.5 minutes (14.53…)
A1
σ2 = 3133 ÷ 13 − (189 ÷ 13)2 = 29.63…
M1
σ = 5.44 minutes
A1
Part (d)
13/2 = 6.5 → 7th value = 14 minutes
B1
Part (e)
removing one value changed the mean from 20 to 14.5 and the s.d. from 20.4 to 5.44, but the median only from 14.5 to 14
B1
so the median and IQR are better: they are not distorted by the extreme value of 91
B1
Common error: Calling 20 minutes a typical wait. Only four of the 14 patients waited 20 minutes or more; the mean is pulled up by the single value of 91.
The times, t minutes, taken by 150 people to complete a crossword are summarised in the histogram in Figure 1. No scale is shown on the frequency density axis. The bar for 15 ≤ t < 20 represents 30 people.
(a)
Find the frequency of each class.
(3)
(b)
Estimate the mean time.
(2)
(c)
Given that Σft2 = 105 675, estimate the standard deviation of the times.
(2)
(d)
Use interpolation to estimate the median time.
(2)
(e)
Estimate the interquartile range of the times.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
the bar for 15–20 covers 5 × 10 = 50 grid squares, so one square represents 30 ÷ 50 = 0.6 people
Exam tip: With no scale on the vertical axis, use the bar you are told about to find how many people one grid square represents. Frequency is proportional to area, so count squares, not heights.
Figure 1 shows the scores, x, of the 12 golfers in group 1 for the first round of a competition. The scores are coded using y = x − 72.
(a)
Show that Σy = 10 and Σy2 = 116.
(2)
(b)
Hence find the mean and the standard deviation of the scores in group 1.
(3)
(c)
The 8 golfers in group 2 had a mean score of 74.5 and a standard deviation of 2.5. Find the mean and the standard deviation of the scores of all 20 golfers.
(4)
(d)
Player E’s score included a 2-stroke penalty that was given in error, and it is corrected to 78. Find the corrected mean and standard deviation of the scores of all 20 golfers.
Common error: Combining the groups with Σy2 instead of Σx2. Coded sums from different codings cannot be added; rebuild each group’s Σx2 from Σx2 = n(σ2 + x2) first.
Figure 1 shows the number of visitors, x, to a museum on each of 12 consecutive days. Each point lies on a gridline intersection. The values are coded using y = x − 300⁄10.
(a)
Find Σy and Σy2.
(2)
(b)
Hence find the mean and the standard deviation of the number of visitors per day.
(4)
(c)
Find the median number of visitors per day.
(2)
(d)
The number of visitors on the 13th day is then recorded. The mean for the 13 days is 325. Find the number of visitors on the 13th day.
(2)
(e)
Find the standard deviation of the number of visitors for the 13 days.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
y values −2, 1, 5, −4, 0, 3, 11, −1, 2, −3, 4, 8
M1
Σy = 24 and Σy2 = 270
A1
Part (b)
y = 24 ÷ 12 = 2
B1
σy2 = 270 ÷ 12 − 22 = 18.5
M1
x = 10 × 2 + 300 = 320
A1
σx = 10 × √18.5 = 43.0
A1
Part (c)
in order 260, 270, 280, 290, 300, 310, 320, …; the median is the mean of the 6th and 7th values
M1
(310 + 320) ÷ 2 = 315
A1
Part (d)
13 × 325 = 4225 and the first 12 days total 12 × 320 = 3840
M1
4225 − 3840 = 385 visitors
A1
Part (e)
y = 8.5 for day 13: Σy = 32.5 and Σy2 = 270 + 72.25 = 342.25
Exam tip: Coding with y = (x − 300) ÷ 10 turns three-digit values into small whole numbers. The mean decodes as 10y + 300, but the standard deviation is only multiplied by 10.
A factory inspects 40 batches made by shift A and 40 batches made by shift B, and records x, the number of faulty items in each batch. Figure 1 shows the results.
(a)
Find the mean and the standard deviation of the number of faulty items per batch for shift A.
(4)
(b)
Find the mean and the standard deviation for shift B.
(3)
(c)
Find the median for each shift.
(2)
(d)
Compare the number of faulty items per batch for the two shifts.
(2)
(e)
Find the mean and the standard deviation of the number of faulty items per batch for all 80 batches.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
Σfx = 0 + 12 + 20 + 21 + 16 + 5 = 74
M1
mean = 74 ÷ 40 = 1.85
A1
Σfx2 = 0 + 12 + 40 + 63 + 64 + 25 = 204
M1
σ2 = 204 ÷ 40 − 1.852 = 1.6775, so σ = 1.30
A1
Part (b)
Σfx = 56, so mean = 1.4
B1
Σfx2 = 0 + 10 + 28 + 45 + 48 + 25 = 156
M1
σ2 = 3.9 − 1.96 = 1.94, so σ = 1.39
A1
Part (c)
A: cumulative frequencies 6, 18, 28; the 20th and 21st values are both 2, so median = 2
B1
B: cumulative frequencies 14, 24; the 20th and 21st values are both 1, so median = 1
B1
Part (d)
shift B has fewer faulty items on average (mean 1.4 against 1.85, median 1 against 2)
B1
the spreads are similar (standard deviations 1.39 and 1.30)
Exam tip: To combine the shifts, add the totals Σfx and Σfx² for the two groups and use n = 80. Averaging the two standard deviations does not give the standard deviation of the combined data.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy
EssentialNeeded for the site and your cookie choice to work.
Always on
AnalyticsAnonymous visit statistics that help improve the free resources.
MarketingMeasures whether ads reach the right families. Off unless you turn it on.