WAEC 2009 · Paper 2 · Q15

The table shows the corresponding values of two variables xx and yy.

xx 33 31 28 25 23 22 19 17 16 14
yy 4 6 4 10 12 10 14 15 18 22
  1. (a)

    Plot a scatter diagram to represent the data.

    Model answer
    101520253035510152025xy

    Plot the ten points (33,4),(31,6),(28,4),(25,10),(23,12),(22,10),(19,14),(17,15),(16,18),(14,22)(33, 4), (31, 6), (28, 4), (25, 10), (23, 12), (22, 10), (19, 14), (17, 15), (16, 18), (14, 22) with xx across and yy up. Don't join them. The points fall from left to right.

  2. (b)

    Calculate xˉ\bar x, the mean of xx, and yˉ\bar y, the mean of yy.

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

  3. (c)

    Draw the line of best fit to pass through (xˉ,yˉ)(\bar x, \bar y).

    Model answer
    101520253035510152025xy(22.8, 11.5)10.5

    Plot the mean point (22.8,11.5)(22.8, 11.5) and draw a straight line through it that follows the trend, with about as many points above it as below. A good line passes near (14,19.2)(14, 19.2) and (33,2.6)(33, 2.6).

  4. (d)

    From your graph in (c), determine the: (i) relationship between xx and yy; (ii) value of yy when xx is 24.

Worked solution (try it first)

(a)

  1. Plot each (x,y)(x, y) pair as a point.
  2. The pattern is a downward trend.

(b)

  1. ∑x=228\sum x = 228, so xˉ=22810=22.8\bar x = \dfrac{228}{10} = 22.8.
  2. ∑y=115\sum y = 115, so yˉ=11510=11.5\bar y = \dfrac{115}{10} = 11.5.

(c)

  1. Mark (22.8,11.5)(22.8, 11.5) and draw a straight line through it with the points spread evenly on either side.

(d)(i)

  1. Read two points on your line, for example (14,19.2)(14, 19.2) and (33,2.6)(33, 2.6).
  2. The gradient is 2.6−19.233−14≈−0.88\dfrac{2.6 - 19.2}{33 - 14} \approx -0.88.
  3. Using (22.8,11.5)(22.8, 11.5): c=11.5+0.88×22.8≈31.5c = 11.5 + 0.88 \times 22.8 \approx 31.5.
  4. So y≈−0.88x+31.5y \approx -0.88x + 31.5: as xx increases, yy decreases.

(ii)

  1. Read up from x=24x = 24 to the line and across: y≈10.5y \approx 10.5.

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