(a) The distribution of junior workers in an... | WAEC SSCE - General Mathematics - 1992 (Essay)

Question 1 Report

(a) The distribution of junior workers in an institution is as follows: Clerks - 78, Drivers - 36, Typists - 44, Messengers - 52, Others - 30. Represent the above information by a pie chart.

(b) The table below shows the frequency distribution of marks scored by 30 candidates in an aptitude test.

Marks	4	5	6	7	8	9
No of candidates	5	8	5	6	4	2

Find the mean score to the nearest whole number.

Answer

(a) Pie chart of junior workers

First find the total number of workers:

\[ 78+36+44+52+30 = 240 \]

The angle of each sector is a fraction of the whole circle \(\left(360^\circ\right)\):

\[ \text{Sector angle} = \frac{\text{Number in category}}{240}\times 360^\circ \]

Category	Number	Sector angle
Clerks	78	\(\frac{78}{240}\times360^\circ=117^\circ\)
Drivers	36	\(\frac{36}{240}\times360^\circ=54^\circ\)
Typists	44	\(\frac{44}{240}\times360^\circ=66^\circ\)
Messengers	52	\(\frac{52}{240}\times360^\circ=78^\circ\)
Others	30	\(\frac{30}{240}\times360^\circ=45^\circ\)
Total	240	\(360^\circ\)

The five sector angles add up to \(117^\circ+54^\circ+66^\circ+78^\circ+45^\circ=360^\circ\), confirming the calculation. Drawn to scale with a protractor, the pie chart is:

Pie chart of junior workers: Clerks 117°, Drivers 54°, Typists 66°, Messengers 78°, Others 45° (total 360°).

(b) Mean score of the aptitude test

Multiply each mark \(x\) by its frequency \(f\) to obtain \(fx\):

Marks \((x)\)	4	5	6	7	8	9	Total
No. of candidates \((f)\)	5	8	5	6	4	2	30
\(fx\)	20	40	30	42	32	18	182

\[ \sum f = 30, \qquad \sum fx = 20+40+30+42+32+18 = 182 \] \[ \text{Mean} = \frac{\sum fx}{\sum f} = \frac{182}{30} = 6.07 \]

To the nearest whole number, the mean score is 6 marks.

Read lesson note on Representation Of Data (JAMB)

Read lesson note on Areas (WAEC)

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