The table below shows the marks obtained by forty... | WAEC SSCE - General Mathematics - 1996 (Essay)

Question 1 Report

The table below shows the marks obtained by forty pupils in a Mathematics test.

Marks	0 - 9	10 - 19	20 - 29	30 - 39	40 - 49	50 - 59
No of pupils	4	5	6	12	8	5

(a) Draw a histogram for the mark distribution ;

(b) Use your histogram to estimate the mode ;

Answer

(a) Histogram

Use continuous class boundaries. Since all class widths are 10 marks, the bar heights are the frequencies.

Marks	Class boundaries	Frequency	Cumulative frequency
0 - 9	-0.5 - 9.5	4	4
10 - 19	9.5 - 19.5	5	9
20 - 29	19.5 - 29.5	6	15
30 - 39	29.5 - 39.5	12	27
40 - 49	39.5 - 49.5	8	35
50 - 59	49.5 - 59.5	5	40

graph — Histogram with touching bars of equal width 10 marks. The bar heights are 4, 5, 6, 12, 8 and 5 respectively.

(b) Mode

The modal class is \(30-39\), with class boundaries \(29.5-39.5\).

\[\text{Mode}=L+\frac{d_1}{d_1+d_2}\times c\]

\[=29.5+\frac{12-6}{(12-6)+(12-8)}\times10\]

\[=29.5+\frac{6}{10}\times10=\boxed{35.5\text{ marks}}\]

(c) Median

\[N=40,\qquad \frac{N}{2}=20\]

The 20th value lies in the class \(30-39\), since the cumulative frequency rises from \(15\) to \(27\) in this class.

\[\text{Median}=L+\frac{\left(\frac{N}{2}-\text{cumulative frequency before median class}\right)}{f_m}\times c\]

\[=29.5+\frac{20-15}{12}\times10\]

\[=29.5+4.1667=\boxed{33.67\text{ marks}}\]

Read lesson note on Representation Of Data (JAMB)

Read lesson note on Probability (WAEC)

Read lesson note on Measures Of Location (JAMB)

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