Edexcel IAL Maths
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Mean, median and mode for raw lists, discrete frequency tables and grouped data
Quartiles and percentiles of a list, the range and the interquartile range
Variance and standard deviation, including from Σx, Σx² and Σfx, Σfx²
Estimating the mean and standard deviation of grouped data from midpoints
Interpolation for the median, quartiles and percentiles of grouped data, with class boundaries
Coding: how adding, subtracting, multiplying and dividing change the mean and standard deviation
Comparing data sets; combining groups; adding, removing or correcting a value
1. A data set is summarised by a measure of location (mean, median or mode) and a measure of spread (range, IQR or standard deviation), and two data sets are compared using one of each, in context.
2. Mean x̄=Σx÷n (or Σfx÷Σf); median = middle value; mode = most common value (modal class for grouped data).
3. Lists: for the median and quartiles find n/2, n/4, 3n/4. Not a whole number → round up; a whole number k → average the kth and (k+1)th values.
4. Grouped data: interpolate. Median = lower boundary + (n/2-cf before)/(class frequency) × class width.
5. Variance σ²=Σx²/n-x̄² ("mean of the squares minus square of the mean"); σ=√(variance).
6. Coding y=(x-a)/b: ȳ=(x̄-a)/b and σ_y=σₓ/b. Shifting does not change the spread.
Measure of location.
A single value that represents the centre, or a typical value, of a set of data. x̄=Σx/n x̄=Σfx/Σf where f is the frequency of the value x (for grouped data, x is the class midpoint) and Σf=n.
The median is the middle value when the data are in order; for n values it is the (n+1)/2th value, which is the same as the rule "n/2: round up, or average the two values either side if n/2 is a whole number". The mode is the value with the highest frequency; a data set can have no mode or several. For grouped data quote the modal class (highest frequency density if the widths differ).
The three behave differently. The mean uses every value, so one extreme value can drag it a long way. The median and mode do not care how extreme the largest or smallest values are. That difference decides which average to use.
WORKED EXAMPLE · QUESTION
A courier records the number of parcels, x, delivered to one office on each of 40 days.
| Parcels, x | 0 | 1 | 2 | 3 | 4 | 5 |
| Frequency, f | 6 | 11 | 9 | 7 | 4 | 3 |
Find (a) the mode, (b) the median, (c) the mean number of parcels per day.
SOLUTION
(a) The highest frequency is 11, so mode = 1.
(b) n/2=20 is a whole number, so average the 20th and 21st values. Cumulative frequencies are 6, 17, 26, …, so the 18th to 26th values are all 2: median = 2.
(c) Σfx=0×6+1×11+2×9+3×7+4×4+5×3=81, so x̄=(81/40)= 2.025 parcels per day. The mean need not be a possible data value.
Quartiles Q₁, Q₂, Q₃ are the values a quarter, a half and three quarters of the way through the ordered data. Positions in a list of n ordered values: Q₁: n/4, Q₂: n/2, Q₃: 3n/4, rth percentile: rn/100. Not a whole number → round up and take that value; a whole number k → the mean of the kth and (k+1)th values.
Three measures of spread come from these positions. The range (largest − smallest) is quick but decided entirely by the two most extreme values. The interquartile range IQR=Q₃-Q₁ is the spread of the middle half. An interpercentile range, such as P₉₀-P₁₀, is the spread of the middle 80%. Always put the data in order first; with a frequency table, use cumulative frequencies to find a position.
WORKED EXAMPLE · QUESTION
The numbers of minutes that 15 trains arrived late were 0, 1, 1, 2, 3, 3, 4, 5, 6, 6, 8, 9, 12, 15, 27. (a) Find the median and the interquartile range. (b) Find the 10th and 90th percentiles. (c) A 16th train arrives 20 minutes late. Find the new median.
SOLUTION
(a) n=15. Median: 15/2=7.5→ 8th value; Q₁: 15/4=3.75→ 4th; Q₃: 45/4=11.25→ 12th. Median = 5, Q₁=2, Q₃=9, so median 5 min, IQR 7 min.
(b) 10×15/100=1.5→ 2nd value;
90×15/100=13.5→ 14th value: P₁₀=1, P₉₀=15.
(c) n=16, 16/2=8 is a whole number, so average the 8th and 9th values of the new list: (5+6)/2= 5.5 min.
Common error
Taking the 4th value as Q₁ for 16 values. 16/4=4 is a whole number, so Q₁ is the mean of the 4th and 5th values; in part (c) that is (2+3)÷2=2.5.
Standard deviation.
A measure of the typical distance of the values from their mean, in the same units as the data. With divisor n: σ²=(Σ(x-x̄)²)/n=Σx²/n-x̄²=Sₓₓ/n, σ=√(σ²) For frequency tables σ²=Σfx²/Σf-x̄², and Sₓₓ=Σx²-(Σx)²/n.
The first form explains the idea: measure each value's deviation from the mean, square it so that negatives do not cancel, and average. The second form is the one you calculate with. Variance is in squared units (years², cm²); the standard deviation is back in the units of the data.
The order matters: Σx²÷n first, then subtract x̄².
A variance can never be negative. A negative answer means a slip, usually (Σx)² in place of Σx².
Keep x̄ at full accuracy before squaring it; rounding first can change the third significant figure of σ.
Adding a value equal to the mean leaves the mean unchanged and makes the standard deviation smaller (the same total spread shared among more values).
Using your calculator
In statistics mode, enter the data (with a frequency column if needed) and read off x̄, σₓ, Σx and Σx². Use σₓ (divisor n), not sₓ.
Still write the sums down: if a later part adds or corrects a value, you need them.
WORKED EXAMPLE · QUESTION
The ages, x years, of the 25 members of a choir give Σx=1150 and Σx²=54 610. (a) Find the mean and the standard deviation of the ages. (b) A new member aged 46 joins. Find the new mean and standard deviation.
SOLUTION
(a) x̄=(1150/25)=46 years. σ²=54 610/25-46²=2184.4-2116=68.4, so mean 46 years, σ=8.27 years (3 s.f.).
(b) Update the sums: Σx=1196, Σx²=54 610+46²=56 726, n=26. x̄=(1196/26)=46; σ²=56 726/26-46²=65.8, so mean 46 years, σ=8.11 years.
Check by Sₓₓ: 54 610-1150²/25=1710 and 1710÷25=68.4. The new value adds 0 to Sₓₓ but 1 to n.
In grouped data the individual values are lost, so assume every value in a class sits at the class midpoint, then use Σfx÷Σf and Σfx²÷Σf-x̄². The answers are estimates; say so. The midpoint is halfway between the class boundaries: for 0≤t<10 it is 5; for ages "20–29" in whole years the boundaries are 20 and 30, midpoint 25; for lengths "20–29" to the nearest mm the boundaries are 19.5 and 29.5, midpoint 24.5.
WORKED EXAMPLE · QUESTION
The journey times, t minutes, of 70 commuters are summarised below. Estimate the mean and the standard deviation of the journey times.
| Time, t (min) | Frequency | Midpoint | Cumulative frequency |
|---|---|---|---|
| 0≤t<10 | 8 | 5 | 8 |
| 10≤t<15 | 14 | 12.5 | 22 |
| 15≤t<20 | 21 | 17.5 | 43 |
| 20≤t<30 | 17 | 25 | 60 |
| 30≤t<50 | 10 | 40 | 70 |
SOLUTION
Σfx=40+175+367.5+425+400=1407.5, so x̄=(1407.5/70)=20.107…
Σfx²=8×5²+14×12.5²+21×17.5²+17×25²+10×40²=35 443.75
σ²=35 443.75/70-20.107…²=102.04, so mean ≈ 20.1 min, σ ≈ 10.1 min.
Exam tip — add columns
Add a midpoint column and an fx column to the table, and write the totals Σfx and Σfx² in your answer. If the final value slips, those totals still show your method.
Linear interpolation in a grouped frequency table.
value=L+(p-C)/f_c×w p = position (n/2 for the median, n/4 and 3n/4 for the quartiles, rn/100 for the rth percentile); L = lower class boundary of the class containing position p; C = cumulative frequency up to L; f_c = frequency of that class; w = class width.
Interpolation assumes the values are spread evenly through the class, which is why the answer is an estimate. This page uses n/2, not (n+1)/2, for grouped data. The formula is not something to memorise blindly: it simply says "go (p-C)/f_c of the way through the class".
WORKED EXAMPLE · QUESTION
Use the journey-time table in Worked example 4. (a) Estimate the median and the interquartile range. (b) Estimate the 10th to 90th interpercentile range.
SOLUTION
(a) Median position 70/2=35, in 15≤t<20 (cumulative frequencies 22, 43): Q₂=15+(35-22)/21×5=18.1 min.
Q₁: position 17.5, in 10≤t<15: Q₁=10+(17.5-8)/14×5=13.39. Q₃: position 52.5, in 20≤t<30: Q₃=20+(52.5-43)/17×10=25.59. Median 18.1 min, IQR = 25.59 − 13.39 = 12.2 min.
(b) P₁₀: position 7, in 0≤t<10: 0+(7/8)×10=8.75. P₉₀: position 63, in 30≤t<50: 30+(63-60)/10×20=36. Range 36 − 8.75 = 27.25 min.
Common error
Using the stated class limits instead of the boundaries. Lengths recorded to the nearest cm as "30–34" and "35–39" have boundaries 29.5, 34.5, 39.5, so L=34.5 and w=5 for the class 35–39, not L=35 and w=4.
Coding y=(x-a)/b.
ȳ=(x̄-a)/b, σ_y=σₓ/b, σ_y²=σₓ²/b² Reverse it to decode: x̄=bȳ+a and σₓ=bσ_y (for b>0).
Adding or subtracting moves every value by the same amount: the centre moves, the spread does not. Multiplying or dividing scales every distance, so it changes both. The same rules hold for the median and quartiles (the IQR divides by b; subtracting a does not change it).
WORKED EXAMPLE · QUESTION
The daily output, x kg, of a machine is recorded on 20 days. The data are coded using y=(x-200)/5, giving Σy=34 and Σy²=182. (a) Find the mean and standard deviation of y. (b) Hence find the mean and standard deviation of x. (c) Find Σx and Σx².
SOLUTION
(a) ȳ=(34/20)=1.7; σ_y²=(182/20)-1.7²=9.1-2.89=6.21; σ_y=2.4920…
(b) Decode with x=5y+200: x̄=5×1.7+200= 208.5 kg; σₓ=5×2.4920…= 12.5 kg.
(c) Σx=20×208.5= 4170. σₓ²=25×6.21=155.25, so Σx²=20×(155.25+208.5²)= 872 550.
Common error
Adding 200 back on to the standard deviation, or multiplying the variance by 5 instead of 25. Only the scaling factor affects spread: σₓ=5σ_y, σₓ²=25σ_y².
A comparison needs one measure of location and one measure of spread, each in context: "the median time for group A is higher, so group A were generally slower". Use median and IQR when the data contain extreme values or are not symmetrical; use mean and standard deviation when the data are roughly symmetrical with no extreme values, because they use every value.
Combining two groups: never average the two means. Recover each group's Σx=nx̄ and Σx²=n(σ²+x̄²), add, then use the formulae on the totals. For example, 30 values with mean 62 and s.d. 5, and 20 values with mean 70 and s.d. 4: Σx=1860+1400=3260, Σx²=116 070+98 320=214 390, so the combined mean is 3260÷50=65.2 and the combined s.d. is √(214 390÷50-65.2²)=√36.76=6.06.
Correcting a value: if a value p was recorded as q, replace it in both sums: Σx-q+p and Σx²-q²+p².
WORKED EXAMPLE · QUESTION
Two groups of 12 people each completed a puzzle. Their times, in seconds, were
A: 38, 41, 44, 45, 47, 48, 50, 52, 53, 55, 58, 61
B: 29, 33, 36, 40, 42, 46, 49, 55, 60, 64, 70, 84
(a) For group A, find the median and the interquartile range. (b) For group B the median is 47.5 s and the interquartile range is 24 s. Compare the two groups. (c) Find the mean time of all 24 people.
SOLUTION
(a) n=12: 12/2 = 6, 12/4 = 3 and 36/4 = 9 are whole numbers, so average pairs. Median =(48+50)/2=49; Q₁=(44+45)/2=44.5; Q₃=(53+55)/2=54. Median 49 s, IQR 9.5 s.
(b) The medians are similar (49 s and 47.5 s), so a typical person in each group took about the same time. Group B's IQR (24 s) is much larger than group A's (9.5 s), so group B's times were much more varied.
(c) Add the totals, not the means: Σx=592+608=1200, so mean =(1200/24)= 50 s.
Group B contains one very slow time (84 s). The median and IQR are not affected by how extreme it is, which is why they are the better pair to compare here.
Longer questions combine a grouped table, coding for the mean and standard deviation, interpolation for the median and quartiles, and an interpretation. Keep full accuracy in the middle of the working and round only the final answers.
WORKED EXAMPLE · QUESTION
The masses, m grams, of 80 apples from an orchard are summarised in the table.
| Mass, m (g) | 100≤m<120 | 120≤m<130 | 130≤m<140 | 140≤m<150 | 150≤m<170 | 170≤m<200 |
| Frequency | 7 | 15 | 24 | 19 | 11 | 4 |
The midpoints of the classes are coded using y=(m-135)/5. (a) Use the coding to estimate the mean and the standard deviation of the masses. (b) Use interpolation to estimate the median mass. (c) Estimate the interquartile range. (d) Estimate the number of apples whose masses lie within one standard deviation of the mean.
SOLUTION
(a) Midpoints 110, 125, 135, 145, 160, 185 give y=-5,-2,0,2,5,10. Σfy=-35-30+0+38+55+40=68; Σfy²=175+60+0+76+275+400=986. ȳ=(68/80)=0.85; σ_y²=(986/80)-0.85²=12.325-0.7225=11.6025, σ_y=3.4062… Mean =135+5×0.85= 139.25 g (139 g to 3 s.f.); s.d. =5×3.4062…= 17.0 g.
(b) Position 40; cumulative frequencies 7, 22, 46, so the median is in 130≤m<140: Q₂=130+(40-22)/24×10= 137.5 g.
(c) Q₁: position 20, in 120≤m<130:
120+(20-7)/15×10=128.67. Q₃: position 60, in 140≤m<150: 140+(60-46)/19×10=147.37. IQR = 18.7 g.
(d) Mean ± s.d. =139.25±17.03, from 122.22 g to 156.28 g. Interpolate at both ends: (130-122.22)/10×15+24+19+(156.28-150)/20×11=11.67+43+3.45=58.1, so about 58 apples. That is about 73% of 80; for a roughly symmetrical, single-peaked distribution about two thirds usually lie within one standard deviation, so this is reasonable.
Exam tip 1 — write the sums down
Before any variance, write Σx (or Σfx) and Σx² (or Σfx²) as numbers. They are the method, and you need them again if a later part adds, removes or corrects a value.
Exam tip 2 — check the position rule twice
For a list, decide first whether n/4, n/2, 3n/4 are whole numbers: round up if not, average two values if so. For grouped data, use n/2, n/4, 3n/4 directly in the interpolation, with class boundaries.
Exam tip 3 — decode carefully
Write the coding the other way round before using it: y=(x-200)/5 becomes x=5y+200. The mean follows the whole rule; the standard deviation only feels the ×5.
Exam tip 4 — compare in context, with two measures
"Group B has a higher IQR" is not enough on its own. Say which measure, which group, and what it means: "the IQR for B is larger, so B's times are less consistent", paired with a comparison of medians or means.
Exam tip 5 — round only at the end
Keep x̄ at full accuracy when you square it, give final answers to 3 significant figures unless told otherwise, and describe results from grouped data as estimates.
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Variance = (Σx)²÷n-x̄². | (Σx)² is the square of the total, not the total of the squares. | σ²=Σx²÷n-x̄². |
| The mean of two combined groups is the mean of their means. | That only works when the groups are the same size. | Add the totals nx̄ and divide by the combined n. |
| Adding 10 to every value adds 10 to the standard deviation. | Every value moves by the same amount, so distances from the mean are unchanged. | The mean goes up by 10; the s.d. is unchanged. |
| The median of grouped data is the midpoint of the median class. | The median need not be in the middle of its class. | Interpolate: L+(p-C)/f_c×w. |
| The modal class has the highest frequency, whatever the widths. | A wide class can collect many values without being dense. | With unequal widths, compare frequency densities. |
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| The range is the best measure of spread because it is easy. | It depends only on the two most extreme values. | Use the IQR (resistant) or the standard deviation (uses all values). |
S1.1 Representing data
Quartiles and the IQR build box plots and the outlier rule; comparing mean, median and mode, or the quartiles, describes skewness.
S1.4 Correlation and regression
Sₓₓ=Σx²-(Σx)²/n is the same quantity used for the correlation coefficient and the regression line, and coding appears there too.
S1.5 Discrete random variables and S1.6 The Normal distribution
Var(X)=E(X²)-[E(X)]² is the probability version of "mean of squares minus square of mean"; E(aX+b) and Var(aX+b) follow the coding rules.
| Formula | Use | In the formulae list? |
|---|---|---|
| x̄=Σx÷n; x̄=Σfx÷Σf | mean of raw data; of a frequency table (midpoints for grouped data) | No – learn it |
| Q₁, Q₂, Q₃ at n/4, n/2, 3n/4 | quartile positions (round up, or average two values if whole) | No – learn it |
| L+(p-C)/f_c×w | interpolation for grouped data | No – derive it |
| σ²=Σx²÷n-x̄²; σ²=Σfx²÷Σf-x̄² | variance of data (divisor n); σ=√(σ²) | No – learn it |
| Sₓₓ=Σx²-(Σx)²÷n | variance =Sₓₓ÷n | Yes |
| y=(x-a)÷b: ȳ=(x̄-a)÷b, σ_y=σₓ÷b | coding and decoding | No – learn it |
| IQR=Q₃-Q₁; range = max − min | measures of spread | No – learn it |
1. Find the mean and standard deviation of 3, 7, 8, 10, 12.
2. Eight values have Σx=96 and Σx²=1240. Find the mean and standard deviation.
3. Data coded with y=(x-50)÷2 have ȳ=3 and σ_y=1.5. Find x̄ and σₓ.
4. Why is the median usually preferred to the mean for the incomes of people in a town?
5. Lengths are recorded to the nearest cm in classes 10–14, 15–19, 20–24. What are the boundaries and width of the class 15–19?
1. Σx=40, so x̄=8. Σx²=366, so σ²=73.2-64=9.2 and σ=3.03.
2. x̄=12; σ²=155-144=11; σ=3.32.
3. x=2y+50, so x̄=2×3+50=56 and σₓ=2×1.5=3.
4. A few very high incomes pull the mean upwards, so it is not typical of most people. The median is not affected by how extreme those values are.
5. 14.5 and 19.5; width 5. Its midpoint is 17.
Check understanding
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The number of books, x, borrowed in one month by each of 50 members of a library is shown in the table.
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]
| Books, x | 0 | 1 | 2 | 3 | 4 |
| Frequency | 9 | 17 | 12 | 8 | 4 |
Mark scheme click to reveal each step
mode = 1 book
50/2 = 25, so the median is the mean of the 25th and 26th values; cumulative frequencies 9, 26, …
both are 1, so median = 1 book
Σfx = 0 + 17 + 24 + 24 + 16 = 81
mean = 81 ÷ 50 = 1.62 books
The heights, in mm, of 11 seedlings grown in a tray are
23 31 18 27 35 29 22 40 26 33 24
(a) Find the median and the lower and upper quartiles of the heights. [3]
(b) Find the interquartile range and the range. [1]
(c) The height recorded as 40 mm should have been 14 mm. Find the correct median. [2]
Mark scheme click to reveal each step
ordered: 18, 22, 23, 24, 26, 27, 29, 31, 33, 35, 40
median = 6th value = 27 mm
Q₁ = 3rd value = 23 mm, Q₃ = 9th value = 33 mm
IQR = 33 − 23 = 10 mm and range = 40 − 18 = 22 mm, ft their quartiles
new order 14, 18, 22, 23, 24, 26, 27, …; the median is still the 6th value
median = 26 mm
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