June 2024 Paper 3 Q5
5. The records for a school athletics club show that the height, \(H\) metres, achieved by students in the high jump is normally distributed with mean 1.4 metres and standard deviation 0.15 metres.
The records also show that the time, \(T\) seconds, to run 1500 metres is normally distributed with mean 330 seconds and standard deviation 26 seconds.
The school’s Head would like to use these distributions to estimate the proportion of students from the school athletics club who can jump higher than 1.6 metres and can run 1500 metres in less than 5 minutes.
Students in the school athletics club also throw the discus.
The random variable \(D \sim \mathrm{N}\left(\mu, \sigma^2\right)\) represents the distance, in metres, that a student can throw the discus.
Given that \(\mathrm{P}(D \lt 16.3) = 0.30\) and \(\mathrm{P}(D \gt 29.0) = 0.10\)
| Scheme | Marks | AO |
|---|---|---|
| [\(\mathrm{P}(H \gt 1.6) =\)] \(0.091211\ldots =\) awrt 0.0912 | B1 | 1.1b |
| (1) |
Notes
B1 for awrt 0.0912 (from calculator)
| Scheme | Marks | AO |
|---|---|---|
| Need \(H\) and \(T\) to be independent or events \(\{H \gt 1.6\}\) and \(\{T \lt 300\}\) are independent | B1 | 2.4 |
| (1) |
Notes
B1 for a suitable reason mentioning or implying \(H\) and \(T\) are independent
Allow: e.g. “they”/ “each event”/ “\(\mathrm{P}(H)\) and \(\mathrm{P}(T)\)”/ “the variables” and “independent”
B0 for “the results” / “the values” are independent.
Ignore other comments that are not incorrect or contradictory.
| Scheme | Marks | AO |
|---|---|---|
| [\(\mathrm{P}(T \lt 300) =\)] \(0.124(2816\ldots)\) | M1 | 3.4 |
| Prob both is: \(\text{``}0.0912\ldots\text{''} \times \text{``}0.124\ldots\text{''}\) | M1 | 1.1b |
| \(= 0.011335\ldots =\) awrt 0.0113 | A1 | 1.1b |
| (3) |
Notes
1st M1 for using model for \(T\) to attempt to find \(\mathrm{P}(T \lt 300)\) e.g. sight of 0.124 or better or sight of \(\pm\left(\dfrac{300 - 330}{26}\right)\) or \(\pm\left(\dfrac{5 - 5.5}{0.433\ldots}\right)\) or \(Z = \pm\,1.15(3\ldots)\)
2nd M1 for multiplying their two probabilities together ft part (a) and their \(\mathrm{P}(T \lt 300)\) provided both values are probabilities. NB M0M1 is possible here
A1 for awrt 0.0113 [Correct answer with no incorrect working 3/3]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{16.3 - \mu}{\sigma} = -0.5244(0051\ldots)\), \(\dfrac{29 - \mu}{\sigma} = 1.2816\) (calc: 1.28155156…) | M1M1 | 3.1a 1.1b |
| e.g. \(29 - 16.3 = \sigma(\text{``}1.2816\text{''} - \text{``}{-0.5244}\text{''})\) | M1 | 1.1b |
| \(\sigma = 7.032115\ldots =\) awrt 7.03 | A1 | 1.1b |
| \(\mu = 19.9876\ldots = \underline{\mathbf{19.95}} \leqslant \underline{\mu} \leqslant \underline{\mathbf{20.0}}\) | A1 | 3.2a |
| (5) | ||
| (10 marks) |
Notes
1st M1 for standardising 16.3 and setting equal to \(z\) value where \(0.5 \lt |z| \lt 0.6\)
2nd M1 for standardising 29 and setting equal to \(z\) value where \(1 \lt |z| \lt 1.5\)
3rd M1 dep on 1st or 2nd M1 for solving their two linear eq’ns – reach an eq’n in one variable
May be implied by sight of \(\sigma = 7\) (or better) or \(\mu = 20\) (or better)
For 1st A mark we must also see one of \(-0.5244\) or 1.2816 (or better) used in their equ’ns OR both \(z\) values correct to 3dp i.e. \(-0.524\) and 1.282
1st A1 for \(\sigma =\) awrt 7.03 (but see 3rd case below)
2nd A1 for \(\mu =\) in [19.95, 20.0] (i.e shouldn’t see something rounding down to 20.0)
Allow 20 from equations with suitable \(z\) values (see examples below)
NB Use of \(-0.524\) and 1.28 [would give 7.0399… and 19.988…] and scores M3A0A1
Use of \(-0.524\) and 1.2816 [would give 7.033… and 19.99…] and scores M3A1A1
Use of \(-0.5244\) and 1.28 [would give 7.038… and 19.99 …] and scores M3A1A1
Both \(z\) values correct to 3dp i.e. \(-0.524\) and 1.282 [should give 7.032 and 19.984] scores A1A1