June 2022 Paper 2 Q9
9 At the beginning of the academic year, all the pupils in year 12 at a college take part in an assessment. Summary statistics for the marks obtained by the 2021 cohort are given below.
\(n = 205\quad \sum x = 23\,042\quad \sum x^2 = 2\,591\,716\)
Marks may only be whole numbers, but the Head of Mathematics believes that the distribution of marks may be modelled by a Normal distribution.
- The mean mark
- The variance of the marks
One candidate in the cohort scored less than 105.
| Scheme | Marks | AO |
|---|---|---|
| mean 112.4 isw or 112 isw | B1 | 1.1 |
| variance 8.8 or \(\sqrt{8.8^2}\) cao isw | B1 | 1.1 |
| [2] |
Notes
B1: B0 for 8.757 explicitly rounded to 8.8
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{N}(\)their 112.4, their 8.8\()\) | M1 | 3.3 |
| \(\mathrm{N}(a, b)\) | A1 | 1.1 |
| [2] |
Notes
M1: allow M1 for \(8.8^2\) or \(\sqrt{8.8}\)
A1: \(a = 112.4\) or 112 and \(b = 8.8\) or \(2.97^2\)
| Scheme | Marks | AO |
|---|---|---|
| P(mark \(\lt 104.5\)) or P(mark \(\lt 105\)) found from their distribution in part (b) | M1 | 3.4 |
| \(205 \times\) their non-zero 0.00387 | M1 | 3.1a |
| 0.79 to 0.794 or 1.17 to 1.175 so consistent oe | A1 | 3.5a |
| [3] |
Notes
M1: may see \(\mathrm{N}(-\infty, 104.5, 112.4, \sqrt{8.8})\)
NB 0.00387 or 0.0063(06) implies M1
NB 0.00573 or 0.00914 implies M1
NB 0.00379(69..) or 0.00619(81…) may imply M1 FT use of variance = 8.757
NB 0.200(199…) and 0.184(665…) may imply M1 FT use of sd = 8.8
if probability is correctly found to be 0 eg from use of \(\mathrm{N}\left(112.4, \frac{8.8}{205}\right)\) allow M1 only – no further marks available
M1: or compare \(\frac{1}{205}\ (\approx 0.00488)\) with their non-zero 0.00387
A1: or probabilities similar so consistent oe
Alternatively
| Scheme | Marks |
|---|---|
| \(\mathrm{InvNorm}\left(\frac{1}{205}, 112.4, \sqrt{8.8}\right)\) or \(\mathrm{InvNorm}\left(\frac{1}{205}, 112, \sqrt{8.8}\right)\) used to find their mark | M1 |
| compares their mark with 105 | M1 |
| 104.7 or 104.3 is close to 105 so good fit | A1 |
M1: FT their distribution
| Scheme | Marks | AO |
|---|---|---|
| P(mark between 114.5 and 115.5) found | M1 | 3.4 |
| 18.75 to 18.77 so allow 18 or 19 or 16.5 to 16.534 so allow 16 or 17 | A1 | 3.5a |
| [2] |
Notes
M1: NB awrt 0.0915 or awrt 0.0807 implies M1
A1: unsupported answers score M0