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1996 WAEC Mathematics Theory The table below shows the marks obtained by forty pupils in a Mathematics test. Marks...

Mathematics
WAEC 1996

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 ;

(c) Calculate the median of the distribution.

 

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Explanation

(a) 

Marks Class boundary Freq Cum freq
0 - 9 0 - 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

 

(b) Mode = 35.5

(c) Median = \(L_{1} + \frac{(\frac{N}{2} - \sum f_{p}) \times c}{f_{m}}\)

where \(L_{1}\) = lower class boundary of median class = 29.5

 \(N = \sum f = 40\) ; \(\sum f_{p}\) = cumulative frequency before median class = 15.

\(f_{m}\) = frequency of median class = 12, c = class interval = 10.

 Median = \(29.5 + \frac{(\frac{40}{2} - 15) \times 10}{12}\)

= \(29.5 + \frac{(20 - 15) \times 10}{12}\)

= \(29.5 + 4.17 = 33.67\)


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