S1 June 2014 Q7
7. The heights of adult females are normally distributed with mean 160 cm and standard deviation 8 cm.
Any adult female whose height is greater than 170 cm is defined as tall.
An adult female is chosen at random. Given that she is tall,
Half of tall adult females have a height greater than \(h\) cm.
| Scheme | Marks |
|---|---|
| The random variable \(H \sim\) height of females \(\mathrm{P}(H \gt 170) = \mathrm{P}\left(Z \gt \dfrac{170 - 160}{8}\right) \quad [= \mathrm{P}(Z \gt 1.25)]\) | M1 |
| \(= 1 - 0.8944\) | M1 |
| \(= 0.1056\) (calc 0.1056498…) awrt 0.106 (accept 10.6%) | A1 |
| (3) |
Notes
1st M1 for attempt at standardising with 170, 160 and 8. Allow \(\pm\) i.e. for \(\pm\dfrac{170 - 160}{8}\)
2nd M1 for attempting \(1 - p\) where \(0.8 \lt p \lt 1\). Correct answer only 3/3
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(H \gt 180) = \mathrm{P}\left(Z \gt \dfrac{180 - 160}{8}\right) \quad [= 1 - 0.9938]\) | M1 |
| \(= 0.0062\) (calc 0.006209…) awrt 0.0062 or \(\frac{31}{5000}\) | A1 |
| \([\mathrm{P}(H \gt 180 \mid H \gt 170)] = \dfrac{0.0062}{0.1056}\) | M1 |
| \(= 0.0587\) (calc 0.0587760…) awrt 0.0587 or 0.0588 | A1 |
| (4) |
Notes
1st M1 for standardising with 180, 160 and 8
1st A1 for 0.0062 seen, maybe seen as part of another expression/calculation.
2nd M1 using conditional probability with denom = their (a) and num < their denom. Values needed.
2nd A1 for awrt 0.0587 or 0.0588. Condone 5.87% or 5.88% or \(\frac{31}{528}\). Correct answer only 4/4
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(H \gt h \mid H \gt 170)\ (= 0.5)\) or \(\dfrac{\mathrm{P}(H \gt h)}{\mathrm{P}(H \gt 170)}\ (= 0.5)\) | M1 |
| \([\mathrm{P}(H \gt h)] = 0.5\times\text{"}0.1056\text{"} = 0.0528\) (calc 0.0528249…) or \([\mathrm{P}(H \lt h)] = 0.9472\) | A1ft |
| \(\dfrac{h - 160}{8} = 1.62\) (calc 1.6180592…) | M1 B1 |
| \(h\) = awrt 173 cm awrt 173 | A1 |
| (5) | |
| (12 marks) |
Notes
1st M1 for a correct conditional probability statement. Either line and don’t insist on 0.5, ft (a)
1st A1ft for \([\mathrm{P}(H \gt h)] = 0.5\times\text{their}(a)\). Award M1A1ft for correct evaluation of \(0.5\times\text{their}(a)\) or sight of 0.0528 or better
2nd M1 for attempt to standardise (\(\pm\)) with 160 and 8 and set equal to \(\pm z\) value (\(1.56 \lt |z| \lt 1.68\))
B1 for \((z =)\) awrt \(\pm 1.62\) (seen)
2nd A1 for awrt 173 but dependent on both M marks.