A Level Mathematics 9709 Paper 53, Oct/Nov 2023

$2k+6k+12k+20k=1,[k=\frac{1}{40}]$

Table with correct $X$ values and correct probabilities.

$[\text{E}(X)=][\text{E}(X)=]\frac{1\times 2+2\times 6+3\times 12+4\times 20}{40}=\frac{2+12+36+80}{40}$

$[\text{Var}(X)=\frac{1^2\times 2+2^2\times 6+3^2\times 12+4^2\times 20}{40}-(\text{their E}(X){)}^2=]$ $\frac{1\times 2+4\times 6+9\times 12+16\times 20}{40}-(\text{their }\frac{13}{4}{)}^2$ $[\frac{2+24+108+320}{40}-(\text{their }\frac{13}{4}{)}^2]$

$\text{E}(X)=\frac{13}{4},3\frac{1}{4},3.05\text{Var}(X)=\frac{63}{80},0.7875$

$[\text{P}(1.42<X<1.52)=]\text{P}(\frac{1.42-1.5}{0.05}<Z<\frac{1.52-1.5}{0.05})$

$[=\text{P}(-1.6<Z<0.4)=\mathrm{\Phi }(0.4)+\mathrm{\Phi }(1.6)-1]$ $=0.6554+0.9452-1\text{ or }0.6554-0.0548$

$=0.601$

$[\text{P}(X<0.9)=\text{P}(Z<\frac{0.9-0.75}{\sigma })=0.68]$

$\frac{0.9-0.75}{\sigma }=0.468$

$\sigma =0.321,\frac{25}{78}$

Either $\frac{2}{6}\times \frac{8}{15}\times \frac{7}{14}$ or $\frac{4}{6}\times \frac{6}{15}\times \frac{5}{14}$ seen

$\frac{q}{6}\times \frac{r}{15}\times \frac{r-1}{14}+\frac{6-q}{6}\times \frac{s}{15}\times \frac{s-1}{14}$

$[=\frac{56}{630}+\frac{60}{630}=\frac{4}{45}+\frac{2}{21}]=\frac{58}{315}\text{ or }0.184$

$\text{P}(\text{W \& R from bag B})=\frac{2}{3}\times \frac{6}{15}\times \frac{7}{14}+\frac{2}{3}\times \frac{7}{15}\times \frac{6}{14}\text{ or }2\times \frac{2}{3}\times \frac{6}{15}\times \frac{7}{14}[=\frac{4}{15}\text{ or }0.267]$

$\text{P}(\text{WR or RW})=\text{P}(\text{W \& R from bag A})+\text{P}(\text{W \& R from bag B})$ $=a\times \frac{2}{6}\times \frac{8}{15}\times \frac{4}{14}+a\times \frac{4}{6}\times \frac{6}{15}\times \frac{7}{14}\text{ or }=a\times \frac{2}{6}\times \frac{8}{15}\times \frac{4}{14}+\text{their P}(\text{W \& R from bag B})$

$\frac{\text{their identified P}(\text{W \& R from bag B})}{\text{their identified P}(\text{WR or RW})}$

$=\frac{168}{232},\frac{21}{29}\text{ or }0.724$

At least 3 points plotted accurately at class upper end points (40,0) (60,14) (65,38) (70,60) (85,106) (100,120). Linear cumulative frequency scale $0⩽\text{cf}⩽120$ and linear weight scale $40⩽\text{weight(kg)}⩽100$ with at least 3 values identified on each axis. Condone scale reversed.

All points plotted correctly, curve drawn (within tolerance) and joined to (40,0). Axes labelled cumulative frequency (cf), weight (w) and kg (kilograms) – or a suitable title.

$[120\times 0.65=]78\text{ seen}$

$76\text{ [kg]}$

Frequencies: [0] 14 24 22 46 14

Midpoints: 20 50 62.5 67.5 77.5 92.5

$\text{Mean}=\frac{0\times 20+14\times 50+24\times 62.5+22\times 67.5+46\times 77.5+14\times 92.5}{120}$

$=71.2$

$\text{Variance}=\frac{0\times {20}^2+14\times {50}^2+24\times {62.5}^2+22\times {67.5}^2+46\times {77.5}^2+14\times {92.5}^2}{120}-{71.2}^2$

$\text{Standard deviation}=11.8$

Method 1: $[\text{P}(2⩽X⩽6)=\text{P}(X⩽6)-\text{P}(X⩽1)=]1-(0.7{)}^6-(1-0.7)$ OR Method 2: $\text{P}(X=2,3,4,5,6)=0.7\times 0.3+{0.7}^2\times 0.3+{0.7}^3\times 0.3+{0.7}^4\times 0.3+{0.7}^5\times 0.3$

$=0.582$

$3\frac{1}{3}$

One term seen ${}^5{\text{C}}_x(p{)}^x(1-p{)}^{5-x},0<p<1,x\mathrm{\ne }0,5$.

Correct expression, accept unsimplified, no terms omitted leading to final answer.

$=0.163,\frac{4077}{25000}$

$[\text{Mean}=75\times 0.3=]22.5$ $[\text{Var}=75\times 0.3\times 0.7=]15.75$

Substituting their $\mu$ and $\sigma$ into $\pm$ standardisation formula (any number for 20.5), not ${\sigma }^2$ not $\sqrt{\sigma }$.

Using continuity correction 19.5 or 20.5 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.

$=0.693$

Method 1: $6!\times a$, $a$ integer > 1. OR Alternative: $c\times d\times e\times f\times g\times h$, $2⩽c,d,e,f,g,h\text{ (different integers)}⩽12$

Method 1: $b\times 2^6$, $b$ integer $⩾1$. OR Alternative: Correct unsimplified.

$=46080$

$5!\times 5!\times k$, $k$ positive integer, 1 may be implied (no adding/subtracting).

$=57600$

Correct probability / value for one identified scenario.

Method 1: Denominator $12\times 11$ for all probabilities, (1, 2 or 3 scenarios). OR Method 2: ${}^{12}{\text{C}}_a{\times }^{12-a}{\text{C}}_b$, $a=3,4,5;b=3,4,5(a\mathrm{\ne }b)$

Method 1: 3 correct probabilities, accept unsimplified. OR Method 2: 27720 Seen alone or as denominator of probability – accept unsimplified.

Method 1: Adding probabilities for 3 correct scenarios. OR Method 2: Values of 3 correct scenarios added, accept unsimplified – or correct.

$\frac{19}{66},0.288$