Foundation June 2019 Paper 1 Q20
20 Luke is an office receptionist.
Each day, for 60 days, he records the number of people visiting the office.
| Number of people, (\(n\)) | Frequency | ||
|---|---|---|---|
| \(0 \leqslant n \leqslant 5\) | 20 | ||
| \(5 \lt n \leqslant 10\) | 14 | ||
| \(10 \lt n \leqslant 20\) | 11 | ||
| \(20 \lt n \leqslant 40\) | 15 |
(a) Calculate an estimate of the mean number of people visiting the office. [4]
(b) Luke says the range is 40.
Explain why he may be wrong. [1]
| Answer | Marks | Part marks and guidance |
|---|---|---|
| 12.8[3....] | 4 | B1 for at least 3 mid-points seen (from 2.5, 7.5, 15, 30) or implied by products 50,105,165,450 or 770 M1 for \(\Sigma mf\) where \(m\) is a value within each group. Allow use of boundaries; allow one error in calculation. If no midpoints seen may be implied by their mf M1 dep on previous M1 for their \(770 \div 60\) |
| Answer | Marks | Part marks and guidance |
|---|---|---|
| The highest number may not have been 40 or the lowest number may not have been 0. | 1 | See appendix |
Appendix
Exemplar responses for Q20(b)
| Response | Mark |
|---|---|
| Because you don’t know the exact biggest and smallest numbers of people | 1 |
| Exact number of people is not known, they are just boundaries where they could be | 1 |
| No exact numbers given/don’t have exact values | 1 |
| 40 – 5 = 35 so he could be wrong | 1 |
| You don’t know the exact weight | 0 |
| Not exactly 40 people visited the office | 0 |
| Because we are using midpoints so won’t get accurate result | 0 |
| 40 – 5 = 35 therefore he is wrong | 0 |
| There could be less than 40 | 0 |