If it is given that the interquartile range (IQR) of the data is 45%, calculate the minimum percentage of learners who passed in this centre.
About this question
Question 2.2 of Mathematics Grade 12 Paper 2, NSC Examination (May/June) (DBE, May/June 2026). It tests the Term 3 ATP topic Statistics (Regression and Correlation).
- TermTerm 3
- ATP topicStatistics (Regression and Correlation)
- PaperPaper 2
- Marks3 marks for the method (formula and substitution) and the correct final answer
2.1 Marks (as a %) of Grade 12 learners at a school in the June 2025 Mathematics examination:
| Class interval (%) | Frequency |
|---|---|
| \(20 \lt x \leq 30\) | 8 |
| \(30 \lt x \leq 40\) | 14 |
| \(40 \lt x \leq 50\) | 16 |
| \(50 \lt x \leq 60\) | 21 |
| \(60 \lt x \leq 70\) | 16 |
| \(70 \lt x \leq 80\) | 9 |
| \(80 \lt x \leq 90\) | 6 |
| \(90 \lt x \leq 100\) | 2 |
2.2 A box and whisker diagram shows the marks (as a %) of learners at another school in the same examination. The five values labelled from left to right are: minimum \((x - 2)\%\), lower quartile \(\left(\frac{3x - 3}{2}\right)\%\), median \((2x + 5)\%\), upper quartile \((3x)\%\) and maximum \(91\%\).
\(27\%\)
More Questions On Statistics (Regression and Correlation)
Plot two points from the equation, for example:
\(x = 90\):
\[\hat{y} = -43{,}72 + 2{,}36(90) = 168{,}68\]\(x = 150\):
\[\hat{y} = -43{,}72 + 2{,}36(150) = 310{,}28\]Join the points with a straight line that covers the data range, from about \(x = 85\) to \(x = 160\).
Median position:
\[\frac{1}{2} \times 60 = 30\text{th value}\]Reading across from 30 to the ogive and down: median \(\approx 65\) minutes
The tallest bars are on the left and the tail stretches to the right.
The data is skewed to the right (positively skewed).
From the ogive, 26 employees travel less than 60 minutes.
Employees who qualify:
\[60 - 26 = 34\]Percentage:
\[\frac{34}{60} \times 100\]\(\approx 56{,}67\%\)