The frequency distribution of the weight of 100... | WAEC SSCE - General Mathematics - 2010 (Essay)

Question 1 Report

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 - 69	70 - 79
Number of Participants	10	18	22	25	16	9

(a) Construct the cumulative frequency table.

(b) Draw the cumulative frequency curve.

(c) From the curve, estimate the : (i) median ; (ii) semi- interquartile range ; (iii) probability that a participant chosen at random weighs at least 60 kg.

Answer

(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.

graph — Ogive plotting cumulative frequency against upper class boundaries. Reading across from cf = 50 gives median ≈ 49.5 kg; cf = 25 gives Q1 ≈ 38.5 kg; cf = 75 gives Q3 ≈ 59.5 kg.

(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 \(100 - 75 = 25\).

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

Read lesson note on Probability (WAEC)

Read lesson note on Linear Inequalities (WAEC)

Read lesson note on Modular Arithmetic (WAEC)

Read lesson note on Sets (JAMB)

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.