S1 June 2010 Q7
7. The distances travelled to work, \(D\) km, by the employees at a large company are normally distributed with \(D \sim \mathrm{N}(30, 8^2)\).
An outlier is defined as any value of \(D\) such that \(D \lt h\) or \(D \gt k\) where
\[h = Q_1 - 1.5 \times (Q_3 - Q_1) \quad \text{and} \quad k = Q_3 + 1.5 \times (Q_3 - Q_1)\]An employee is selected at random.
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(D \gt 20) = \mathrm{P}\left(Z \gt \dfrac{20 - 30}{8}\right)\) | M1 |
| \(= \mathrm{P}(Z \gt -1.25)\) | A1 |
| \(= 0.8944\) awrt 0.894 | A1 |
| (3) |
Notes
M1 for an attempt to standardise 20 or 40 using 30 and 8.
1st A1 for \(z = \pm\,1.25\)
2nd A1 for awrt 0.894
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(D \lt Q_3) = 0.75\) so \(\dfrac{Q_3 - 30}{8} = 0.67\) | M1 B1 |
| \(Q_3 =\) awrt 35.4 | A1 |
| (3) |
Notes
M1 for \(\dfrac{Q_3 - 30}{8}\) = to a \(z\) value
M0 for 0.7734 on RHS.
B1 for (\(z\) value) between 0.67~0.675 seen.
M1B0A1 for use of \(z\) = 0.68 in correct expression with awrt 35.4
| Scheme | Marks |
|---|---|
| 35.4 - 30= 5.4 so \(Q_1 = 30 - 5.4\) = awrt 24.6 | B1ft |
| (1) |
Notes
Follow through using their of quartile values.
| Scheme | Marks |
|---|---|
| \(Q_3 - Q_1 = 10.8\) so \(1.5(Q_3 - Q_1) = 16.2\) so \(Q_1 - 16.2 = h\) or \(Q_3 + 16.2 = k\) | M1 |
| \(h\)=8.4 to 8.6 and \(k\)= 51.4 to 51.6 both | A1 |
| (2) |
Notes
M1 for an attempt to calculate 1.5(IQR) and attempt to add or subtract using one of the formulae given in the question - follow through their quartiles
| Scheme | Marks |
|---|---|
| \(2\mathrm{P}(D \gt 51.6) = 2\mathrm{P}(Z \gt 2.7)\) | M1 |
| \(= 2[1 - 0.9965]\) = awrt 0.007 | M1 A1 |
| (3) | |
| (12 marks) |
Notes
1st M1 for attempting \(2\mathrm{P}(D \gt \text{their } k)\) or ( \(\mathrm{P}(D \gt \text{their } k) + \mathrm{P}(D \lt \text{their } h)\))
2nd M1 for standardising their \(h\) or \(k\) (may have missed the 2) so allow for standardising \(\mathrm{P}(D \gt 51.6)\) or \(\mathrm{P}(D \lt 8.4)\)
Require boths Ms to award A mark.