A Level Further Mathematics 9231 Paper 43, May/June 2026

$$\bar{x}=\frac{192.2}{9}$$

$$s^2=\frac{1}{8}(4127.64-\frac{{192.2}^2}{9})$$

$$t=\frac{\frac{192.2}{9}-20}{\sqrt{\frac{s^2}{9}}}$$

2.39

'2.39' > 2.306, reject ${\text{H}}_0$ / significant.

There is sufficient evidence to support that $\mu >20$.

Population of differences may not be normally distributed.

$${\text{H}}_0\text{: difference in population medians}=0$$ $${\text{H}}_1\text{: difference in population medians}\mathrm{\ne }0$$

Differences: 4, 8, 15, 3, –1, –2, 5, –7, 11, 6

Signed ranks: 4, 8, 10, 3, –1, –2, 5, –7, 9, 6

[$P=45,Q=10$] $T=10$

Critical value = 8

'10' > '8', accept ${\text{H}}_0$ / do not reject ${\text{H}}_0$ / not significant.

Insufficient evidence to support that there is (on average) a difference between the marks (awarded by the teachers).

$$G_X^{\mathrm{\prime }}(t)=0.3+0.8t+0.6t^2$$

$$\text{E}(X)=1.7$$

$$G_X^{\mathrm{\prime }\mathrm{\prime }}(t)=0.8+1.2t$$

$$\text{Var}(X)=2+1.7-{1.7}^2$$

0.81

Correct coefficients in a 4-term polynomial.

Correct powers of t.

$${\int }_{1.5}^2\frac{2}{x^3}\text{d}x={[-\frac{1}{x^2}]}_{1.5}^2=\frac{7}{36}$$

$$a=\frac{7}{36}\times 200=38.89$$

5.89 or 5.90

$${\text{H}}_0\text{: f is a good fit for the data.}$$ $${\text{H}}_1\text{: f is not a good fit for the data.}$$

Combine 5th and 6th columns: (16, 9.72) OR Combine final two columns: (20, 16.33)

5th and 6th columns combined: $$\frac{(97-111.11{)}^2}{111.11}+\cdots +\frac{(16-9.72{)}^2}{9.72}+\frac{(14-12.50{)}^2}{12.50}$$ $$[=1.792+0.389+1.389+2.786+4.057+0.18]$$ Final 2 columns combined: $$\frac{(97-111.11{)}^2}{111.11}+\cdots +\frac{(10-5.89{)}^2}{5.89}+\frac{(20-16.33{)}^2}{16.33}$$ $$[=1.792+0.389+1.389+2.786+2.868(2.849)+0.825]$$

5th and 6th columns combined: 10.6 Final two columns combined: 10.0 or 10.1

'10.6' > 9.236, reject ${\text{H}}_0$ / significant.

Sufficient evidence to suggest that f is not a good fit (for the data).

$${\text{H}}_0:{\mu }_x={\mu }_y{\text{H}}_1:{\mu }_x<{\mu }_y$$

$$s_x^2=\frac{1}{43}[1.21-\frac{{6.63}^2}{44}][=4.9065\times {10}^{-3}]$$ or $$s_y^2=\frac{1}{51}[1.65-\frac{{8.56}^2}{52}][=4.7234\times {10}^{-3}]$$

$$\frac{s_x^2}{44}+\frac{s_y^2}{52}=\frac{4.9065\times {10}^{-3}}{44}+\frac{4.7234\times {10}^{-3}}{52}$$

$$2.023\times {10}^{-4}$$

$$z=\frac{\frac{6.63}{44}-\frac{8.56}{52}}{\sqrt{\frac{s_x^2}{44}+\frac{s_y^2}{52}}}=-0.98(0)$$

Accept –0.977 to –0.980.

'–0.980' > –1.282, accept ${\text{H}}_0$ / do not reject ${\text{H}}_0$ / not significant. OR $$\text{P}(Z<-0.980)=0.1635>0.1$$, accept ${\text{H}}_0$ / do not reject ${\text{H}}_0$ / not significant.

Insufficient evidence to support Rhiannon’s claim.

$$(\frac{6.63}{44}-\frac{8.56}{52})\pm 1.645\sqrt{\frac{s_x^2}{44}+\frac{s_y^2}{52}}$$

1.645

$$[-0.0373,0.00947]$$

$$\text{F}(L)=0.25\text{, so }\frac{1}{12}(x^2+4x+4)=\frac{1}{4}$$

$$L=\sqrt{3}-2$$

$$[\text{f}(x)=]\{\begin{array}{ll}\frac{1}{6}x+\frac{1}{3}& [-2⩽x<0]\\ \frac{1}{5}{x}^2+\frac{1}{15}& [0⩽x⩽2]\\ [0& \text{otherwise}]\end{array}$$

$$[\text{E}(X^2)=]{\int }_{-2}^0{x}^2(\frac{1}{6}x+\frac{1}{3})\text{d}x+{\int }_0^2{x}^2(\frac{1}{5}{x}^2+\frac{1}{15})\text{d}x={[\frac{1}{24}{x}^4+\frac{1}{9}{x}^3]}_{-2}^0+{[\frac{1}{25}{x}^5+\frac{1}{45}{x}^3]}_0^2$

$$[\text{E}(X^4)=]{\int }_{-2}^0{x}^4(\frac{1}{6}x+\frac{1}{3})\text{d}x+{\int }_0^2{x}^4(\frac{1}{5}{x}^2+\frac{1}{15})\text{d}x={[\frac{1}{36}{x}^6+\frac{1}{15}{x}^5]}_{-2}^0+{[\frac{1}{35}{x}^7+\frac{1}{75}{x}^5]}_0^2$

$$[\text{Var}(X^2)=]\frac{6992}{1575}-{\left(\frac{42}{25}\right)}^2$

1.62

$$\text{P}(X^2<1)=\text{P}(-1<X<1)=\text{F}(1)-\text{F}(-1)[=\frac{7}{15}-\frac{1}{12}]$$

$$\frac{23}{60}$$