WAEC 2013 · Paper 2 · Q13

Number of days (xx) 10 20 30 40 50 60 70 80
Height (yy m) 1.0 1.1 1.2 1.4 1.6 1.8 2.0 2.3

The table gives the relationship between the height, in metres, of a plant and the number of days it is left to grow.

  1. (a)

    Using a scale of 2 cm to 0.5 units on the yy-axis and 2 cm to 10 units on the xx-axis, draw the scatter diagram for the information.

    Model answer
    10203040506070800.511.52Days (x)Height (y m)

    Plot the eight points, using 2 cm to 10 days across and 2 cm to 0.5 m up; don't join them. The points rise steadily, curving up slightly at the end.

  2. (b)

    Find xˉ\bar x, the mean of xx, and yˉ\bar y, the mean of yy, and plot (xˉ,yˉ)(\bar x, \bar y) on the diagram.

    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) and (10,1)(10, 1).

    Model answer
    10203040506070800.511.52Days (x)Height (y m)(45, 1.55)(10, 1)

    Rule a straight line through (10,1)(10, 1) and the mean point (45,1.55)(45, 1.55), extending it across the graph. Its gradient is 0.5535≈0.016\frac{0.55}{35} \approx 0.016, so the line is y−1=0.016(x−10)y - 1 = 0.016(x - 10). At 75 days it gives a height of about 2.02 m.

  4. (d)

    From the graph, find the: (i) equation of the line of best fit; (ii) height of the plant in 75 days.

Try it on a graph

Scatter points, the mean point (purple) and the line of best fit.

Worked solution (try it first)

(a)

  1. Plot the eight points with the given scales.

(b)

  1. xˉ=3608=45\bar x = \dfrac{360}{8} = 45 and yˉ=12.48=1.55\bar y = \dfrac{12.4}{8} = 1.55.
  2. Plot (45,1.55)(45, 1.55).

(c)

  1. Draw the line through (45,1.55)(45, 1.55) and (10,1)(10, 1).

(d)(i)

  1. Gradient =1.55−145−10= \dfrac{1.55 - 1}{45 - 10}
    =11700= \dfrac{11}{700}, so y=1+11700(x−10)y = 1 + \dfrac{11}{700}(x - 10), about y=0.016x+0.84y = 0.016x + 0.84.

(ii)

  1. At x=75x = 75: y=1+11700×65y = 1 + \dfrac{11}{700} \times 65
    ≈2.02\approx 2.02 m.

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