Hint
State the full support and put \(\displaystyle f(x)=0\) outside it. Integrate the proposed density over its complete support.
Worked solution
(a)
\(\displaystyle k\frac{4^{3}}{3}=1\implies k=\frac{3}{64}\)
Answer: \(\displaystyle \frac{3}{64}\).
Answer: \(\displaystyle \frac{3}{64}\)
(b)
\(\int_0^L xf(x)\,dx=3\)
Answer: \(3\).
Answer: \(3\)
(c)
\(\displaystyle E(X^2)-E(X)^2=\frac{3}{5}\)
Answer: \(\displaystyle \frac{3}{5}\).
Answer: \(\displaystyle \frac{3}{5}\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.
(a) · 2 marks
- 1 mark: Integrate the proposed density over its complete support.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not read the height f(a) as \(\displaystyle P(X=a)\).
(b) · 2 marks
- 1 mark: Integrate x times the normalised density.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: A median solves an area equation, not \(\displaystyle f(m)= \frac{1}{2}\).
(c) · 2 marks
- 1 mark: Compute both moments before subtracting.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not read the height f(a) as \(\displaystyle P(X=a)\).
Total: 6 marks.