The table gives the distribution of marks of 60... | WAEC SSCE - Further Mathematics - 2010

Question 1 Report

The table gives the distribution of marks of 60 candidates in a test.

Marks	23-25	26-28	29-31	32-34	35-37	38-40
Frequency	3	7	15	21	10	4

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

(b) From your curve, estimate the : (i) 80th percentile ; (ii) median ; (iii) semi-interquartile range.

Answer

(a) Cumulative frequency table

Marks	Class boundaries	Frequency	Cumulative frequency
23–25	22.5–25.5	3	3
26–28	25.5–28.5	7	10
29–31	28.5–31.5	15	25
32–34	31.5–34.5	21	46
35–37	34.5–37.5	10	56
38–40	37.5–40.5	4	60

Plot the upper class boundaries against the cumulative frequencies, including the starting point (22.5,0). The resulting less-than ogive is:

graph — Cumulative frequency curve obtained by plotting upper class boundaries against cumulative frequencies.

(b) Estimates from the ogive

(i) 80th percentile

The 80th percentile is the 48th value:

\[P_{80}=34.5+\frac{48-46}{10}\times 3=35.1\]

Hence, \(\boxed{P_{80}\approx35.1\text{ marks}}\).

(ii) Median

The median is the 30th value:

\[\text{Median}=31.5+\frac{30-25}{21}\times 3=32.2\]

Therefore, \(\boxed{\text{Median}\approx32.2\text{ marks}}\).

(iii) Semi-interquartile range

\[Q_1=28.5+\frac{15-10}{15}\times3=29.5\]

\[Q_3=31.5+\frac{45-25}{21}\times3=34.4\]

\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{34.4-29.5}{2}=2.45\]

Thus, \(\boxed{\text{semi-interquartile range}\approx2.4\text{ marks}}\).

Read lesson note on Statistics (WAEC)

Answer Details