CAIE A-Level Further Math AS 4.1 Continuous Random Variables Questions

$$\begin{aligned} \mathrm{P}(X \leqslant 2)=\frac{1}{3}, \quad \text { so }\left[\frac{m x^{2}}{2}\right]_{0}^{2}=\frac{1}{3} m\left(\frac{4}{2}\right)=\frac{1}{3}, \quad m=\frac{1}{6} \end{aligned}$$
AG
Reasoning required.

$\int_{2}^{6}\left(\frac{k}{x^{2}}+c\right) \mathrm{d} x=\frac{2}{3} \quad$ so $\quad\left[-\frac{k}{x}+c x\right]_{2}^{6}=4 c+\frac{1}{3} k=\frac{2}{3} \quad$ or $\quad k+12 c=2$
Attempt to integrate and correct use of correct limits to
form linear equation in k and c.

Also, $2 m=\frac{k}{4}+c$ or $\frac{1}{3}=\frac{k}{4}+c$ or 3 k+12 c=4
Matching at x=2.

Solve, $k=1, \quad c=\frac{1}{12}$
4

LQ: $\int_{0}^{L} \frac{1}{6} x \mathrm{~d} x=\frac{1}{4}, \quad L=\sqrt{3}$
UQ: $\frac{1}{3}+\int_{2}^{U}\left(\frac{k}{x^{2}}+c\right) \mathrm{d} x=\frac{3}{4}$
Or find and use CDF. May be in terms of m.

$$\begin{aligned} -\frac{k}{U}+U c+\frac{k}{2}-2 c=\frac{5}{12} U^{2}-U-12=0 \end{aligned}$$
Attempt at integral and correct use of correct limits.
Simplify to quadratic equation, may be in terms of k and
c.

U=4
$\mathrm{IQR}=4-\sqrt{3}$

A1 FT

FT their UQ - their LQ, exact values only.

5

$\mathrm{F}(x)=\frac{3}{28}\left(2 \mathrm{e}^{\frac{1}{2} x}-8 \mathrm{e}^{-\frac{1}{2} x}(+c)\right)$
+c not required.

$\mathrm{F}(x)=\frac{3}{28}\left(2 \mathrm{e}^{\frac{1}{2} x}-8 \mathrm{e}^{-\frac{1}{2} x}\right)+\frac{9}{14}$
Correct ranges including $0 \leqslant x \leqslant 2 \ln 3$ associated with their F(x) F(x)=0 for x<0 and 1 for $x>2 \ln 3$
Correct ranges including $0 \leqslant x \leqslant 2 \ln 3$ associated with their F(x)
F(x)=0 for x<0 and 1 for $x>2 \ln 3$
No gaps in range.

3

$\mathrm{G}(y)=\frac{3}{28}\left(2 y-\frac{8}{y}+6\right)$
y substituted into their F.

$1 \leq y \leq 3$
Seen anywhere, condone $1 \leqslant y \leqslant \mathrm{e}^{\ln 3}$.

$\mathrm{g}(y)=\frac{3}{28}\left(2+\frac{8}{y^{2}}\right)$
Correct expression, not containing logs.

3

$\frac{3}{28}\left(2 y-\frac{8}{y}+6\right)=\frac{3}{10}$
Their $\mathrm{G}(y)=\frac{3}{10}$$5 y^{2}+8 y-20=0$
Obtain from G(y) and solve quadratic equation to
find y.

y=1.35
Single correct answer.

3

$\mathrm{E}\left(Y^{4}\right)=\int_{1}^{3} \frac{3}{28}\left(2+\frac{8}{y^{2}}\right) y^{4} \mathrm{~d} y=\int_{1}^{3} \frac{3}{28}\left(2 y^{4}+8 y^{2}\right) \mathrm{d} y=\frac{3}{28}\left[\frac{2}{5} y^{5}+\frac{8}{3} y^{3}\right]$

1

Integrate $y^{4} \times$ their g(y). Limits must be 1 and 3.

17.8

A1

$\frac{89}{5}$

2