Unit 5
4 marks · Normal4. The heights (h) of 30 saplings are grouped in a histogram. The table below shows the grouped data:
| Height (cm) | Frequency |
|---|---|
| 0 < h ≤ 10 | 3 |
| 10 < h ≤ 20 | 8 |
| 20 < h ≤ 30 | 12 |
| 30 < h ≤ 40 | 7 |
The heights are said to be transformed using the function y = 2h + 5.
Calculate an estimate for the mean height of the saplings.
Find the value of y for a sapling at the midpoint of the tallest group (30 < h ≤ 40).
Explain what the transformation y = 2h + 5 does to the distribution of heights.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) Estimated mean = (5×3 + 15×8 + 25×12 + 35×7) / 30 = (15 + 120 + 300 + 245)/30 = 680 / 30 = 22.7 cm
(b) y = 2×35 + 5 = 75
(c) The heights are doubled and then increased by 5 cm.
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Finds correct midpoints and applies mean formula
A1 (a) · 1 mark · Obtains correct value for mean (22.7 cm)
B1 (b) · 1 mark · Calculates value using correct formula and midpoint
B2 (c) · 1 mark · Clearly describes transformation effect
Worked solution
a) Midpoints: 5, 15, 25, 35. Estimated mean:
c) The distribution is stretched by a factor of 2 and shifted up by 5 cm.