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WAEC 2008 · Paper 2 · Q5

The deviations of a set of numbers from 55 are −5,−3,−1,0,1,3,5-5, -3, -1, 0, 1, 3, 5 and 77. Calculate the:

  1. (a)

    mean of the numbers;

  2. (b)

    variance of the numbers.

Worked solution (try it first)

(a)

  1. There are n=8n = 8 deviations.
  2. Their sum is −5−3−1+0+1+3+5+7=7-5 - 3 - 1 + 0 + 1 + 3 + 5 + 7 = 7.
  3. Mean deviation: dˉ=78=0.875\bar d = \frac78 = 0.875.
  4. Mean =55+dˉ=55.875= 55 + \bar d = 55.875.

(b)

  1. Square the deviations: 25,9,1,0,1,9,25,4925, 9, 1, 0, 1, 9, 25, 49.
  2. Their sum is ∑d2=119\sum d^2 = 119.
  3. Variance =∑d2n−dˉ 2= \dfrac{\sum d^2}{n} - \bar d^{\,2}
    =1198−0.8752= \dfrac{119}{8} - 0.875^2.
  4. 14.875−0.765625=14.10937514.875 - 0.765625 = 14.109375, so the variance is 14.1114.11 to two decimal places.

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