Data Presentation and Interpretation

This lesson covers methods of presenting and interpreting data as required by the Edexcel A-Level Mathematics specification (9MA0), Paper 3 Section A -- Statistics. You must be able to construct and interpret box plots, cumulative frequency diagrams, histograms (including those with unequal class widths), stem-and-leaf diagrams, and compare distributions.

---

Box Plots (Box-and-Whisker Diagrams)

A box plot displays the five-number summary of a data set:

1. Minimum value (or lowest value that is not an outlier)
2. Lower quartile (Q1) -- the value below which 25% of the data falls
3. Median (Q2) -- the middle value (50th percentile)
4. Upper quartile (Q3) -- the value below which 75% of the data falls
5. Maximum value (or highest value that is not an outlier)

Structure of a box plot

• A box is drawn from Q1 to Q3. The length of the box is the interquartile range (IQR) = Q3 - Q1.
• A line inside the box marks the median.
• Whiskers extend from the box to the smallest and largest values within the non-outlier range.
• Outliers are plotted as individual points (crosses or dots) beyond the whiskers.

Identifying outliers

An outlier is typically defined as any value that lies:

• Below Q1 - 1.5 x IQR, or
• Above Q3 + 1.5 x IQR

Exam Tip: Always state the outlier rule you are using. If the question says "an outlier is defined as a value more than 1.5 x IQR beyond the nearest quartile", use that definition exactly.

Interpreting box plots

• The position of the median within the box tells you about the symmetry of the distribution. If the median is in the centre of the box, the distribution is approximately symmetrical. If the median is closer to Q1, the distribution has a positive (right) skew. If the median is closer to Q3, the distribution has a negative (left) skew.
• A long whisker on one side indicates the data is more spread out in that direction.
• A narrow box means the middle 50% of the data is tightly clustered.

---

Cumulative Frequency Diagrams

A cumulative frequency diagram shows the running total of frequencies up to each class boundary.

How to draw a cumulative frequency diagram

1. Create a cumulative frequency table: for each class, add the frequency of that class to the total of all previous classes.
2. Plot each cumulative frequency against the upper class boundary of its class.
3. Join the points with a smooth curve (or straight line segments).

Reading values from a cumulative frequency diagram

• Median: read across from the point where cumulative frequency = n/2 (where n is the total frequency).
• Lower quartile (Q1): read across from cumulative frequency = n/4.
• Upper quartile (Q3): read across from cumulative frequency = 3n/4.
• Interquartile range: IQR = Q3 - Q1.
• Percentiles: the p-th percentile is found at cumulative frequency = (p/100) x n.

Example

Weight (kg)	Frequency	Cumulative Frequency
50 ≤ w < 55	8	8
55 ≤ w < 60	15	23
60 ≤ w < 65	22	45
65 ≤ w < 70	18	63
70 ≤ w < 75	12	75
75 ≤ w < 80	5	80

Total n = 80. Median is at cumulative frequency 40. This falls in the 60 ≤ w < 65 class.

Exam Tip: Always plot cumulative frequency against the upper class boundary, never the midpoint.

---

Histograms (Including Unequal Class Widths)

A histogram is a bar chart where the area of each bar represents the frequency. This is particularly important when the classes have unequal widths.

Frequency density

The vertical axis of a histogram is frequency density, calculated by:

Frequency density = Frequency / Class width

Why frequency density matters

If class widths are equal, you can use frequency directly. But when class widths are unequal, using frequency directly would be misleading because a wider class would appear to have more data simply because its bar is wider. By using frequency density, the area of each bar equals the frequency, making the chart fair.

Example

Time (minutes)	Frequency	Class Width	Frequency Density
0 ≤ t < 5	10	5	10/5 = 2.0
5 ≤ t < 10	15	5	15/5 = 3.0
10 ≤ t < 20	30	10	30/10 = 3.0
20 ≤ t < 40	24	20	24/20 = 1.2
40 ≤ t < 60	6	20	6/20 = 0.3

To find the frequency from a histogram: Frequency = Frequency density x Class width (i.e. the area of the bar).

Reading a histogram

• The tallest bar does not necessarily represent the highest frequency -- always check the width.
• To find the frequency, calculate the area of each bar (height x width).
• The modal class is the class with the highest frequency density.

Exam Tip: A very common exam question gives you a histogram and asks you to find the frequency for a particular class, or to complete a frequency table. Use Area = Frequency density x Class width.

---

Stem-and-Leaf Diagrams

A stem-and-leaf diagram displays data by splitting each value into a stem (the leading digit(s)) and a leaf (the final digit).

Rules for stem-and-leaf diagrams

• Leaves must be written in ascending order on each row.
• There must be a key to explain what the stem and leaf represent (e.g. "3 | 5 means 35").
• Each leaf represents exactly one data value.

Example

Data: 23, 25, 27, 31, 34, 34, 36, 38, 41, 43, 45, 52

Stem	Leaf
2	3 5 7
3	1 4 4 6 8
4	1 3 5
5	2

Key: 2 | 3 means 23.

Back-to-back stem-and-leaf diagrams

Used to compare two data sets. One data set has leaves extending to the left of the stem; the other extends to the right. Leaves are still in ascending order (moving away from the stem in each direction).

Advantages

• Shows the shape of the distribution.
• Retains the actual data values (unlike histograms and box plots).
• Easy to find the median and quartiles by counting.

---

Comparing Distributions

When comparing two (or more) data sets, you should comment on:

1. A measure of central tendency (location)

Compare the means or medians. State which data set has a higher/lower average and what this means in context.

Example: "The median score for Class A (67) is higher than the median score for Class B (52), suggesting students in Class A performed better on average."

2. A measure of spread (dispersion)

Compare the ranges, interquartile ranges, or standard deviations. State which data set is more or less spread out.

Example: "The IQR for Class A (15) is smaller than the IQR for Class B (28), indicating that scores in Class A are more consistent."

3. Skewness (if appropriate)

Comment on the shape of the distribution. A distribution may be:

• Symmetrical: mean approximately equals median, box plot symmetric.
• Positively skewed (right skew): mean > median, longer right tail / whisker.
• Negatively skewed (left skew): mean < median, longer left tail / whisker.

Comparing with box plots

When given two box plots side by side, compare:

• Medians (which is higher?)
• IQRs (which is more spread?)
• Overall range (which has wider spread?)
• Skewness (symmetric, positive, or negative?)
• Outliers (does one data set have more outliers?)

Exam Tip: Always make comparisons in context. Do not just write "the median is higher". Write "the median height of boys (172 cm) is greater than the median height of girls (163 cm), suggesting boys tend to be taller."

---

Cleaning Data

Before analysing data, it is important to consider:

• Errors in the data (e.g. a height recorded as 17.2 m instead of 1.72 m).
• Missing values -- how should they be handled?
• Outliers -- are they genuine extreme values or errors?

At A-Level, you are expected to identify anomalies and comment on their possible impact on statistical measures.

---

Summary

• Box plots display the five-number summary and are useful for comparing distributions and identifying outliers (Q1 - 1.5 x IQR to Q3 + 1.5 x IQR).
• Cumulative frequency diagrams allow you to estimate the median, quartiles and percentiles by reading off the graph.
• Histograms use frequency density on the y-axis; area of each bar = frequency. This is essential when class widths are unequal.
• Stem-and-leaf diagrams retain actual data values and show the distribution shape. Back-to-back versions compare two data sets.
• When comparing distributions, always comment on a measure of location, a measure of spread, and (if appropriate) skewness -- and always in context.

A-Level Deep Dive: Data Presentation and Interpretation

Spec mapping

Edexcel 9MA0-03 specification, Paper 3 — Statistics and Mechanics, Section 2 (Data presentation and interpretation) covers interpret diagrams for single-variable data, including histograms with unequal class widths, frequency polygons, box-and-whisker plots and cumulative-frequency diagrams; identify outliers from a data set using a stated rule; compare distributions using appropriate measures of central tendency and spread (refer to the official specification document for exact wording). This sub-strand is examined alongside Section 1 (Statistical sampling) and Section 3 (Probability), with cross-links into Section 4 (Statistical distributions) and Section 5 (Hypothesis testing). The Edexcel formula booklet does not list the outlier rule $Q_1 - 1.5 \times \text{IQR}$Q1​−1.5×IQR / $Q_3 + 1.5 \times \text{IQR}$Q3​+1.5×IQR — it must be memorised, and the question stem will normally state it explicitly because alternative rules (such as $\pm 2\sigma$±2σ from the mean) are also acceptable in different contexts.

---

Worked example with full mark scheme

Question (8 marks):

The mass, $m$m grams, of 80 apples from an orchard is summarised by the histogram below (described): class boundaries $80 \le m < 100$80≤m<100 (frequency density $0.4$0.4), $100 \le m < 110$100≤m<110 ($1.6$1.6), $110 \le m < 120$110≤m<120 ($2.4$2.4), $120 \le m < 140$120≤m<140 ($0.8$0.8), $140 \le m < 200$140≤m<200 ($0.1$0.1), with an additional class $60 \le m < 80$60≤m<80 (frequency $10$10).

(a) Estimate the number of apples with mass less than $115$115 g. (3)

(b) The lower quartile is $Q_1 = 105$Q1​=105 g and the upper quartile is $Q_3 = 125$Q3​=125 g. An outlier is defined as any value more than $1.5 \times \text{IQR}$1.5×IQR below $Q_1$Q1​ or above $Q_3$Q3​. Determine the boundaries beyond which an apple's mass would be classified as an outlier, and state, with reasoning, whether the heaviest class $140 \le m < 200$140≤m<200 contains any potential outliers. (5)

Solution with mark scheme:

(a) Step 1 — recover frequencies from frequency densities.

Frequency $=$= frequency density $\times$× class width:

• $80 \le m < 100$80≤m<100: $0.4 \times 20 = 8$0.4×20=8
• $100 \le m < 110$100≤m<110: $1.6 \times 10 = 16$1.6×10=16
• $110 \le m < 120$110≤m<120: $2.4 \times 10 = 24$2.4×10=24
• $120 \le m < 140$120≤m<140: $0.8 \times 20 = 16$0.8×20=16
• $140 \le m < 200$140≤m<200: $0.1 \times 60 = 6$0.1×60=6

Together with the stated class $60 \le m < 80$60≤m<80 (frequency $10$10), the total is $10 + 8 + 16 + 24 + 16 + 6 = 80$10+8+16+24+16+6=80, matching the sample size.

M1 — multiplying frequency density by class width, not reading frequency directly off the vertical axis. The single most common error on histogram questions is treating frequency density as frequency.

Step 2 — use linear interpolation across the class containing $115$115.

The class $110 \le m < 120$110≤m<120 contains 24 apples. Assuming uniform distribution within the class, the proportion below $115$115 is $\dfrac{115 - 110}{120 - 110} = 0.5$120−110115−110​=0.5, contributing $0.5 \times 24 = 12$0.5×24=12 apples.

M1 — correct linear interpolation set-up.

Step 3 — sum cumulative frequency below $115$115.

$10 + 8 + 16 + 12 = 46$10+8+16+12=46 apples.

A1 — fully correct cumulative count.

(b) Step 1 — compute IQR.

$\text{IQR} = Q_3 - Q_1 = 125 - 105 = 20$IQR=Q3​−Q1​=125−105=20.

B1 — correct IQR.

Step 2 — compute outlier boundaries.

Lower boundary: $Q_1 - 1.5 \times \text{IQR} = 105 - 30 = 75$Q1​−1.5×IQR=105−30=75 g.

Upper boundary: $Q_3 + 1.5 \times \text{IQR} = 125 + 30 = 155$Q3​+1.5×IQR=125+30=155 g.

M1 A1 — M1 for the $1.5 \times \text{IQR}$1.5×IQR structure, A1 for both numerical boundaries.

Step 3 — interpret in context.

The class $140 \le m < 200$140≤m<200 has 6 apples. Any apple with mass $> 155$>155 g is a potential upper outlier. The class spans $140$140 to $200$200, so apples with mass between $155$155 and $200$200 are outliers. Assuming uniform distribution within this class, the expected number of outliers is $\dfrac{200 - 155}{200 - 140} \times 6 = \dfrac{45}{60} \times 6 = 4.5$200−140200−155​×6=6045​×6=4.5, so we estimate between 4 and 5 outliers in this class.

M1 A1 — M1 for comparing class boundaries against $155$155, A1 for stating that the class does contain outliers with a contextual estimate.

Total: 8 marks (M5 A2 B1, split as shown).

---

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): A box-and-whisker plot summarises the daily rainfall ($r$r mm) in two locations, $A$A and $B$B, over the same 60-day period.

• Location $A$A: minimum $0$0, $Q_1 = 2$Q1​=2, median $5$5, $Q_3 = 9$Q3​=9, maximum $22$22.
• Location $B$B: minimum $1$1, $Q_1 = 4$Q1​=4, median $6$6, $Q_3 = 8$Q3​=8, maximum $14$14.

(a) For each location, determine which (if any) data values would be classified as outliers using the rule $1.5 \times \text{IQR}$1.5×IQR beyond the quartiles. (3)

(b) Compare the two distributions, referring to both centre and spread. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — $\text{IQR}_A = 7$IQRA​=7, $\text{IQR}_B = 4$IQRB​=4; upper fence $A = 9 + 10.5 = 19.5$A=9+10.5=19.5, upper fence $B = 8 + 6 = 14$B=8+6=14.
• A1 (AO1.1b) — Location $A$A has an outlier at the maximum ($22 > 19.5$22>19.5); lower fence $A = 2 - 10.5 = -8.5$A=2−10.5=−8.5, no lower outlier.
• A1 (AO2.4) — Location $B$B maximum is $14 =$14= upper fence, so by the strict inequality "more than", $B$B has no outliers (or by the inclusive form, exactly one borderline value).

(b)

• B1 (AO2.2b) — comparing centre: medians $5$5 vs $6$6, so location $B$B has slightly higher typical rainfall.
• B1 (AO2.2b) — comparing spread: $\text{IQR}_A = 7$IQRA​=7 vs $\text{IQR}_B = 4$IQRB​=4, so location $A$A is more variable.
• B1 (AO3.5a) — interpreting in context: location $A$A experiences more extreme rainfall events (presence of an outlier; larger range $22$22 vs $13$13), suggesting weather patterns at $A$A are less stable than at $B$B.

Total: 6 marks split AO1 = 1, AO2 = 4, AO3 = 1. Note the AO2 dominance — comparing distributions is interpretive work, and Edexcel rewards explicit linking of statistical measures to contextual claims.

---

Synoptic links

Connects to:

1. Section 1 — Measures of location and spread: medians, quartiles and IQR feed directly into both box plots and the outlier rule. The choice between mean/SD and median/IQR pairs depends on whether the distribution is symmetric or skewed — a decision that recurs throughout descriptive statistics and motivates robust estimation.

2. Section 4 — Correlation and regression: scatter plots and residual plots are the bivariate analogues of histograms and box plots. Identifying influential points in regression uses an outlier-style rule (residuals more than $2$2 standard deviations from $0$0), structurally identical to the $1.5 \times \text{IQR}$1.5×IQR rule for univariate data.