A Level Mathematics 9709 Paper 52, Oct/Nov 2025

$[(0.6{)}^4(0.4)=]0.05184,\frac{162}{3125}$

$[1-(1-{0.6}^6)=]{0.6}^6$ Or $1-(0.4+0.4\times 0.6+0.4\times {0.6}^2+0.4\times {0.6}^3+0.4\times {0.6}^4+0.4\times {0.6}^5)$

$=0.0467,\frac{729}{15625}$

Table with correct $x$ values and at least 1 correct probability.

A second correct probability correctly linked to the correct $x$ value, need not be in table, accept un-simplified.

Two more correct probabilities correctly linked to the correct $x$ values, need not be in table, accept un-simplified.

5 correct probabilities linked with correct $x$ values.

$[\text{E}(X)=]1\times \frac{5}{15}+2\times \frac{4}{15}+3\times \frac{3}{15}+4\times \frac{2}{15}+5\times \frac{1}{15}[\frac{5+8+9+8+5}{15}=\frac{35}{15},\frac{7}{3}]$

$[\text{Var}(X)=]1^2\times \frac{5}{15}+2^2\times \frac{4}{15}+3^2\times \frac{3}{15}+4^2\times \frac{2}{15}+5^2\times \frac{1}{15}-(\text{their }\frac{35}{15}{)}^2[\frac{[1\times ]5+4\times 4+9\times 3+16\times 2+25[\times 1]}{15}-\frac{49}{9}]$

$=\frac{14}{9},1.56$

$\text{P}(\frac{182.4-187.4}{6.4}<Z<\frac{192.4-187.4}{6.4})$

$[=\text{P}(-0.78125<Z<0.78125)=2\mathrm{\Phi }(0.7813)-1]=2\times 0.7826-1$ Or $0.7826-(1-0.7826)$ Or $0.7826-0.2174$ Or $(0.7826-0.5)+(0.7826-0.5)$ Or $2\times 0.2826$

$0.5652$

$[\text{Expected number}=124\times 0.5652=70.08,]70$

$0.7385<z<0.7395$ or $-0.7395<z<-0.7385$ seen.

$\frac{170.3-172.7}{\sigma }=-0.739$ or $\frac{172.7-170.3}{\sigma }=0.739$

$\sigma =3.25$

Either $\frac{6}{18}\times \frac{x-12}{x}$ or $\frac{10}{18}\times \frac{9}{x}$ seen.

Correct equation formed. Fractions may be simplified.

Rearrange probabilities to form quadratic equation and solve to find a value for $x$.

$[x=]42$

Two identified un-simplified outcomes with their $x$ substituted.

Add probabilities, $0<p<1$, for 3 correct scenarios, no incorrect/repeated scenarios.

$=0.365,\frac{23}{63}$

At least 4 correct frequency densities (F/CW, e.g. $\frac{12}{10},\frac{38}{10},\frac{68}{5},\frac{76}{5},\frac{46}{20}$). Accept un-simplified. May be read from graph if scale sufficiently accurate.

All heights correct on graph. Daylight rule applied full width of bar. Minimum scale: FD axis uses at least ½ the grid.

Bar ends at 0.5, 10.5, 20.5, 25.5, 30.5, 50.5, with a linear scale, 0.5 to 50.5, and at least 3 values indicated 'linearly'. Daylight rule applied $0\text{ cm}⩽\text{FD}⩽1\text{ cm}$ vertically.

Axes labelled frequency density (OE e.g. fd) and time, minutes (OE e.g. t, min). FD scale starts at 0 with a linear scale and at least 3 values indicated 'linearly'. (condone 0.5 on time scale and no value on FD scale). Minimum: time scale use at least ½ the grid. Axes can be reversed.

At least 4 correct midpoints: Midpoints 5.5, 15.5, 23, 28, 40.5. May be seen by data table, accept un-simplified.

Correct mean formula using their 5 midpoints (must be within class, not upper bound, not lower bound), condone 1 error.

$=25.875$

$[(\frac{5}{7}{)}^{10}=]0.0346,\frac{9765625}{282475249}$

One term ${}^{10}{\text{C}}_x(p{)}^x(1-p{)}^{10-x}$. With $0<p<1,x\mathrm{\ne }0$ or 10.

Correct un-simplified expression. Allow 10 for ${}^{10}{\text{C}}_1$.

0.838

56 and 48 seen, allow un-simplified, may be seen in the standardisation formula.

Substituting their $\mu$ and positive $\sigma$ into the $\pm$ standardising formula (any number for 65.5), allow ${\sigma }^2$ or $\sqrt{\sigma }$.

Use continuity correction 64.5 or 65.5 in $\pm$standardisation formula.

Appropriate area $\mathrm{\Phi }$, from final process, must be a probability.

= 0.0851 final answer

$\frac{8!}{2!}$ seen alone (not multiplied/divided). (Method 1) OR $6!\times e,1⩽e⩽42$. (Method 2)

$b-7!,5040<b$. (Method 1) OR $f\times \frac{7\times 6}{2},1<f$ accept ${}^7{\text{C}}_2$ or $\frac{{}^7{\text{P}}_2}{2}$. (Method 2)

$\frac{8!}{c!}-\frac{7!}{c!d!},c=1,2,d=1,3$. (Method 1) OR $\frac{6!}{g!}\times \frac{7\times 6}{h},g=1,3,h=1,2,3$. (Method 2)

= 15120

$\frac{8!}{3!}\times i,i⩾1$. (Method 1) OR ${}^8{\text{P}}_5\times k,k⩾1$. (Method 2)

$\frac{8!}{j!}\times 4,j=1,2,3$. (Method 1) OR ${}^8{\text{P}}_m\times \frac{4!}{3!}$ or ${}^8{\text{P}}_m\times 4,m=3,4,5$ or ${}^8{\text{C}}_5\times \frac{4!}{3!}$ or ${}^8{\text{C}}_5\times 4$. (Method 2)

26880

$\frac{3}{10}\times \frac{2}{9}+\frac{2}{10}\times \frac{1}{9}$ seen (Method 1) OR $\frac{3}{d}\times \frac{7}{d}+\frac{2}{d}\times \frac{8}{d}+\frac{5}{d}\times \frac{9}{d},d=9\text{ or }10$ (Method 2) OR ${}^3{\text{C}}_2+{}^2{\text{C}}_2$ seen (Method 3) OR ${}^3{\text{C}}_1\times {}^2{\text{C}}_1+{}^3{\text{C}}_1\times {}^5{\text{C}}_1+{}^5{\text{C}}_1\times {}^2{\text{C}}_1+{}^5{\text{C}}_2$ seen (Method 4).

$1-(\frac{3}{d}\times \frac{2}{d}+\frac{2}{d}\times \frac{1}{d}),d=9\text{ or }10$ (Method 1) OR $1-\frac{f}{{}^{10}{\text{C}}_2},1⩽f<45$ (Method 3) OR $\frac{g}{{}^{10}{\text{C}}_2},1⩽g<45$ (Method 4).

$=\frac{41}{45},0.911$