The frequency distribution of the weight of 100 participants in a high jump competition is as shown below : Weight (kg) 20 - 29 30 - 39 40 - 49 50 - 59 60 -...

Answer Details

(a) Cumulative frequency table.

Total number of participants \(N = 10+18+22+25+16+9 = 100\). The cumulative frequency is plotted against the upper class boundary of each class.

Weight (kg)	Frequency	Upper class boundary	Cumulative frequency
20 - 29	10	29.5	10
30 - 39	18	39.5	28
40 - 49	22	49.5	50
50 - 59	25	59.5	75
60 - 69	16	69.5	91
70 - 79	9	79.5	100

(b) Cumulative frequency curve (ogive).

The curve begins at the lower boundary of the first class, \((19.5,\,0)\), and passes through each point \((\text{upper boundary},\ \text{cumulative frequency})\); the points are joined with a smooth curve.

(c) Estimates read from the curve.

(i) Median \((Q_2)\). The median is the weight corresponding to \(\dfrac{N}{2} = \dfrac{100}{2} = 50\) on the cumulative-frequency axis. Drawing a horizontal line from cumulative frequency \(50\) to the curve and dropping down to the weight axis:

\[\text{Median} \approx 49.5\ \text{kg}.\]

(ii) Semi-interquartile range. The lower quartile \(Q_1\) is read at \(\dfrac{N}{4} = \dfrac{100}{4} = 25\), and the upper quartile \(Q_3\) at \(\dfrac{3N}{4} = \dfrac{300}{4} = 75\). From the curve:

\[Q_1 \approx 38.5\ \text{kg}, \qquad Q_3 \approx 59.5\ \text{kg}.\]\[\text{Semi-interquartile range} = \tfrac{1}{2}\,(Q_3 - Q_1) = \tfrac{1}{2}\,(59.5 - 38.5) = \tfrac{1}{2}\,(21) = 10.5\ \text{kg}.\]

(iii) Probability of weighing at least 60 kg. A weight of at least \(60\) kg corresponds to the class boundary \(59.5\) kg. From the table the cumulative frequency at \(59.5\) kg is \(75\), so the number weighing at least \(60\) kg is \(100 - 75 = 25\).

\[P(\text{weight} \ge 60\ \text{kg}) = \frac{25}{100} = \frac{1}{4} = 0.25.\]