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WAEC 2010 · Paper 2 · Q13

The table shows the marks obtained by 40 students in a test.

Marks 1–10 11–20 21–30 31–40 41–50 51–60 61–70 71–80 81–90 91–100
Frequency 2 4 4 6 8 6 5 2 2 1
  1. (a)

    Using an assumed mean of 45.545.5, calculate, correct to two decimal places, the: (i) mean; (ii) variance; of the distribution.

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

Worked solution (try it first)
  1. The class marks are 5.5,15.5,25.5,…,95.55.5, 15.5, 25.5, \dots, 95.5.
  2. Deviations from A=45.5A = 45.5: d=−40,−30,−20,−10,0,10,20,30,40,50d = -40, -30, -20, -10, 0, 10, 20, 30, 40, 50.
  3. Multiply by the frequencies: fd=−80,−120,−80,−60,0,60,100,60,80,50fd = -80, -120, -80, -60, 0, 60, 100, 60, 80, 50, so ∑fd=10\sum fd = 10.

(i)

  1. Mean =A+∑fd∑f= A + \dfrac{\sum fd}{\sum f}
    =45.5+1040= 45.5 + \dfrac{10}{40}
    =45.75= 45.75.

(ii)

  1. Square the deviations and multiply by ff: fd2=3200,3600,1600,600,0,600,2000,1800,3200,2500fd^2 = 3200, 3600, 1600, 600, 0, 600, 2000, 1800, 3200, 2500, so ∑fd2=19 100\sum fd^2 = 19\,100.
  2. Variance =∑fd2∑f−(∑fd∑f)2= \dfrac{\sum fd^2}{\sum f} - \left(\dfrac{\sum fd}{\sum f}\right)^2
    =19 10040−(1040)2= \dfrac{19\,100}{40} - \left(\dfrac{10}{40}\right)^2.
  3. Work it out: 477.5−0.0625=477.44477.5 - 0.0625 = 477.44 (2 d.p.).

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