A Level Mathematics 9709 Paper 52, May/June 2025

Table with correct values of $x$ and at least one non-zero probability correct.

Two more correct non-zero probabilities linked with correct outcome (3 correct probabilities present).

Four correct probabilities linked with the correct outcomes.

112 and 33.6 (CAO) seen, allow un-simplified.

Substituting their 112 and their 33.6 into the $\pm$standardising formula (any number for 120.5), allow ${\sigma }^2$ or $\sqrt{\sigma }$.

Use continuity correction 119.5 or 120.5 in their standardisation formula.

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

Final answer AWRT.

OE. $\frac{4}{16}\times \frac{4}{16}$ and either $\frac{12}{16}\times \frac{11}{15}$ or $\frac{12}{16}\times \frac{11}{16}$ seen.

ISW. CAO, (0.6125, $\frac{49}{80}$ must be seen).

$\frac{4}{16}\times \frac{12}{16}$ seen as a numerator of a single fraction.

$\left\{\text{their}(\frac{4}{16}\times \frac{12}{16})\text{ or correct}\right\}+\left\{\text{their}(\frac{12}{16}\times \frac{11}{15})\text{ from part (a) or correct}\right\}\text{ or }0.7375\text{ or }\frac{59}{80}$ seen as the denominator of a single fraction.

0.254237… to at least 3 SF.

OE. $0.3\times 0.25\times 0.45\times k$, $k$ an integer $>1$. E.g. $6=3!={}^3{\text{P}}_3$ etc.

CAO exact answer.

$1-{0.7}^d$, $d=8,9$, OR $0.3+0.3(0.7)+0.3(0.7{)}^2+0.3(0.7{)}^3+0.3(0.7{)}^4+0.3(0.7{)}^5+0.3(0.7{)}^6+0.3(0.7{)}^7[+0.3(0.7{)}^8]$

$0.942⩽p<0.9425$.

$(p{)}^2(1-p{)}^5\times k$, $0<p<1$, $k=5\text{ or }6$.

AWRT.

At least 5 points plotted accurately at class upper end points: (10, 34), (20, 86), (30, 142), (40, 208), (60, 265), (90, 300). Linear cf scale $0⩽\text{cf}⩽300$ and linear time scale $0⩽\text{time}⩽90$ with at least 3 values identified on each.

All points plotted correct, curve drawn, no line segments and joined to (0,0) passing within ½ square of points. Axes labelled cumulative frequency (cf), time (t) and minutes (min or m) – or a suitable title. Curve must be <300 for $0⩽t<90$.

Use of graph must be seen. Must be an increasing graph. Annotate this mark on 5(a) grid.

Strict FT, reading at their curve condone use of $t$.

Frequencies: [34,] 52, 56, 66, 57, 35. May be un-simplified and/or in variance calculation.

At least 5 correct midpoints seen, may be un-simplified, may be in calculation, may be by data table.

Correct un-simplified mean formula with their 6 midpoints (not upper bound, lower bound, upper limits, lower limits, class width, frequency density, frequency or cumulative frequency and must be within class) and their frequencies (not cw, cf, or fd), accept un-simplified. Condone 1 value error on numerator. The ‘table’ approach may be used with multiplications and additions implied by appropriate values.

WWW. $33.78⩽\text{mean}⩽33.8,\frac{2027}{60},33\frac{47}{60}$.

Correct un-simplified Variance formula with their midpoints and their frequencies – their mean${}^2$ for variance or standard deviation. Condone one value error on numerator. Condone their $\sum f$ from denominator of mean calculation if not 300.

$20.4⩽\sigma <20.45$ WWW.

$4!\times m,m\in \mathbb{Z},m⩾1$ OR $\frac{8!}{4!}\times r-5!\times m,m\in \mathbb{Z},m⩾3,r=1,2$.

$n\times {}^5{\text{C}}_1$ or $n\times {}^5{\text{C}}_4$ or $n\times 5$, allow $n\times {}^5{\text{P}}_1$ or $n\times \frac{{}^5{\text{P}}_4}{4!},n\in \mathbb{Z},n>1$ OR $k-5!\times 13\times r,k\in \mathbb{Z},k>1560,r=1,2$.

120

$\frac{8!}{4!}\times b,b\in \mathbb{Z},b⩾1$ OR $\frac{6!}{4!}\times b,b\in \mathbb{Z},b⩾1$.

$\frac{c!}{d}\times 6,c=8,9,10,d=1,2!,4!\text{ or }2!\times 4!$ OR $\frac{c!}{d}\times {}^8{\text{P}}_3,c=6,7,8,d=1,2!,4!\text{ or }2!\times 4!$.

10080

One identified outcome correct (excluding MMAAA or MAAAA in Method 1, or MAAAA in Method 2), accept un-simplified.

Five correct scenarios added (values do not need to be correct) [Method 1] OR Three correct scenarios added (values do not need to be correct) [Method 2].

[Total] 16

Use of $\pm$standardisation formula once with 74.8, 3.2 and either 72.4 or 76.3. No continuity correction, not ${\sigma }^2$, not $\sqrt{\sigma }$.

Calculating the appropriate probability area, must be a probability (assume a 4SF value in the calculation it is a probability, assume a 3SF value in the calculation it is not a probability). (leading to their final answer, expect < 0.5).

AWRT 0.454 WWW.

Strict FT their 4-figure probability $\times 120$ (check with calculator). One integer answer.

$0.841<z_1<0.843$ or $-0.843<z_1<-0.841$ seen.

$0.582<z_2<0.584$ or $-0.584<z_2<-0.582$ seen.

Use of the $\pm$standardisation formula once with $\mu ,\sigma$ equating to a $z$-value (not 0.20, 0.80, 0.28, 0.72, 0.5793, 0.4207, 0.6103, 0.3897, 0.7881, 0.2119, 0.7642, 0.2358, –0.417, 0.417, -0.158 0.158, etc). No continuity correction, not ${\sigma }^2$, not $\sqrt{\sigma }$.

Solve 2 equations in $\mu$ and $\sigma$ with an attempt at the elimination method, substitution method or other appropriate approach to obtain values for both $\mu$ and $\sigma$.

AWRT $\mu =0.192,\sigma =0.0119$.

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

Correct un-simplified expression.

$0.677<p⩽0.678$.