1a.
Simple random sampling.
b.
Advantage: Cheaper/faster than a census.
Disadvantage: Sample might not perfectly represent the population due to random variation.
2a.
Stem and leaf diagram drawn correctly:
1 | 4
2 | 1 1 1 6 8
3 | 3 4 5 8 9
4 | 2 4 5
5 | 1
Key: 1 | 4 represents 14 marks
c.
Lower quartile ($Q_1$) = $21$, Upper quartile ($Q_3$) = $42$
d.
Interquartile Range (IQR) = $21$
3a.
Lower boundary is $19$. Since $8 < 19$, it is an outlier. Upper boundary is $59$, and since $58 < 59$, it is not an outlier.
b.
Box plot drawn correctly:
4a.
Mean $\bar{x} = 46.5 \text{ m}$
b.
$s = 2.33 \text{ m}$ (to 3 s.f.)
5a.
A histogram is appropriate because the data is continuous and grouped into classes of unequal widths.
b.
Width = $8\text{ cm}$, Height = $6\text{ cm}$
6a.
Median $\approx 28.5\text{ mins}$
b.
IQR $\approx 16.5\text{ mins}$
c.
$9$ patients received an apology.
7a.
Mean = $36\text{ cm}$, $s_A \approx 8.06\text{ cm}$
b.
Combined mean = $37.25\text{ cm}$
c.
Combined $s = 6.40\text{ cm}$ (to 3 s.f.)
8a.
$\bar{x} = 15$, $\bar{y} = 70$
b.
Correlation does not imply causation.
c.
The gradient ($2.8$) means for every additional hour studied per week, the predicted test score increases by $2.8$ marks.
d.
Unreliable due to extrapolation ($30$ hours is outside the original data range) and predicting an impossible score ($112$ out of $100$).