The table shows the frequency distribution of... | WAEC SSCE - Further Mathematics - 2011

Question 1 Report

The table shows the frequency distribution of marks scored by some candidates in an examination.

Marks	0-9	10-19	20-29	30-39	40-49	50-59	60-69	70-79	80-89	90-99
Freq	2	5	8	18	20	15	5	4	2	1

(a) Draw the cumulative frequency curve of the distribution.

(b) Use your graph to estimate the :

(i) semi-interquartile range of the distribution; (ii) percentage of candidates who passed with distinction if the least mark for distinction was 72.

Answer

(a) Cumulative frequency table

Marks	Frequency	Upper class boundary	Cumulative frequency
0–9	2	9.5	2
10–19	5	19.5	7
20–29	8	29.5	15
30–39	18	39.5	33
40–49	20	49.5	53
50–59	15	59.5	68
60–69	5	69.5	73
70–79	4	79.5	77
80–89	2	89.5	79
90–99	1	99.5	80

The total number of candidates is \(N=80\). Plot cumulative frequency against the upper class boundaries and draw a smooth increasing ogive.

(b)(i) Semi-interquartile range

\[Q_1\text{ is the }\frac{N}{4}=\frac{80}{4}=20\text{th value.}\]
From the ogive, \(Q_1\approx 32.5\).

\[Q_3\text{ is the }\frac{3N}{4}=\frac{3(80)}{4}=60\text{th value.}\]
From the ogive, \(Q_3\approx 53.0\).

\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{53.0-32.5}{2}=\boxed{10.25\text{ marks}}\]

(b)(ii) Percentage passing with distinction

From the ogive, the cumulative frequency at a mark of \(72\) is approximately \(74\).

\[\text{Number scoring at least }72=80-74=6\]

\[\text{Percentage with distinction}=\frac{6}{80}\times100\%=\boxed{7.5\%}\]

Answer Details