A Level Further Mathematics 9231 Paper 43, May/June 2025

${\text{H}}_0$: uniform distribution fits the data. ${\text{H}}_1$: uniform distribution does not fit the data.

Expected frequencies are 40, 40, 40.

$\frac{(38-40{)}^2}{40}+\frac{(52-40{)}^2}{40}+\frac{(30-40{)}^2}{40}[=0.1+3.6+2.5]$ OR $\frac{{38}^2}{40}+\frac{{52}^2}{40}+\frac{{30}^2}{40}-120$

6.2

[2 degrees of freedom, so critical value is 5.991.] '6.2' > 5.991, reject ${\text{H}}_0$ (significant).

Sufficient evidence to suggest that the scientist’s claim is incorrect. OR Sufficient evidence eye colour is not uniformly distributed. OR Sufficient evidence to suggest that uniform distribution is not a good fit.

$s_x^2=\frac{1}{39}(720-\frac{{168}^2}{40})=0.36923[=\frac{24}{65}]$ $s_y^2=\frac{1}{59}(900-\frac{{228}^2}{60})=0.56949[=\frac{168}{295}]$

Both unbiased estimates correct to 3sf or better.

$s^2=\frac{\text{’0.36923’}}{40}+\frac{\text{’0.56949’}}{60}$

$s^2=0.0187[=\frac{359}{19175}]$

$\text{CI}=(\frac{168}{40}-\frac{228}{60})\pm 1.645\text{’s’}$

1.645 (accept 1.64 or 1.65) used in CI formula.

$[0.175,0.625]$

${\int }_0^{1}kx\text{d}x+{\int }_1^{8}k(8-x)\text{d}x=1[{\left[\frac{1}{2}k{x}^2\right]}_0^1+{[8kx-\frac{1}{2}k{x}^2]}_1^8=1]$ $\frac{1}{2}k+(64k-32k)-(8k-\frac{1}{2}k)=1$ OR $\frac{1}{2}\times 1\times k+\frac{1}{2}\times 7\times 7k=1$

$25k=1$, leading to $k=\frac{1}{25}$.

${\int }_0^{1}kx\text{d}x+{\int }_1^{m}k(8-x)\text{d}x=\frac{1}{2}$ OR $\frac{1}{2}\times (8-m)\times k(8-m)=\frac{1}{2}$

$m^2-16m+39=0$ or $(8-m{)}^2=25$

$m=3$

$\text{F}(x)=\{\begin{array}{ll}\frac{1}{50}{x}^2& [0⩽x<1]\\ \frac{8}{25}x-\frac{1}{50}{x}^2-\frac{7}{25}& [1⩽x⩽8]\end{array}$ $[\text{F}(x)=0\text{ for }x<0\text{ and F}(x)=1\text{ for }x>8]$

For $-\frac{7}{25}$.

$\text{G}(y)=\{\begin{array}{ll}\frac{1}{50}{y}^6& [0⩽y<1]\\ \frac{8}{25}{y}^3-\frac{1}{50}{y}^6-\frac{7}{25}& [1⩽y⩽2]\end{array}$ $[\text{G}(y)=0\text{ for }x<0\text{ and G}(y)=1\text{ for }y>2]$

Differentiation of their $\text{G}(y)$.

$\text{g}(y)=\{\begin{array}{ll}\frac{3}{25}{y}^5& 0⩽y<1\\ \frac{24}{25}{y}^2-\frac{3}{25}{y}^5& 1⩽y⩽2\\ 0& \text{otherwise.}\end{array}$

${\text{H}}_0$: population medians are equal for under 25s and over 50s (or $m_{<25}=m_{>50}$). ${\text{H}}_1$: population median for under 25s < population median time for over 50s (or $m_{<25}<m_{>50}$).

Attempt at ranks using the integers 1 to 14 (ranks may be reversed).

Sum of ranks for sample of size 6: $R_6=60$

$6(6+8+1)-\text{’}{R}_6\text{’}[=30]$

$[W=\min (60,30)=]30$

Critical value for $n=6,m=8$ is 31.

'30' $⩽31$, reject ${\text{H}}_0$ (significant).

Sufficient evidence to suggest that older people take longer to react. OR Sufficient evidence to support researcher’s claim.

${\text{H}}_0:{\mu }_B-{\mu }_A=k,{\text{H}}_1:{\mu }_B-{\mu }_A>k$

Differences, $d$: 19, 17, 8, 16, 9, 23, 6, 5.

$\sum d=103,\sum d^2=1641,$ $\bar{d}=12.875,s_d^2=\frac{1}{7}(1641-\frac{1}{8}({103}^2))=44.982[=\frac{2519}{56}]$

$t=\frac{12.875-k}{\sqrt{\frac{s_d^2}{8}}}$

Correct evaluation of test statistic formula.

$\frac{12.875-k}{\sqrt{\frac{s_d^2}{8}}}⩾1.895$

$[0⩽]k⩽8.38$

Population of differences are normally distributed.

Probabilities: $\frac{1}{15},\frac{7}{15},\frac{7}{15}$ for 0, 1, 2 heads.

${\text{G}}_X(t)=\frac{1}{15}+\frac{7}{15}t+\frac{7}{15}{t}^2$

Probabilities: $\frac{8}{15},\frac{7}{15}$ for 1, 2 colours.

${\text{G}}_Y(t)=\frac{8}{15}t+\frac{7}{15}{t}^2$

$X$ and $Y$ are not independent.

Correct 3 cases identified correctly where probabilities are not zero from 1,2 colours: $(x,y)=(0,1),(1,2)\text{ or }(2,1)$.

At least one non-zero probability correct. Allow for $\frac{1}{15},\frac{7}{15},$ or $\frac{14}{15}$.

All probabilities correct.

${\text{G}}_Z(t)=\frac{1}{15}t+\frac{14}{15}{t}^3$

Attempt to differentiate ${\text{G}}_Z(t)$ and find ${\text{G}}_Z^{\mathrm{\prime }}(1)$. FT their ${\text{G}}_Z(t)$.

$\frac{43}{15}[=2.87]$