Statistics involves collecting, organising, representing, analysing and interpreting data. In examinations, learners must calculate accurately, choose suitable graphs and explain what the results mean in context.
Measures of central tendency describe the centre or typical value of a data set.
The mean formula is:
\overline{x}=\frac{\sum x}{n}For a frequency table:
\overline{x}=\frac{\sum fx}{\sum f}Question: Calculate the mean, median and mode of the marks:
4,\ 7,\ 7,\ 8,\ 9Given information:
n=5Mean formula:
\overline{x}=\frac{\sum x}{n}Substitute:
\overline{x}=\frac{4+7+7+8+9}{5}Simplify:
\overline{x}=\frac{35}{5} \overline{x}=7The third value is the median:
\text{Median}=7The most frequent value is:
\text{Mode}=7Common Mistake: Do not find the median before arranging the data in ascending order.
Measures of dispersion describe how spread out the data values are.
The formulas are:
\text{Range}=\text{Maximum}-\text{Minimum} \text{IQR}=Q_3-Q_1 \text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}Given:
2,\ 4,\ 5,\ 7,\ 8,\ 10,\ 12Calculate the range:
\text{Range}=12-2 \text{Range}=10The lower half is:
2,\ 4,\ 5Therefore:
Q_1=4The upper half is:
8,\ 10,\ 12Therefore:
Q_3=10Calculate the interquartile range:
\text{IQR}=10-4 \text{IQR}=6Remember: The interquartile range is less affected by extreme values than the range.
Standard deviation shows how far data values generally lie from the mean. A small standard deviation indicates that the values are clustered near the mean. A large standard deviation indicates greater variation.
The population standard deviation is:
\sigma=\sqrt{\frac{\sum(x-\overline{x})^2}{n}}Question: Calculate the standard deviation of:
2,\ 4,\ 4,\ 6Step 1: Calculate the mean.
\overline{x}=\frac{2+4+4+6}{4} \overline{x}=\frac{16}{4} \overline{x}=4Step 2: Write the formula.
\sigma=\sqrt{\frac{\sum(x-\overline{x})^2}{n}}Step 3: Substitute.
\sigma=\sqrt{\frac{(2-4)^2+(4-4)^2+(4-4)^2+(6-4)^2}{4}}Step 4: Simplify.
\sigma=\sqrt{\frac{4+0+0+4}{4}} \sigma=\sqrt{2}Step 5: Final answer.
\sigma\approx1{,}41Exam Tip: Use the population standard deviation function on your calculator unless the question instructs otherwise. Do not round intermediate answers too early.
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The five number summary provides the main values needed to describe the distribution and draw a box and whisker plot.
It consists of:
Given:
3,\ 5,\ 6,\ 8,\ 9,\ 11,\ 14,\ 16There are eight values. The median is the mean of the two middle values:
Q_2=\frac{8+9}{2} Q_2=8{,}5The lower half is:
3,\ 5,\ 6,\ 8 Q_1=\frac{5+6}{2} Q_1=5{,}5The upper half is:
9,\ 11,\ 14,\ 16 Q_3=\frac{11+14}{2} Q_3=12{,}5The five number summary is:
3;\ 5{,}5;\ 8{,}5;\ 12{,}5;\ 16A box and whisker plot displays the five number summary on a number line.
Using the five number summary:
3;\ 5{,}5;\ 8{,}5;\ 12{,}5;\ 16Draw a number line and mark these five values. Draw the box from:
Q_1=5{,}5\quad\text{to}\quad Q_3=12{,}5Draw the median at:
Q_2=8{,}5Extend the whiskers to:
\text{Minimum}=3,\quad\text{Maximum}=16Exam Tip: Box and whisker plots being compared must use the same scale.
A histogram represents continuous grouped data. The bars touch because the intervals are continuous.
Important facts:
Suppose the following grouped marks are given:
\begin{array}{c|c} \text{Mark interval}&\text{Frequency}\\ 0\leq x<10&2\\ 10\leq x<20&5\\ 20\leq x<30&8\\ 30\leq x<40&4 \end{array} [/katex] <p class="wp-block-paragraph">Draw four touching bars with heights:</p> [katex display="true"] 2,\ 5,\ 8,\ 4The modal class is the interval with the greatest frequency:
20\leq x<30 [/katex] <p class="wp-block-paragraph">Common Mistake: Do not leave spaces between the bars of a histogram.</p> <div style="height:50px" aria-hidden="true" class="wp-block-spacer"></div> <figure class="wp-block-image size-large"><a href="https://sqooltutors.co.za/sign-up-b/"><img src="https://lmxddlwowqlsyucleuyv.supabase.co/storage/v1/object/public/pdf_ads/batch%201/Doc%20Image%20Aug%2022,%202026,%2004_22_20%20PM%20(1).png" alt=""/></a></figure> <h2 class="wp-block-heading">Ogives</h2> <p class="wp-block-paragraph">An ogive is a cumulative frequency curve. It shows how many observations are less than or equal to particular upper class boundaries.</p> <p class="wp-block-paragraph">To draw an ogive:</p> <ul class="wp-block-list"> <li>Calculate cumulative frequencies.</li> <li>Plot upper class boundaries against cumulative frequencies.</li> <li>Include the lower starting boundary with cumulative frequency zero.</li> <li>Join the points using a smooth increasing curve.</li> </ul> <div style="height:50px" aria-hidden="true" class="wp-block-spacer"></div> <h3 class="wp-block-heading">Example</h3> <p class="wp-block-paragraph">Using frequencies:</p> [katex display="true"] 2,\ 5,\ 8,\ 4Calculate cumulative frequencies:
2 2+5=7 7+8=15 15+4=19Plot the points:
(0;0),\ (10;2),\ (20;7),\ (30;15),\ (40;19)The median corresponds to half the total frequency:
\frac{19}{2}=9{,}5Locate the cumulative frequency of approximately 9{,}5 on the vertical axis and read the estimated median from the horizontal axis.
A scatter plot represents paired numerical data as points on the Cartesian plane. It helps identify a relationship between two variables.
Consider the points:
(1;9),\ (2;7),\ (3;6),\ (4;4),\ (5;2)As x increases, y generally decreases. The data therefore shows a strong negative correlation.
Practical Application: A scatter plot can compare the age of a car with its selling price.
A least squares regression line models the relationship between two variables and can be used for prediction.
Its equation is commonly written as:
\widehat{y}=a+bxHere, b is the gradient and a is the y-intercept.
An examination data set relating car age to selling price produced the regression equation:
\widehat{y}=331\,397{,}20-22\,988{,}32xThe negative gradient shows that the predicted selling price decreases as the car becomes older.
Question: Predict the selling price of a five-year-old car.
Given information:
x=5Write the formula:
\widehat{y}=331\,397{,}20-22\,988{,}32xSubstitute:
\widehat{y}=331\,397{,}20-22\,988{,}32(5)Simplify:
\widehat{y}=331\,397{,}20-114\,941{,}60Final answer:
\widehat{y}=\text{R}216\,455{,}60The gradient means that the predicted selling price decreases by approximately \text{R}22\,988{,}32 per year.
Common Mistake: Do not use the regression equation to predict far outside the range of the original data. This is called extrapolation and may be unreliable.
The correlation coefficient, represented by r, measures the strength and direction of a linear relationship.
-1\leq r\leq1Given:
r=-0{,}95The negative sign indicates negative correlation. Since the value is close to -1, the relationship is strong. The points should lie close to a downward regression line, making predictions within the observed range reasonably reliable.
Remember: Correlation does not prove that one variable causes the other variable to change.
Interpretation requires learners to explain calculations and graphs in the context of the question.
Important comparisons include:
Class A has:
\overline{x}=65,\quad \sigma=4Class B has:
\overline{x}=65,\quad \sigma=11Both classes have the same mean, so their average performance is equal. Class A has the smaller standard deviation, so its marks are more consistent and lie closer to the mean. Class B’s marks are more widely spread.
Exam Tip: Do not merely write “Class A is better”. State whether you are comparing average performance, consistency or spread, and support the conclusion with the relevant statistic.