Statistics: Data Analysis and Representation | IGCSE Mathematics (0580) Extended Revision Notes

Introduction

Alright, let's talk about stats. Before you scroll away, this isn't just about boring charts. Statistics is the secret language behind everything you do online – from your Spotify Wrapped playlist to your gaming stats and even how Instagram decides what to show you on your feed. It's the ultimate life hack for making sense of a chaotic world. In this chapter, we're going to become data detectives. First, we'll figure out how to find a 'typical' value in a sea of numbers with Averages and Spread (is your average screen time really that high?). Then, we'll learn how to present that data without putting everyone to sleep using powerful Data Representation techniques. We'll even see if there's a connection between two things, like hours spent studying and your exam grade, using Scatter Diagrams. Finally, we'll level up with Cumulative Frequency graphs to see where you rank in a group and master Histograms, the pro version of bar charts. Let's get it!

Measures of Central Tendency and SpreadE9.3

Alright, let's break down 'Averages and Spread'. Think of it as getting the main character energy and the overall vibe of a set of data. First up, the measures of central tendency – these are your 'averages', which tell you the typical value. The Mean is the one you know best: add everything up and divide by how many there are ($ \frac{\sum x}{n} $). It's like finding the average score for your squad in a game. But be careful! The mean is super sensitive to outliers (extreme values). If one friend gets a god-tier score, it can pull the whole team's average up, making it seem like everyone did better than they actually did. Next, the Median. This is the literal middle value when you line up all your data in order. If you have an even number of data points, you just find the mean of the two middle ones. The median doesn't care about outliers; it's the chill, unbothered center. It’s a better representation when your data has some wild high or low values. Finally, the Mode is the easiest of all: it's just the value that shows up the most. Like the one song that's on repeat on everyone's Spotify playlist. A dataset can have one mode, more than one, or no mode at all. Now for the measures of spread – this is all about how consistent or spread out your data is. The Range is the simplest: just subtract the smallest value from the biggest. It tells you the total spread, but like the mean, it's easily skewed by one random high or low value. A much better measure is the Interquartile Range (IQR). This sounds complex, but it's just the range of the middle 50% of your data. First, find the median (which is also the second quartile, $Q2$). Then, find the median of the lower half of the data – that's your lower quartile, $Q1$. The median of the upper half is your upper quartile, $Q3$. The IQR is simply $Q3 - Q1$. A diagram showing a sorted list of numbers, with Q1, Median (Q2), and Q3 marked, and a bracket indicating the IQR between Q1 and Q3. The IQR ignores the extreme highs and lows, giving you a solid idea of where the 'main bulk' of your data lies. A small IQR means your data is consistent (like getting similar lap times in a race), while a large IQR means it's all over the place. When data is grouped into a frequency table, we can't find the exact mean, but we can estimate it. You find the midpoint of each group, multiply it by the frequency, sum those up, and divide by the total frequency. The Modal Class is even easier – it's just the group with the highest frequency. Easy win!

Estimating Mean and Identifying Modal Class from Grouped DataE9.3

Problem The table shows the time (in minutes) 50 students spent on their part-time job shifts one weekend. Estimate the mean time and identify the modal class. | Time, $t$ (hours) | Frequency, $f$ | |---|---| | $0 < t \le 2$ | 4 | | $2 < t \le 4$ | 15 | | $4 < t \le 6$ | 22 | | $6 < t \le 8$ | 9 |

Step 1 First, let's find the modal class. This is the easiest mark you'll get! Just look for the class interval with the highest frequency. The highest frequency is 22, which corresponds to the class $4 < t \le 6$. *Modal Class = $4 < t \le 6$ hours*

Step 2 To estimate the mean, we need a single value to represent each class. We use the midpoint ($x$) of each interval. Let's add columns for the midpoint ($x$) and for frequency times midpoint ($f \times x$). Midpoints: For $0 < t \le 2$, midpoint is $(0+2)/2 = 1$ For $2 < t \le 4$, midpoint is $(2+4)/2 = 3$ For $4 < t \le 6$, midpoint is $(4+6)/2 = 5$ For $6 < t \le 8$, midpoint is $(6+8)/2 = 7$

Step 3 Now, calculate the $f \times x$ column by multiplying the frequency of each class by its midpoint. This weights each midpoint by how many students are in that group. $f \times x$ values: $4 \times 1 = 4$ $15 \times 3 = 45$ $22 \times 5 = 110$ $9 \times 7 = 63$

Step 4 Time to find the totals. We need the sum of the frequencies, $\sum f$, (which should be 50, as stated in the problem) and the sum of the $f \times x$ column, $\sum(f \times x)$. $$\sum f = 4 + 15 + 22 + 9 = 50$$ $$\sum(f \times x) = 4 + 45 + 110 + 63 = 222$$

Step 5 Finally, we can estimate the mean using the formula. Divide the total of the $f \times x$ column by the total frequency. $$\text{Estimated Mean} = \frac{\sum(f \times x)}{\sum f} = \frac{222}{50} = 4.44$$ So, the estimated mean time spent on a shift is 4.44 hours.

Data Representation: Charts and DiagramsE9.3

Alright, let's talk about data. It's literally everywhere – from your Spotify Wrapped stats to your K/D ratio in a game. Raw data is just a massive list of numbers, which is kinda boring and hard to understand. That's where data representation comes in. It's all about giving that data a visual 'glow up' so we can see patterns and trends instantly. Think of it as turning a messy spreadsheet into a fire Instagram infographic. First up, the basics: a simple frequency distribution is just a fancy name for a table that organizes your data. It lists the items or groups and their frequency (how many times they appear). From there, we can build some awesome charts. • Bar Charts: You know these. They use bars to show frequencies of different categories, like the most popular phone brand in your year group. But we can level them up! A dual bar chart places bars side-by-side to compare two sets of data, like comparing your screen time on TikTok vs. Snapchat each day. A dual bar chart comparing the number of likes on posts for two different social media accounts over a week. A composite (or stacked) bar chart stacks the data on top of each other in a single bar. This is perfect for showing how a total is broken down. Imagine a bar showing your total part-time job earnings for a month, stacked to show how much came from weekday vs. weekend shifts. • Pie Charts: These are all about showing proportions of a whole, like slicing up a pizza. Each slice represents a category, and the size of the slice shows what percentage of the total it is. To find the angle for each slice, you use the formula: Angle = $(\frac{\text{frequency}}{\text{total frequency}}) \times 360^\circ$. Super useful for things like showing the breakdown of your monthly spending. • Stem-and-Leaf Diagrams: These are a super clever way to show the shape of your data and keep all the original numbers. It's like a bar chart on its side. The 'stem' is the first part of the number (e.g., the tens digit) and the 'leaf' is the last digit. Two rules you absolutely CANNOT forget: 1) The leaves must be ordered (smallest to largest). 2) You MUST include a key to explain what it means (e.g., Key: 7 | 2 means 72). Forgetting the key is a classic exam mistake, so don't let that be you! An ordered stem-and-leaf diagram showing the scores of 20 students on a test, with a clear key below it.

Worked Example: Stem-and-Leaf Diagrams and Composite Bar ChartsE9.3

Problem A survey records the number of hours 12 students spent driving in one month after getting their license. The results were: 15, 8, 21, 34, 16, 25, 9, 15, 22, 30, 11, 21 a) Construct an ordered stem-and-leaf diagram for this data. b) The students were from two different driving schools: 'AutoDrive' and 'GoFast'. The data is split as follows: • AutoDrive: 15, 21, 9, 15, 30, 11 • GoFast: 8, 34, 16, 25, 22, 21 Group the data into classes 0-9, 10-19, 20-29, 30-39 and draw a composite bar chart to compare the two schools.

Step 1 First, let's create the ordered stem-and-leaf diagram for all 12 students. The 'stem' will be the tens digit (0, 1, 2, 3) and the 'leaf' will be the units digit. It's best to do a rough, unordered version first, then write out the neat, ordered final version. Unordered: 0 | 8 9 1 | 5 6 5 1 2 | 1 5 2 1 3 | 4 0 Now, let's put the leaves in order for our final answer: $$ \begin{array}{c|l} \text{Stem} & \text{Leaf} \\ \hline 0 & 8 \quad 9 \\ 1 & 1 \quad 5 \quad 5 \quad 6 \\ 2 & 1 \quad 1 \quad 2 \quad 5 \\ 3 & 0 \quad 4 \end{array} $$

Step 2 Crucial step! Don't forget the key. Without it, the diagram is meaningless. This tells us how to read the numbers. Key: 1 | 5 means 15 hours

Step 3 For part (b), we need to organize the data for the two driving schools into a frequency table using the given classes. We just count how many students from each school fall into each time bracket. Frequency Table: | Hours | AutoDrive | GoFast | |---------|-----------|--------| | 0-9 | 1 | 1 | | 10-19 | 3 | 1 | | 20-29 | 1 | 2 | | 30-39 | 1 | 1 |

Step 4 Now we plan the composite bar chart. The horizontal axis (x-axis) will be our 'Hours' classes. The vertical axis (y-axis) will be the 'Frequency' (number of students). Each bar will represent an hour bracket, and we'll stack the frequencies for 'AutoDrive' and 'GoFast' on top of each other. We also need a legend to show which color represents which school. Plan: - X-axis: Classes (0-9, 10-19, 20-29, 30-39) - Y-axis: Frequency (from 0 up to at least 4, as the max height for a stack is 3+1=4 for the 10-19 bar) - Legend: One color/pattern for AutoDrive, another for GoFast.

Step 5 Let's describe the final chart. For the '10-19' class, you'd draw the 'AutoDrive' bar up to a frequency of 3, and then stack the 'GoFast' bar (with a height of 1) on top of it, making the total bar height 4. You do this for all four classes. Remember to give your chart a title, like 'Driving Hours for Different Schools'. A completed composite bar chart with 'Hours' on the x-axis and 'Frequency' on the y-axis. Each of the four bars is stacked to show the number of students from AutoDrive and GoFast, with a clear legend distinguishing the two schools.

Bivariate Data and Scatter DiagramsE9.3

Alright, let's talk about Bivariate Data. It sounds complex, but 'bi' just means two. So, we're looking at the relationship between two different variables at once. Think: does the number of hours you spend revising affect your exam score? To see this, we use a scatter diagram. It's a graph where you plot points (as little 'x's) for your two variables. Each 'x' represents one piece of data, like one student's revision time and their score. Once plotted, you look for the vibe between them, which we call correlation. There are three types: • Positive Correlation: As one variable increases, the other tends to increase too. Like, the more hours you work, the more money you earn. The points trend upwards. A scatter diagram showing points in a clear upward trend from left to right, indicating positive correlation. • Negative Correlation: The opposite. As one goes up, the other goes down. For example, the more data you use streaming shows, the less battery life you have. The points trend downwards. A scatter diagram showing points in a clear downward trend from left to right, indicating negative correlation. • Zero Correlation: No connection at all. The points are scattered randomly. Think shoe size vs. your gaming score – they're unrelated. A scatter diagram with randomly scattered points, indicating zero correlation. If there is a correlation, we can draw a line of best fit. This is a single, straight line you draw with a ruler that goes through the middle of your data cloud. It's drawn by eye, aiming to have roughly the same number of points above and below the line along its whole length. This line is awesome for making predictions, like estimating a score for a certain amount of revision time.

Worked Example: Revision Time vs. Test ScoresE9.3

Problem The table below shows the number of hours 10 students spent revising for a maths test and the score they achieved. Hours (h): 1, 2, 2.5, 4, 4.5, 5, 6, 7, 7.5, 9 Score (%): 45, 50, 62, 65, 71, 74, 80, 88, 85, 95 (a) Draw a scatter diagram for this data. (b) Describe the correlation. (c) Draw a line of best fit. (d) Use your line to estimate the score for a student who revised for 3.5 hours.

Step 1 First, set up your axes. Put 'Hours Revising (h)' on the horizontal x-axis and 'Test Score (%)' on the vertical y-axis. Make sure your scales are sensible and cover all the data values. Then, carefully plot each pair of values as a small 'x'. For example, the first point is at (1, 45). A scatter diagram with correctly labelled axes and all 10 points plotted as 'x's.

Step 2 Look at the pattern of the points. They generally go up from left to right. This means as the number of hours revising increases, the test score also tends to increase. So, we have a strong positive correlation. The data shows a strong, positive correlation.

Step 3 Now, take a clear ruler and draw a single straight line that goes through the middle of the points. You want to 'balance' the points, so there are roughly an equal number of 'x's on either side of the line along its entire length. It doesn't have to pass through any of the points exactly. The same scatter diagram as before, but now with a straight line of best fit ruled through the points.

Step 4 To estimate the score for 3.5 hours of revision, find 3.5 on the horizontal (hours) axis. From there, draw a vertical line up until you hit your line of best fit. Then, draw a horizontal line across to the vertical (score) axis and read the value. Your answer may vary slightly depending on your line, but it should be reasonable. Find $h=3.5$ on the x-axis. Follow the line up to the line of best fit, then across to the y-axis. The estimated score is approximately 64%. (Note: Your answer might be between 62% and 66% and still be correct).

Cumulative Frequency Tables and DiagramsE9.6

Alright, let's talk about Cumulative Frequency. The name sounds way more complicated than it is, I promise! Think of it like your XP in a video game or your follower count on Insta – it’s a running total that only ever goes up. 'Cumulative' just means 'how much so far?'. First, we build a cumulative frequency table. You start with a normal grouped frequency table (like 'time spent gaming') and add a new column. For the first group, the cumulative frequency is just its frequency. For the next group, you add its frequency to the previous cumulative total. You keep doing this, adding as you go, until the last entry in your cumulative frequency column equals the total number of data points. Easy. Next, we draw the graph, which has a cool name: an ogive. This is the S-shaped curve you've probably seen. Here’s the key move: you plot the cumulative frequency (y-axis) against the UPPER bound of each class interval (x-axis). This is a classic spot to lose marks, so don't forget it! Plot your points with a little 'x' and then join them with a smooth curve. Your curve should start from the lower bound of the first group at a frequency of 0. An S-shaped cumulative frequency curve showing cumulative frequency on the y-axis and a continuous variable like 'time' on the x-axis. Points are marked with 'x' at the upper boundary of each class interval and joined with a smooth curve. So why bother? Because this graph is a powerhouse for estimating values. To find the median (the middle value, or Q2), you find the halfway point on your y-axis ($0.5 \times \text{total frequency}$), read across to your curve, and then drop down to the x-axis. Boom, estimated median. You can do the same for the lower quartile (Q1) at the 25% mark ($0.25 \times \text{total}$) and the upper quartile (Q3) at the 75% mark ($0.75 \times \text{total}$). The interquartile range (IQR) is just $Q3 - Q1$, showing the spread of the middle 50% of your data. You can even find any percentile – wanna know the top 10%? Find the value at the 90% mark! This is super useful for seeing where you rank in test scores or how your part-time job hours compare to everyone else's.

Worked Example: Analysing Streaming TimesE9.3

Problem The table shows the time, $t$ (in hours), that 120 students spent streaming music in a week. Use this data to estimate the median, the interquartile range, and the number of students who streamed for more than 13 hours.

Step 1 First, we need to create a cumulative frequency table. We'll add a new column and create a running total of the frequencies. Remember to use the upper bounds for plotting later! Here is the completed table: Time ($t$ hours) | Frequency | Cumulative Frequency ----------------|-----------|--------------------- $0 < t \le 4$ | 12 | 12 $4 < t \le 8$ | 30 | 42 (12 + 30) $8 < t \le 12$ | 48 | 90 (42 + 48) $12 < t \le 16$ | 22 | 112 (90 + 22) $16 < t \le 20$ | 8 | 120 (112 + 8)

Step 2 To find the median (Q2), we find the position of the middle value. The total frequency is 120, so the median is at the ($120 \div 2 = 60$)th position. On your graph, you would go to 60 on the cumulative frequency axis, read across to the curve, and then down to the time axis. $$\text{Median position} = \frac{1}{2} \times 120 = 60$$ Reading from the graph, this gives an estimated median time of approximately 9.8 hours.

Step 3 Now for the quartiles. The lower quartile (Q1) is at the 25% mark and the upper quartile (Q3) is at the 75% mark. We find their positions and read the corresponding times from our graph. $$\text{Q1 position} = \frac{1}{4} \times 120 = 30$$ From the graph, Q1 is approx. 7.0 hours. $$\text{Q3 position} = \frac{3}{4} \times 120 = 90$$ From the graph, Q3 is exactly 12 hours.

Step 4 The interquartile range (IQR) is the difference between the upper and lower quartiles. It tells us the spread of the middle 50% of the students' streaming times. $$\text{IQR} = Q3 - Q1 = 12 - 7.0 = 5.0 \text{ hours}$$

Step 5 Finally, to find the number of students who streamed for more than 13 hours, we first find how many streamed for 13 hours or less. Go to 13 on the time axis, read up to the curve, and then across to the cumulative frequency axis. Then, we subtract this from the total number of students. From the graph, the number of students streaming up to 13 hours is approx. 96. $$\text{Number of students streaming > 13 hours} = \text{Total} - (\text{Number} \le 13) = 120 - 96 = 24$$ So, an estimated 24 students are the super-streamers!

Histograms and Frequency DensityE9.7

Alright, let's talk histograms. You've probably seen bar charts since you were a kid, right? They're great for separate, distinct categories, like your favourite artists on Spotify. But what about continuous data, like the exact time it takes you to get to school? That's where histograms come in. They look like bar charts, but the bars are all squished together because the data flows from one category to the next. Now, here's the plot twist. Sometimes, the groups (or 'class intervals') aren't the same width. Maybe one group is for people who game 0-2 hours a week, but the next is for 2-6 hours. If we just make the bar's height the frequency, the 2-6 hours bar would be twice as wide and look way more important, which is totally misleading. It's like comparing a 2-hour movie with a 30-minute episode of a show; you can't just compare them directly. To make it a fair comparison, we use a secret weapon called Frequency Density. This is the absolute key. Instead of the y-axis being 'Frequency', on a proper histogram, it's always 'Frequency Density'. The formula is your new best friend: $$ \text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}} $$ This genius move standardizes everything. It means that the area of each bar ($ \text{height} \times \text{width} $, or $ \text{Frequency Density} \times \text{Class Width} $) is what actually represents the frequency. This is the golden rule! So, when you look at a histogram, don't just look at how high the bars are; look at their total area to understand the real story of the data. A side-by-side comparison of two graphs. The left graph is an incorrect histogram with unequal class widths where the y-axis is frequency, making a wide bar look disproportionately large. The right graph is the correct histogram of the same data, wit

Worked Example: Calculating Frequency Density and Interpreting a HistogramE9.7

Problem The table shows the time, $t$ (in hours), that 80 students spent on a part-time job in one week. We need to calculate the frequency densities and draw a histogram for this data. | Time, $t$ (hours) | Frequency | |---|---| | $0 < t \le 10$ | 15 | | $10 < t \le 15$ | 25 | | $15 < t \le 20$ | 24 | | $20 < t \le 30$ | 16 |

Step 1 First up, we need to figure out the 'Class Width' for each group. This is just the difference between the upper and lower bounds of the time interval. For $0 < t \le 10$, Class Width = $10 - 0 = 10$ For $10 < t \le 15$, Class Width = $15 - 10 = 5$ For $15 < t \le 20$, Class Width = $20 - 15 = 5$ For $20 < t \le 30$, Class Width = $30 - 20 = 10$

Step 2 Now we can calculate the Frequency Density for each group using our key formula: $\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}$. Let's get these values so we know how high to draw our bars. For $0 < t \le 10$: F.D. = $\frac{15}{10} = 1.5$ For $10 < t \le 15$: F.D. = $\frac{25}{5} = 5.0$ For $15 < t \le 20$: F.D. = $\frac{24}{5} = 4.8$ For $20 < t \le 30$: F.D. = $\frac{16}{10} = 1.6$

Step 3 Time to draw! Set up your axes. The x-axis will be 'Time, $t$ (hours)' and it needs to go up to 30. The y-axis is the crucial one: label it 'Frequency Density'. Based on our calculations, it needs to go up to at least 5. [This step describes the drawing process, which would result in a visual histogram. No calculation here, just the setup for drawing.]

Step 4 Now, draw the bars. The first bar goes from 0 to 10 on the x-axis and has a height of 1.5. The second bar goes from 10 to 15 and has a height of 5.0. Keep going for all the groups. Remember, no gaps between bars! Bar 1: Width (0 to 10), Height 1.5 Bar 2: Width (10 to 15), Height 5.0 Bar 3: Width (15 to 20), Height 4.8 Bar 4: Width (20 to 30), Height 1.6

Step 5 Let's check our work by re-calculating a frequency from the histogram. What's the frequency for the $10 < t \le 15$ group? We find the area of that bar. This confirms that area = frequency. $$ \text{Frequency} = \text{Frequency Density} \times \text{Class Width} = 5.0 \times (15 - 10) = 5.0 \times 5 = 25 $$It matches the table! We nailed it. ✨

Classifying and Tabulating Statistical DataE9.3

Before you can calculate a single average, you need to organise the data. Raw numbers are like an unsorted Spotify playlist — useless until you clean it up. Two key distinctions to drill in: • Qualitative vs quantitative. Qualitative = non-numerical (colours, favourite artist, sport). Quantitative = numerical. • Discrete vs continuous. Discrete = specific separate values: number of siblings, shoe size, goals scored. Continuous = any value in a range: height, weight, time. (If you can 'count' it, it's discrete. If you 'measure' it, usually continuous.) ⚖️ Tally and frequency tables are the simplest move. For small discrete data, list each value, tally it, total it up: | Shoe size | Tally | Frequency | | :---: | :---: | :---: | | 5 | ||| | 3 | | 6 | |||| | 4 | | 7 | || | 2 | Grouped frequency tables are for continuous data or wide ranges. Choose equal class intervals that cover the whole range with no gaps and no overlaps. For continuous data, use $\leq$: e.g. $0 \leq h < 10$, $10 \leq h < 20$. The value $10$ goes in the second class, not both. Boundary rules matter. 🎯 Two-way tables handle two variables at once — e.g. gender × sport — with row and column totals. Great for conditional probability questions later. A side-by-side comparison. Left: a tally-frequency table for 'number of pets' with 5 rows. Right: a two-way table with rows 'male/female/total' and columns 'football/tennis/swimming/total', with sample cells filled.

Worked Example: Building a Grouped Frequency TableE9.3

Problem The heights (cm) of $20$ students are: $152, 161, 148, 170, 165, 158, 172, 155, 163, 168, 150, 174, 159, 166, 151, 162, 169, 157, 164, 171$. Build a grouped frequency table using intervals of width $5$ starting at $145 \leq h < 150$. 📏

Step 1 Spot the range so you know how many classes are needed. Min = $148$, Max = $174$. $$\text{Range} = 174 - 148 = 26$$

Step 2 Set up classes of width $5$ starting at $145$: that's $145\text{–}150,\ 150\text{–}155,\ 155\text{–}160,\ 160\text{–}165,\ 165\text{–}170,\ 170\text{–}175$. Six classes will cover everyone. $$6 \text{ classes}$$

Step 3 Tally each value into its class. Watch the boundary rule: $150$ goes into $150 \leq h < 155$, not the class below. $$145\text{–}150\!: 1,\quad 150\text{–}155\!: 3,\quad 155\text{–}160\!: 4,\quad 160\text{–}165\!: 5,\quad 165\text{–}170\!: 4,\quad 170\text{–}175\!: 3$$

Step 4 Sanity check — frequencies add up to the total count of values. If not, you lost someone. 🕵️ $$1 + 3 + 4 + 5 + 4 + 3 = 20 \text{ ✅}$$

Interpreting Statistical Data and Comparing DistributionsE9.3

Drawing a graph or table is only half the job — examiners want you to actually read it and compare. The magic formula for comparison questions: use one measure of centre (mean, median, or mode) AND one measure of spread (range, IQR). Every time. 💯 Template sentence shape: > The mean score for Class A is higher than Class B, so on average Class A performed better. The range for Class A is smaller than Class B, so Class A's scores are more consistent. Two sentences. One about centre, one about spread. That's a full-mark answer. Caveats that earn bonus credibility: • Tiny sample sizes are unreliable — call it out if $n$ is small. • Outliers mess with the mean; median is often more robust (the median doesn't care about that one weird data point). • Correlation ≠ causation. Two things moving together doesn't mean one causes the other. (Ice-cream sales and shark attacks are correlated — both peak in summer. Ice cream doesn't attract sharks. Probably.) 🦈🍦 • A bar chart with an axis not starting at zero can exaggerate differences. Sneaky. Two side-by-side dot plots or box plots titled 'Class A' and 'Class B' showing clearly different means and different spreads, with arrow annotations reading 'higher mean', 'smaller spread'. Reading charts: pie chart angles are fractions of the whole — $\text{fraction} = \dfrac{\text{angle}}{360°}$. Stem-and-leaf diagrams must have a key (otherwise you have no idea what the numbers mean). Histograms with unequal widths use frequency density for height, not frequency itself. 🎯

Worked Example: Comparing Two Data SetsE9.3

Problem The times (in minutes) taken by $5$ students from two schools to complete the same puzzle:\\ School A: $12, 14, 13, 15, 16$\\ School B: $8, 18, 10, 20, 14$.\\ Compare the performance of the two schools using the mean and the range. 🧩

Step 1 Mean for each = sum ÷ count. Do both. $$\bar{x}_A = \frac{12+14+13+15+16}{5} = \frac{70}{5} = 14 \text{ min}$$ $$\bar{x}_B = \frac{8+18+10+20+14}{5} = \frac{70}{5} = 14 \text{ min}$$

Step 2 Range = biggest − smallest. Do both. $$\text{Range}_A = 16 - 12 = 4 \text{ min}$$ $$\text{Range}_B = 20 - 8 = 12 \text{ min}$$

Step 3 Write the comparison using BOTH measures. Centre + spread = full marks. 🎯 $$\text{On average both schools took the same time (mean = 14 min).}$$ $$\text{School A is more consistent: range of 4 min vs 12 min for School B.}$$

Worked Example: Median, Quartiles and IQR from a Cumulative Frequency GraphE9.6

Problem A cumulative frequency graph has been drawn for the test scores of $80$ students. Reading from the graph: the cumulative frequency reaches $20$ at score $42$, reaches $40$ at score $58$, and reaches $60$ at score $71$. Estimate (a) the median, (b) the lower quartile, (c) the upper quartile, (d) the interquartile range. 📊

Step 1 For $n = 80$ students, the key positions on the cumulative frequency axis are: median at $\tfrac{n}{2} = 40$, lower quartile at $\tfrac{n}{4} = 20$, upper quartile at $\tfrac{3n}{4} = 60$. $$\text{Median position: } 40$$ $$Q_1 \text{ position: } 20$$ $$Q_3 \text{ position: } 60$$

Step 2 Read across from each position to the curve, then down to the score axis. The problem already gave us those lookups. $$\text{Median} \approx 58$$ $$Q_1 \approx 42$$ $$Q_3 \approx 71$$

Step 3 Interquartile range = $Q_3 - Q_1$. That's the spread of the middle $50\%$ of the data. $$\text{IQR} = 71 - 42 = 29$$

Worked Example: Histogram with Unequal Class Widths — Finding FrequencyE9.7

Problem A histogram has class intervals of unequal widths. One class has a width of $5$ and a frequency density of $3.2$. Find the frequency of that class. (Formula: $\text{frequency density} = \dfrac{\text{frequency}}{\text{class width}}$.) 📈

Step 1 The formula works both ways — rearrange to make frequency the subject. $$\text{frequency} = \text{frequency density} \times \text{class width}$$

Step 2 Sub in the values. $$\text{frequency} = 3.2 \times 5 = 16$$

Step 3 Sanity check: frequency density is literally 'how many students per unit of class width' — multiplying back by the width unwinds it. Classic example of why histograms with unequal widths use density, not raw frequency, so the area of each bar represents the frequency. Area = frequency = $16$. ✅ $$\text{Frequency of the class} = 16$$

Worked Example: Scatter Diagram and Line of Best FitE9.3

Problem A scatter diagram shows hours of revision ($x$) and test scores ($y$) for $12$ students. A line of best fit is ruled through the points. The line passes through $(2, 35)$ and $(8, 80)$. (a) Describe the correlation. (b) Find the equation of the line of best fit. (c) Use it to estimate the score of a student who revised for $5$ hours. 📉➡️📈

Step 1 As revision hours go up, scores go up — points trend upward on the scatter. That's positive correlation. Fill in the descriptor. $$\text{Positive correlation}$$

Step 2 For (b), find the gradient using the two given points on the line. $$m = \frac{80 - 35}{8 - 2} = \frac{45}{6} = 7.5$$

Step 3 Sub a known point into $y = mx + c$ and solve for $c$. $$35 = 7.5(2) + c \implies c = 35 - 15 = 20$$ $$y = 7.5x + 20$$

Step 4 For (c), substitute $x = 5$ into the line equation. $$y = 7.5(5) + 20 = 37.5 + 20 = 57.5$$ $$\text{Estimated score} \approx 58$$

Key Points to Remember

Averages (mean, median, mode) give you a central value, while spread (range, IQR) tells you how consistent the data is. Don't forget the formulas for grouped data!
Effective data representation is about choosing the right tool for the job. Always label your axes, include a title, and, if needed, a key.
Bivariate data involves two variables. Use a scatter diagram to identify the type of correlation (positive, negative, or no correlation) and draw a line of best fit to make predictions.
Cumulative frequency graphs (ogives) are S-shaped curves that help you find the median ($Q_2$) and quartiles ($Q_1$, $Q_3$). The interquartile range is $IQR = Q_3 - Q_1$.
For histograms with unequal class widths, the area of the bar represents the frequency. The vertical axis is Frequency Density, calculated as $Frequency Density = \frac{Frequency}{Class Width}$.

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