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WAEC 2016 · Paper 2 · Q9

The weight (in kg) of 50 contestants at a competition is as follows:

65 66 67 66 64 66 65 63 65 68 64 62 66 64 67 65 64 66 65 67 65 67 66 64 65 64 66 65 64 65 66 65 64 65 63 63 67 65 63 64 66 64 68 65 63 65 64 67 66 64

  1. (a)

    Construct a frequency table for the discrete data.

    Model answer
    Weight (kg) 62 63 64 65 66 67 68
    Frequency 1 5 12 14 10 6 2

    One column for each weight (the data are discrete, so do not group them). Tally through the list, then check that the frequencies add up to 50.

  2. (b)

    Calculate, correct to 2 decimal places, the: (i) mean; (ii) standard deviation of the data.

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

Worked solution (try it first)

(a)

  1. The weights are separate whole numbers from 62 to 68, so make one row for each weight (don't group them).
  2. Tally through the list:
  3. Weight (kg) 62 63 64 65 66 67 68
    Frequency 1 5 12 14 10 6 2
  4. The frequencies add up to 50.

(b)(i)

  1. ∑fx=62+315+768+910+660+402+136\sum fx = 62 + 315 + 768 + 910 + 660 + 402 + 136
    =3253= 3253, so the mean is 325350=65.06\frac{3253}{50} = 65.06 kg.

(ii)

  1. Use an assumed mean of 65, with d=x−65=−3,−2,−1,0,1,2,3d = x - 65 = -3, -2, -1, 0, 1, 2, 3.
  2. Then ∑fd=−3−10−12+0+10+12+6\sum fd = -3 - 10 - 12 + 0 + 10 + 12 + 6
    =3= 3 and ∑fd2=9+20+12+0+10+24+18\sum fd^2 = 9 + 20 + 12 + 0 + 10 + 24 + 18
    =93= 93.
  3. Standard deviation =∑fd2∑f−(∑fd∑f)2= \sqrt{\frac{\sum fd^2}{\sum f} - \left(\frac{\sum fd}{\sum f}\right)^2}
    =9350−(350)2= \sqrt{\frac{93}{50} - \left(\frac{3}{50}\right)^2}
    =1.86−0.0036= \sqrt{1.86 - 0.0036}
    =1.8564= \sqrt{1.8564}
    ≈1.36\approx 1.36 kg.

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