A Level Mathematics 9709 Paper 52, May/June 2023

$[3k+3k+8k=1,\text{so }k=]\frac{1}{14}$

Table with correct values of $x$, and at least one correct probability linked with outcome.

The outcomes in the table must be –2, 2 and 3. 2 further correct probabilities in table or 3 correct probabilities not in table linked to outcomes, or 3 correct FT probabilities in table using their $k$, or 3 incorrect probabilities summing to 1 in table if $k$ not stated.

$[\text{E}(X)=-2\times \frac{3}{14}+2\times \frac{3}{14}+3\times \frac{8}{14}=-\frac{6}{14}+\frac{6}{14}+\frac{24}{14}]$

Appropriate variance formula using their $(\text{E}(X){)}^2$ value.

$\text{E}(X)=\frac{12}{7},1.71,1\frac{5}{7}$ $\text{Var}(X)=\frac{192}{49},3.92,3\frac{45}{49}$

$[\text{P}(\text{no rain})=0.6\times (0.8{)}^3=]0.3072,\frac{192}{625}$

$0.6\times 0.8\times 0.2$

$=0.096[0],\frac{12}{125}$

Correct probability for one clearly identified outcome evaluated accept unsimplified.

Add 4 probability values, $0<p<1$, for appropriate identified scenarios. Accept unsimplified.

$0.2112,\frac{132}{625}$

Median = 2710

$2840-2610$

$230$

All 5 key values for $B$ plotted accurately in standard format using a linear scale with 3 identified values. Labelled $B$. Scale at least $1\text{ cm}=\mathrm{\$}100$.

All 5 key values for $A$, FT from (a), plotted accurately in standard format using a linear scale with 3 identified values. Labelled $A$. Scale at least $1\text{cm}=\mathrm{\$}100$

Whiskers not through box for both, not drawn at corners of boxes, single linear scale for the diagram and labelled 'salaries' (oe) and $.

Mean less appropriate than median because of extreme value for company $B$ [at $\mathrm{\$}3090$]. / No, extreme value in company B. / No, $\mathrm{\$}3090$ is an anomaly.

$[\text{P}(X=4)=(0.8{)}^3(0.2)=]0.1024,\frac{64}{625}$

$[\text{P}(X<6)=]1-{0.8}^d,d=5,6$ or $[\text{P}(X<6)=]\left(\frac{1}{5}\right)+\left(\frac{4}{5}\right)\left(\frac{1}{5}\right)+{\left(\frac{4}{5}\right)}^2\left(\frac{1}{5}\right)+{\left(\frac{4}{5}\right)}^3\left(\frac{1}{5}\right)+{\left(\frac{4}{5}\right)}^4\left(\frac{1}{5}\right)$

$=0.672,\frac{2101}{3125}$

$[\text{P}(X>0\cap X\mathrm{\ne }2)=]\frac{14}{25},0.56[0]$ seen as numerator or denominator of conditional probability fraction (or $[\text{Number of outcome }(X>0\cap X\mathrm{\ne }2)=]14$ seen as numerator or denominator).

$[\text{P}(X\mathrm{\ne }2)=]\frac{19}{25},0.76[0]$ seen as denominator of conditional probability fraction (or $[\text{Number of outcome }(X\mathrm{\ne }2)=]19$ seen as denominator).

Final answer $=\frac{14}{19},0.737$

One term ${}^9{\text{C}}_x(p{)}^x(1-p{)}^{9-x},0<p<1,0<x<9$.

$1-({}^9{\text{C}}_0(1-p{)}^9{+}^9{\text{C}}_1(p{)}^1(1-p{)}^8{+}^9{\text{C}}_2(p{)}^2(1-p{)}^7),0<p<1$ or ${}^9{\text{C}}_3(p{)}^3(1-p{)}^6{+}^9{\text{C}}_4(p{)}^4(1-p{)}^5+\cdots {+}^9{\text{C}}_8(p{)}^8(1-p{)}^1{+}^9{\text{C}}_9(p{)}^9,0<p<1$.

$0.371$

Use of $\pm$ standardisation formula once with 16.5, 0.6 and either 15.4 or 16.8 substituted.

Calculating the appropriate probability area (leading to their final answer, expect > 0.5). $0.6915-(1-0.9666)$ or $(0.6915-0.5)+(0.9666-0.5)$ OE are alternatives.

$=0.658$

$[\text{Expected number}=]0.6581\times 150$ $=98,99$

$0.5825<z⩽0.583$ or $-0.583⩽z<-0.5825$ seen.

Use of the $\pm$ standardisation formula with 17.1, 18.4, $\sigma$ and a $z$-value (not 0.28, 0.72, 0.4175, 0.2358, 0.7642, 0.6103, 0.3897, ...). Condone continuity correct $\pm 0.05$, not ${\sigma }^2,\sqrt{\sigma }$.

$\sigma =2.23$

$86.4,84\frac{2}{5}$ and $24\frac{24}{125},24.192$ to at least 3SF seen, allow unsimplified. May be seen in standardisation formula. ($4.918⩽\sigma ⩽4.919$ implies correct variance) Incorrect notation is penalised.

Substituting their mean (not 18.4) and their positive 4.9185 into $\pm$ standardisation formula (any number for 79.5), condone their ${4.918}^2$ and $\sqrt{\text{their }4.918}$.

Using continuity correction 79.5 or 80.5 in their standardisation formula.

Appropriate area $\mathrm{\Phi }$, from final process, must be a probability. Expect final answer < 0.5. Note: correct final answer implies this M1.

$0.0804$

${}^6{\text{C}}_e{\times }^8{\text{C}}_f{\times }^{11}{\text{C}}_g$, with $e+f+g=7$ seen.

Correct outcome/value for 1 identified scenario, accept unsimplified, www.

Add values of 3 correct scenarios. No incorrect scenarios, no repeated scenarios. Condone ${}^6{\text{C}}_e{\times }^8{\text{C}}_f{\times }^{11}{\text{C}}_g$, with $e+f+g=7$ to identify S, C, R.

$[\text{Total}=]26964$

$2!\times 3!\times 4!\times k,k$ an integer > 0. 1 can be implied.

$=1728$

$6!\times k,k$ an integer > 0. 1 can be implied (or for Method 4: $9!-7!r!-q,r$ an integer > 1, $q$ an integer $⩽0$. 0 and 1 may be implied).

$\frac{m!}{a!\times b!}\times 7\times n\times r;6⩽m⩽9;a=1,2;b=1,4;1⩽n,r⩽6,n\mathrm{\ne }r$ (or equivalent step in Methods 2, 3, 4).

$\frac{m!}{a!\times b!}\times 7\times 6\times 5;6⩽m⩽9;a=1,2;b=1,4$ (or equivalent step in Methods 2, 3, 4).

$151200$