S1 June 2012 Q6
6. The heights of an adult female population are normally distributed with mean 162 cm and standard deviation 7.5 cm.
Sarah is a young girl. She visits her doctor and is told that she is at the 60th percentile for height.
The heights of an adult male population are normally distributed with standard deviation 9.0 cm.
Given that 90% of adult males are taller than the mean height of adult females,
| Scheme | Marks |
|---|---|
| \(\left[z =\right] \pm\left(\dfrac{150 - 162}{7.5}\right)\) | M1 |
| \([z =]\ -1.6\) | A1 |
| \(\left[\mathrm{P}(F \gt 150) = \mathrm{P}(Z \gt -1.6) =\right] = 0.9452(0071\ldots)\) awrt 0.945 | A1 |
| (3) |
Notes
M1 for attempting to standardise with 150, 162 and 7.5. Accept \(\pm\). Allow use of symmetry and therefore 174 instead of 150
1st A1 for \(-1.6\) seen. Allow 1.6 seen if 174 used or awrt 0.945 is seen. Sight of 0.945(2) is A1.
2nd A1 for awrt 0.945 Do not apply ISW, if 0.9452 is followed by 1 – 0.9452 then award A0. Correct answer only 3/3
| Scheme | Marks |
|---|---|
| \(z = \pm 0.2533\) (or better seen) | B1 |
| \((\pm)\,\dfrac{s - 162}{7.5} = 0.2533(47\ldots)\) | M1 |
| \(s = 163.9\) awrt 164 | A1 |
| (3) |
Notes
B1 for \((z =) \pm 0.2533\) (or better) seen. Giving \(z = \pm 0.25\) or \(\pm 0.253\) scores B0 here but may get M1A1
M1 for standardising with \(s\) (o.e.), 162 and 7.5, allow \(\pm\), and setting equal to a \(z\) value. Only allow \(0.24 \leqslant z \leqslant 0.26\). Condone e.g. 160 for 162 etc
A1 for awrt 164 (Correct answer only scores B0M1A1)
| Scheme | Marks |
|---|---|
| \(z = \pm 1.2816\) (or better seen) | B1 |
| \(\dfrac{162 - \mu}{9} = -1.2815515\ldots\) | M1 A1 |
| \(\mu = 173.533\ldots\) awrt 174 | A1 |
| (4) | |
| (10 marks) |
Notes
B1 for \((z =) \pm 1.2816\) (or better) seen. Allow awrt \(\pm 1.28\) if B0 scored in (b) for \(z =\) awrt \(\pm 0.25\)
M1 for attempting to standardise with 162, 9 and \(\mu\), and setting equal to a \(z\) value where \(1.26 \lt |z| \lt 1.31\). Allow \(\pm\) here so signs don’t have to be compatible.
1st A1 for a correct equation with compatible signs and \(1.26 \lt |z| \lt 1.31\)
2nd A1 for awrt 174 (Correct answer only scores B0M1A1A1). Dependent on 1st A1
An equation \(\dfrac{162 - \mu}{9} = 1.2816\) leading to an answer of \(\mu = 174\) is A0A0 unless there is clear correct working such as: \(\dfrac{162 - x}{9} = 1.2816 \Rightarrow x = \ldots\ \therefore\ \mu = 162 + (162 - x) = 174\) then award A1A1
NB A common error is: \(\dfrac{162 - \mu}{9} = 1.2816\) followed by \(\mu = 162 + 9\times 1.2816\) = awrt 174. It gets A0A0