Unit 9
10 marks · HardStudents study the ages of participants in a marathon and compare this with a graph transformation.
The table shows the ages of participants.
| Age, $a$ (years) | Frequency |
|---|---|
| 10 < a ≤ 20 | 6 |
| 20 < a ≤ 30 | 22 |
| 30 < a ≤ 45 | 31 |
| 45 < a ≤ 60 | 14 |
| 60 < a ≤ 80 | 7 |
Calculate the class widths and frequency densities for each interval.
Draw a histogram for the data (rough sketch acceptable).
The function $g(x) = (x-30)^2$ is sketched for $x$ in the interval $10 \leq x \leq 80$ on the same axes as your histogram.
Describe the transformation that maps $y = x^2$ onto $y = (x-30)^2$. Explain how this can represent a change in the age profile.
From your histogram, estimate the modal class and comment on how this relates to the minimum point of the curve $g(x)$.
Suppose the data is adjusted so every age is reduced by 10 years. Write down the equation for the new function and state the transformation applied.
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(i)) 10,10,15,15,20; Frequencies:0.6,2.2,2.07,0.93,0.35 · must show Widths:10,10,15,15,20, Frequencies:0.6,2.2,2.07,0.93,0.35
(a(ii)) Histogram with heights matching densities, correct intervals · must show Histogram bars drawn as above
(b) Curve $y=(x-30)^2$ sketched, minimum at $x=30$; overlays histogram. · must show Minimum at $x=30$, Curve drawn on same axes
(c) $y = (x-30)^2$ is $y = x^2$ shifted right by 30 units; this means the central age is shifted to 30. · must show Shift right by 30, Relates to profile centre
(d) Modal class: 30 < a ≤ 45. This matches the minimum of the curve, showing most runners are aged around 30–45. · must show State class, Connection to minimum
(e) $g(x) = (x-20)^2$, shift right by 20. · must show Equation, Transformation
Must show: Widths:10,10,15,15,20, Frequency densities:0.6,2.2,2.07,0.93,0.35, Histogram matches, Transformation: right by 30, Relate modal class and curve minimum, New function $g(x) = (x-20)^2$
Common mistakes: Misplace minimum of curve; Reverse shift direction
Show mark scheme and worked solution
Marking points
M1 (a(i)) · 1 mark · Correct class widths for all intervals.
A1 (a(i)) · 1 mark · Accurate calculation of all frequency densities.
B1 (a(ii)) · 1 mark · Histogram drawn with heights matching frequency densities. (follow-through)
B2 (b) · 2 marks · Curve $y=(x-30)^2$ correctly sketched with minimum at $x=30$, on same axes as histogram.
B3 (c) · 2 marks · Identifies translation by 30 units right and explains centralisation of age profile.
B4 (d) · 2 marks · States modal class and explains alignment with minimum of the curve.
B5 (e) · 1 mark · States correct new function and identifies shift by 10 units left.
Worked solution
a(i) Widths: 10,10,15,15,20. FDs: 6/10=0.6, 22/10=2.2, 31/15≈2.07, 14/15≈0.93, 7/20=0.35.
a(ii) Draw bars: 10–20 at height 0.6, 20–30 at 2.2, etc.
b) Curve $y=(x-30)^2$ passes through minimum at x=30, increases on both sides. Overlay on histogram axes.
c) The transformation is a shift right by 30 units. $y=(x-30)^2$ is $y=x^2$ with the vertex at $x=30$, representing a central age of 30.
d) Modal class is 30 < a ≤ 45; this interval contains most runners. Minimum of the curve is at $x=30$ aligning with this class.
e) If all ages reduce by 10 years, new function is $y=(x-20)^2$, which is a right shift of 20.