Foundation June 2025 Paper 1 Q18
18 The table gives information about the distances 100 adults travel to work.
| Distance (\(d\) km) | Frequency |
|---|---|
| \(0 \lt d \leqslant 5\) | 26 |
| \(5 \lt d \leqslant 10\) | 40 |
| \(10 \lt d \leqslant 15\) | 16 |
| \(15 \lt d \leqslant 20\) | 10 |
| \(20 \lt d \leqslant 25\) | 8 |
(a) Write down the modal class. (1)
(b) Work out an estimate for the mean distance. (4)
| Scheme | Marks |
|---|---|
| \(5 \lt d \leqslant 10\) | B1 |
| (1) |
Notes
B1: allow 5 – 10 or 5 to 10
or \(5 \lt d \lt 10\) or \(5 \leqslant d \leqslant 10\)
or \(5 \leqslant d \lt 10\)
| Scheme | Marks |
|---|---|
| 2.5 × 26 + 7.5 × 40 + 12.5 × 16 + 17.5 × 10 + 22.5 × 8 (= 920) or 65 + 300 + 200 + 175 + 180 (= 920) [lower bound products are: 0, 200, 160, 150, 160] [sum of lower bound products is: 670] [products using 3, 8, 13, 18, 23 are: 78, 320, 208, 180, 184] [sum of products using 3, 8, 13, 18, 23 is: 970] [upper bound products are: 130, 400, 240, 200, 200] [sum of upper bound products is: 1170] | M2 |
| “920” ÷ “100” | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 9.2 | A1 |
| (4) | |
| (5 marks) |
Notes
M2: for at least 4 correct products added (need not be evaluated ie can be in the form 2.5 × 26 + 7.5 × 40 +…)
If not M2 then award:
M1 for consistent use of values within interval (including end points) for at least 4 products added (need not be evaluated ie can be in the form 5 × 26 + 10 × 40 +…)
or
correct midpoints used for at least 4 products and not added
M1: (dep on at least M1)
Allow division by their \(\Sigma f\) provided addition or total under column seen
A1: oe eg \(9\dfrac{1}{5}\) or \(\dfrac{46}{5}\)
SCB2 for answer of 6.7 or 9.7 or 11.7