Numeracy Higher November 2021 Unit 1 Q2
2.
(a) Penystrad is a mountain village.
The daily rainfall for April 2021 is given in the table below.
The daily rainfall for April 2021 is given in the table below.
| Daily rainfall, \(r\) (mm) | Number of days |
|---|---|
| \(0 \leqslant r \lt 6\) | 15 |
| \(6 \leqslant r \lt 12\) | 11 |
| \(12 \leqslant r \lt 18\) | 3 |
| \(18 \leqslant r \lt 24\) | 1 |
(i) Wesley asks,
During April 2021, on how many days did it not rain in Penystrad?
Explain why it is not possible to answer Wesley’s question using the table shown above. [1]
(ii) Calculate an estimate for the mean daily rainfall for the 30 days of April. [4]
(b) Glanwen is a different village.
During the first 25 days of April in Glanwen, the mean daily rainfall was 4·4 mm.
It did not rain in Glanwen during the last 5 days of April.
Calculate the mean daily rainfall in Glanwen for April.
Give your answer correct to 3 significant figures. [4]
| Scheme | Marks | Comments |
|---|---|---|
| (i) Explanation, e.g. ‘data is grouped’, ‘not raw data’, ‘table only gives group information’ ’15 days with less than 6mm of rain, but we don’t know if there was no rain on any of these days’, ‘only results between 0 – 6mm’, ‘doesn’t give days of 0mm rain, it has 0 – 6mm’ | E1 | Allow, e.g. ‘because it shows 0 \(\leqslant\) r < 6 is equal to 15’, ‘doesn’t say if the 15 belongs to 0 or to less than 6’, ‘the table doesn’t give you exactly how many mm in the days’ Do not accept, e.g. ‘Can’t tell’ without further explanation as to why, ‘doesn’t give you enough information’, ‘it’s not accurate enough’, ‘no column with daily rainfall with no rain option’, ‘table only shows daily rainfall, not the number of days without rain’, ‘doesn’t show if it actually rained or not’, ‘no section for 0 rainfall’, ‘doesn’t show a day in the table when there is no rain’, ‘doesn’t say if the 15 belongs to 0 or to the 6’ ‘no record of the number of days it did not rain’ |
| (ii) Mid points 3, 9, 15, 21 | B1 | Note: Check the table |
| 3×15 + 9×11 + 15×3 + 21×1 (45 + 99 + 45 + 21 = 210) | M1 | FT their midpoints provided they fall within the classes including both bounds. FT if 1 slip in one of ‘their midpoints’, used outside the tolerance of bounds for M1, m1 only |
| ÷ 30 | m1 | FT from M1 for intention ‘their 210’/30 |
| 7 (mm) | A1 | Following correct working On FT from incorrect mid points allow rounding or truncation of ‘their final answer’ |
| Scheme | Marks | Comments |
|---|---|---|
| 25 × 4.4 (= 110) | M1 | |
| ÷ 30 | m1 | |
| 3.67 (mm) | A2 | CAO A1 for 3⅔ (mm) or 3.6(66...mm) which allows 3.6(....), 3.7 (mm) Allow A1 for a correct FT from an error in calculating 25 × 4.4 provided rounding to give 3 significant figures required and correct (e.g. 25 × 4.4 as 120 leading to an answer of 4 is A0) |