Foundation November 2020 Paper 3 Q17
17 A record was kept of the number of days that 25 students were absent one term.
The chart represents the results.
(a) Work out the mean number of days absent. [3 marks]
(b) One of the students is chosen at random.
Work out the probability that the student was absent for less than 4 days. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(1 \times 5\) and \(2 \times 6\) and \(3 \times 8\) and \(4 \times 2\) and \(5 \times 4\) or 5 and 12 and 24 and 8 and 20 or 69 | M1 | allow one error |
| \((5 + 12 + 24 + 8 + 20) \div 25\) or \(69 \div 25\) or their \(69 \div 25\) | M1dep | without working their 69 must be the correct sum of their products |
| 2.76 | A1 | oe |
Additional guidance
| Five products or values must be seen for first M1 | |
| Ignore attempt to round after 2.76 seen | M1M1A1 |
| \(69 \div 5\) | M1M0 |
| \(5 + 12 + 24 + 8 + 20 \div 25\) unless recovered | M1M0 |
| Correct products seen with \(25 \div 5\) or \(25 \div 15\) or \(15 \div 5\) | M0 |
| Answer | Mark | Comments |
|---|---|---|
| \(5 + 6 + 8\) or \(25 - (4 + 2)\) or 19 or \(1 - \dfrac{4 + 2}{25}\) | M1 | oe |
| \(\dfrac{19}{25}\) or 0.76 or 76% | A1 | oe |
Additional guidance
| Ignore attempts to simplify or convert a correct fraction | |
| Ignore probability words | |
| 19 out of 25 or 19 in 25 alone on the answer line with a correct answer in working | M1A1 |
| 19 out of 25 or 19 in 25 together with a correct answer on the answer line | M1A1 |
| 19 : 25 with a correct answer together on the answer line | M1A0 |