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WAEC 2021 · Paper 2 · Q10

Mark (%) 1–10 11–20 21–30 31–40 41–50 51–60 61–70 71–80 81–90 91–100
Frequency 2 6 17 30 33 51 29 9 3 0

The table shows the marks scored by some candidates in an examination.

  1. (a)

    Construct a cumulative frequency table.

    Model answer
    Mark (%) Frequency Upper class boundary Cumulative frequency
    1–10 2 10.5 2
    11–20 6 20.5 8
    21–30 17 30.5 25
    31–40 30 40.5 55
    41–50 33 50.5 88
    51–60 51 60.5 139
    61–70 29 70.5 168
    71–80 9 80.5 177
    81–90 3 90.5 180
    91–100 0 100.5 180

    Each cumulative frequency is the running total of the frequencies; the last one equals the total, 180.

  2. (b)

    Draw a cumulative frequency curve.

    Model answer
    0.510.520.530.540.550.560.570.580.590.5100.520406080100120140160180Mark (%)Cumulative frequency

    Plot each cumulative frequency against the upper class boundary of its class, starting from (0.5,0)(0.5, 0) where the cumulative frequency is 0 and ending at (100.5,180)(100.5, 180). Join the points with a smooth rising S-shaped curve (an ogive), not straight lines. Label both axes. Readings from a hand-drawn curve differ a little from person to person; examiners accept a small range, usually about ±1. The last class has frequency 0, so the curve is flat at the end.

    For (c): the median is the 90th mark. Read across from 90: about 51.0. If 60% passed, the bottom 40% (72 candidates) failed, so read across from 72: the pass mark is about 46.1.

  3. (c)

    Using the graph, estimate the: (i) median; (ii) pass mark if 60%60\% of the candidates passed.

    Separate values with commas, e.g. 3, −2

Try it on a graph

The ogive: cumulative frequency plotted at each upper class boundary.

Worked solution (try it first)

(a)

  1. Running totals against the upper class boundaries:
  2. Marks 1–10 11–20 21–30 31–40 41–50 51–60 61–70 71–80 81–90 91–100
    Upper boundary 10.5 20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5 100.5
    Cumulative frequency 2 8 25 55 88 139 168 177 180 180

(b)

  1. Plot the cumulative frequencies at the upper boundaries, starting from (0.5,0)(0.5, 0), and draw a smooth S-shaped curve.

(c)(i)

  1. There are 180 candidates, so the median is at 90.
  2. Go across to the curve and down: about 51.
  3. (Check: 90 is just above 88 at 50.5, and 50.5+90−8851×10≈50.950.5 + \frac{90 - 88}{51} \times 10 \approx 50.9.)

(ii)

  1. If 60%60\% passed, the lowest 40%40\% failed: 0.4×180=720.4 \times 180 = 72 candidates are below the pass mark.
  2. Go across from 72: about 45.5.
  3. (Check: 40.5+72−5533×10≈45.740.5 + \frac{72 - 55}{33} \times 10 \approx 45.7.)

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