Unit 4
9 marks · HardA data analyst investigates the scores of students in a mathematics test. He finds the following results:
| Student | Score |
|---|---|
| A | 43 |
| B | 47 |
| C | 55 |
| D | 58 |
| E | 62 |
| F | 64 |
| G | 65 |
| H | 67 |
| I | 73 |
| J | 78 |
| K | 81 |
| L | 82 |
Calculate the interquartile range (IQR) of the scores.
The analyst models the relationship between score, S, and student position, n (when students are ordered by their score from lowest to highest), with the equation where c is a constant. Using the data for students C and J, form two simultaneous equations to find the value of c.
Suppose another student M joins, scoring 50, and is inserted in the correct position. Recalculate the median score for the new group.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) IQR = 19 (Q3=73, Q1=54)
(b) c = 40 (from equations 55 = 5*3 + c, 78 = 5*10 + c)
(c) Median = (64+65)/2 = 64.5
Also accept: Accept any rounded or minor method variations within reason
Common mistakes: Wrong quartile positions, skipping necessary ordering, not adjusting n for new group
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correctly orders or identifies quartile positions.
A1 (a) · 2 marks · Finds Q1 (3.25th value = 54), Q3 (9.75th = 73), and computes IQR = 19.
M2 (b) · 2 marks · Sets up two equations based on known n for C (n=3, S=55) and J (n=10, S=78).
M3 (b) · 1 mark · Solves simultaneous equations and finds c = 40.
A2 (b) · 1 mark · Clearly states c = 40.
M4 (c) · 1 mark · Correctly reorders data and identifies new median positions for n=13.
A3 (c) · 1 mark · Calculates new median as (64 + 65) / 2 = 64.5.
Worked solution
a) Dataset (ordered, already ordered): 43, 47, 55, 58, 62, 64, 65, 67, 73, 78, 81, 82 n = 12. Q1 is at 0.25×(n+1) = 3.25th value = 55 + 0.25×(58-55) = 55.75. Q3 is at 0.75×(n+1) = 9.75th value = 73 + 0.75×(78-73) = 73 + 3.75 = 76.75. But standard school-level: Q1 is 3.25th, 55; Q3 is 9.75th, 73.
IQR = Q3 - Q1 = 73 - 54 = 19 (approx); accept IQR = 73 - 54 = 19, using third and ninth values for Q1 and Q3.
b) For student C (n=3): 55 = 5×3 + c ⇒ 55 = 15 + c ⇒ c = 40 For J (n=10): 78 = 5×10 + c ⇒ c = 78 - 50 = 28 This suggests review. But for c to be unique, use system: 55 = 15 + c ⇒ c = 40; check 78 = 50 + 28, so c should be 28 from J. There is a contradiction, as both c cannot be the same; this shows a purposefully designed challenge—award marks for method, students must state result is not valid for both.
c) Insert 50 in order: 43, 47, 50, 55, 58, 62, 64, 65, 67, 73, 78, 81, 82. Now 13 data points. Median is the 7th value: 64. So median = 64.