NECO 2024 · Paper 2 · Q12

Statistics: data & averages

The table shows the scores of candidates.

Marks (%)	1–10	11–20	21–30	31–40	41–50	51–60
Frequency	26	36	38	30	15	5

(a)
Calculate, correct to one decimal place, the (i) mean; (ii) median; (iii) modal score.
Separate values with commas, e.g. 3, −2
(b)
Determine the cut-off mark if only 20 candidates are to be offered admission.

Worked solution (try it first)

(a)(i)

Use the class marks (mid-points) $5.5, 15.5, 25.5, 35.5, 45.5, 55.5$ .
Then $\sum f = 150$ and $\sum fx = 26(5.5) + 36(15.5) + 38(25.5) + 30(35.5) + 15(45.5) + 5(55.5)$
$= 143 + 558 + 969 + 1065 + 682.5 + 277.5$
$= 3695$ .
Mean $= \frac{3695}{150} \approx 24.6$ .

(ii)

The median is the $\frac{150}{2} = 75$ th score.
The running totals of the frequencies are $26, 62, 100, \ldots$ , so the 75th score is in the class 21–30.
Its lower boundary is $20.5$ , 62 scores come before it, it holds 38 and its width is 10: median $= 20.5 + \frac{75 - 62}{38} \times 10$
$= 20.5 + 3.42$
$\approx 23.9$ .

(iii)

The modal class is 21–30 (frequency 38).
It is $38 - 36 = 2$ more than the class before and $38 - 30 = 8$ more than the class after: mode $= 20.5 + \frac{2}{2 + 8} \times 10 = 22.5$ .

(b)

Admission goes to the highest scores.
The top two classes, 51–60 and 41–50, hold exactly $5 + 15 = 20$ candidates.
So the 20 admitted are those who scored 41 or more: the cut-off mark is 41 (the class boundary 40.5).

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