WAEC 2020 · Paper 2 · Q10

1. Put the ages in order first: $1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10, 11, 12, 13$ (14 ages).

(a)(i)

1. Range $= 13 - 1 = 12$.

(ii)

1. Number the positions: 1st $= 1$, 2nd $= 2$, 3rd $= 3$, 4th $= 4$, 5th $= 5$, 6th $= 5$, 7th $= 6$, 8th $= 7$, 9th $= 8$, 10th $= 9$, 11th $= 10$, 12th $= 11$, 13th $= 12$, 14th $= 13$.
2. $Q_1$ is at position $\frac{14}{4} = 3.5$: halfway between the 3rd and 4th ages, $\frac{3 + 4}{2} = 3.5$.
3. $Q_3$ is at position $\frac{3 \times 14}{4} = 10.5$: halfway between the 10th and 11th ages, $\frac{9 + 10}{2} = 9.5$.
4. Semi-interquartile range $= \frac{Q_3 - Q_1}{2}$
   $= \frac{9.5 - 3.5}{2}$
   $= 3$.

(iii)

1. The ages add up to 96, so the mean is $\frac{96}{14} \approx 6.86$.

(b)

1. The squares of the ages add up to $1 + 4 + 9 + 16 + 25 + 25 + 36 + 49 + 64 + 81 + 100 + 121 + 144 + 169 = 844$.
2. Standard deviation $= \sqrt{\frac{\sum x^2}{n} - \bar x^2}$
   $= \sqrt{\frac{844}{14} - \left(\frac{96}{14}\right)^2}$
   $= \sqrt{60.286 - 47.020}$
   $= \sqrt{13.265}$
   $\approx 3.64$.