O-Level E-Math 2017 Paper 2, Question 9: Median & quartiles from a CF curve

(a)(i)(a) Median \(=\) the 40th value (half of 80). Reading across from cumulative frequency 40, the median speed is about 43 km/h. (Read from graph.)

(b) Lower quartile \(=\) 20th value \(\approx 38\) km/h; upper quartile \(=\) 60th value \(\approx 45.5\) km/h. Interquartile range \(\approx 45.5 - 38 = 7.5\) km/h. (Read from graph.)

(ii) At a speed of 50 km/h the cumulative frequency is about 76, so about \(80 - 76 = 4\) cars exceeded 50 km/h. Percentage \(= \dfrac{4}{80} \times 100\% = 5\%\). (Read from graph; accept roughly 4 to 7%.)

(iii) Comparing the morning (curve) with the afternoon (box plot, with median \(\approx 41\), \(Q_1 \approx 34\), \(Q_3 \approx 42\)): (1) the afternoon median speed (\(\approx 41\) km/h) is a little lower than the morning median (\(\approx 43\) km/h), so on average cars were slightly slower in the afternoon; (2) the afternoon speeds are at least as spread out as the morning ones (afternoon interquartile range \(\approx 8\) km/h with a range reaching about 55 km/h), so the afternoon speeds are more varied. (Both comparisons read from the graphs.)

(b) Total cars \(= 34 + 17 + 13 + 11 + 5 = 80\).

(i) \(P(\text{no passengers}) = \dfrac{34}{80} = \dfrac{17}{40}\).

(ii) Two cars chosen at random (without replacement). (a) Both had four passengers (5 such cars): \(P = \dfrac{5}{80} \times \dfrac{4}{79} = \dfrac{20}{6320} = \dfrac{1}{316}\). (b) "More than two passengers" means 3 or 4 (\(11 + 5 = 16\) cars); "fewer than two" means 0 or 1 (\(34 + 17 = 51\) cars). The two chosen cars are one from each group, in either order: \[P = 2 \times \frac{16}{80} \times \frac{51}{79} = \frac{2 \times 16 \times 51}{80 \times 79} = \frac{1632}{6320} = \frac{102}{395}.\]