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A Level Maths 9709 · Paper 5 Probability & Statistics 1 · 5.1 · ~5 min read

Representation of data

Stem-and-leaf diagrams, box-and-whisker plots, histograms with unequal class widths and cumulative frequency graphs; median, quartiles and IQR; mean and standard deviation from raw, grouped and coded data, and choosing the right measures.

Stem-and-leafBox plotsHistograms & frequency densityCumulative frequencyMedian & quartilesMean & standard deviationCoded data Σ(x − a)Comparing distributions
RSby Riyaz Sir
mentorriyaz.in
Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Histogram: frequency density = frequency ÷ class width. The area of each bar is the frequency, so bars of unequal width are fair.
  • Standard deviation measures spread about the mean: σ² = Σx²/n − x̄².
  • Coded data: if you know Σ(x − a) and Σ(x − a)², the mean is a + Σ(x − a)/n and the standard deviation is unchanged by the shift.
  • Choose measures to suit the data: median and IQR resist outliers and skew; mean and standard deviation use every value.

By the end you can

  • Draw and interpret stem-and-leaf diagrams, box plots, histograms and cumulative frequency graphs.
  • Find and interpret the median, quartiles and interquartile range.
  • Calculate the mean and standard deviation from raw data, grouped data and Σ(x − a) totals.
  • Choose suitable measures and compare data sets in context.

1Diagrams

Histograms

time (min)frequency density1020304050123450f = 12f = 18f = 22f = 16f = 12Frequency density = frequency ÷ class width, so the AREA of each bar is the frequency.
Unequal class widths: height is frequency density, so each bar's area is its frequency.
  • Class boundaries matter. For "time to the nearest minute, 10–14", the boundaries are 9.5 and 14.5, so the width is 5.
  • Reading a histogram: frequency = frequency density × width. For part of a class, assume values are spread evenly.
Worked example

A histogram has classes 0–10, 10–15, 15–20, 20–30 and 30–50 minutes with frequencies 12, 18, 22, 16, 12. Find the frequency densities, and estimate how many values exceed 25.

  1. densities: 12/10 = 1.2, 18/5 = 3.6, 22/5 = 4.4, 16/10 = 1.6, 12/20 = 0.6 M1 A1
  2. above 25: half of 20–30 gives 8, plus 12 M1
  3. 20 A1
Common mistakePlotting frequency as height for unequal widths. The 2024 Paper 5 report found this was still the most common histogram error.

Box plots and cumulative frequency

time (min)cum. freq.0102030405020406080≈ 26≈ 19≈ 35n = 80: read acrossfrom 40 (median),20 (Q₁) and 60 (Q₃)
Read across at n/2 for the median, n/4 and 3n/4 for the quartiles.
0510152025min 3Q₁ 6median 9Q₃ 13.5max 18IQR = 13.5 − 6 = 7.5. Outliers lie beyond Q₁ − 1.5 × IQR or Q₃ + 1.5 × IQR.
A box plot shows the median, quartiles and extremes at a glance.
  • Stem-and-leaf keeps the raw data and needs a key (e.g. 3 ǀ 7 means 37). A back-to-back diagram compares two sets.
  • Box plots compare two sets quickly: compare medians (average) and IQRs (spread).
  • Skew: if the median is nearer Q₁ than Q₃, the data are positively skewed.

2Measures

Mean, median, standard deviation

x̄ = Σx/n = Σfx/Σf ; σ² = Σx²/n − x̄² = Σfx²/Σf − x̄² FORMULA LIST (MF19)
grouped data use mid-points: the result is an estimate
x̄ = a + Σ(x − a)/n ; σ² = Σ(x − a)²/n − [Σ(x − a)/n]² LEARN IT
coding: subtracting a moves the mean but not the spread
Worked example

For the histogram data above, estimate the mean and the median.

  1. Σfx = 5 × 12 + 12.5 × 18 + 17.5 × 22 + 25 × 16 + 40 × 12 = 1550; n = 80 M1
  2. mean ≈ 19.4 minutes A1
  3. median is the 40th value: 30 values below 15, 22 in 15–20 M1
  4. median ≈ 15 + (10/22) × 5 = 17.3 minutes A1
Worked example

For 20 values, Σ(x − 50) = −36 and Σ(x − 50)² = 1462. Find the mean and standard deviation.

  1. mean = 50 + (−36/20) = 48.2 B1
  2. σ² = 1462/20 − (−1.8)² = 73.1 − 3.24 = 69.86 M1
  3. σ = 8.36 A1
Worked example

Two groups: 12 values with mean 15 and Σx² = 2900; 8 values with mean 20 and Σx² = 3400. Find the mean and standard deviation of all 20 values.

  1. Σx = 180 + 160 = 340 ⇒ mean 17 M1 A1
  2. Σx² = 6300 ⇒ σ² = 6300/20 − 17² = 315 − 289 = 26 M1
  3. σ = 5.10 A1
Common mistakeForgetting to square the mean, or using Σ(x − a)² as if it were Σx². Also, comparing data sets with no context: "the median time is greater for group A, so they usually took longer".
Linking questionProbability from data begins in Probability; spread and the normal model are in The normal distribution.

Test yourself — cover the answers first

Quick check

  1. Find the frequency density for a class of width 4 containing 18 values.
  2. What does the area of a histogram bar represent?
  3. Which measures are best for skewed data?
  4. Σx = 60, n = 12, Σx² = 400. Find the variance.
  5. What does a key like 4 ǀ 2 mean in a stem-and-leaf diagram?

Exam-style

  1. 10 values have Σ(x − 3) = 12 and Σ(x − 3)² = 50. Find the mean and the standard deviation.
  2. The heights in a class of width 20 cm have frequency density 1.5 per cm. How many values are in the class?
  3. Data set A: median 24, IQR 6. Data set B: median 31, IQR 15. Compare.
Show answers
  1. 4.5
  2. frequency
  3. median and IQR
  4. 400/12 − 25 = 8.33
  5. 42 (with the stated units)
  6. mean 4.2; σ² = 5 − 1.44 = 3.56 ⇒ σ = 1.89
  7. 30
  8. B is on average larger and much more spread out than A
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.