A Level Mathematics 9709 Paper 51, Oct/Nov 2024

$[\text{P}(X<8)=]1-{\left(\frac{5}{6}\right)}^7$ or $\frac{1}{6}+\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)+{\left(\frac{5}{6}\right)}^2\left(\frac{1}{6}\right)+{\left(\frac{5}{6}\right)}^3\left(\frac{1}{6}\right)+{\left(\frac{5}{6}\right)}^4\left(\frac{1}{6}\right)+{\left(\frac{5}{6}\right)}^5\left(\frac{1}{6}\right)+{\left(\frac{5}{6}\right)}^6\left(\frac{1}{6}\right)$

= 0.721

${\left(\frac{5}{6}\right)}^6\times {\left(\frac{1}{6}\right)}^2\times 7$

0.0651

[Probs $6k,3k,2k,6k,11k$ so $28k=1,$ ] $k=\frac{1}{28}$

Table with correct outcomes and 2 correct probabilities

Fully correct table

$\text{E}(X)=[-2\times \frac{6}{28}+-1\times \frac{3}{28}+[0\times \frac{2}{28}]+2\times \frac{6}{28}+3\times \frac{11}{28}]=\left[\frac{15}{14}\right]$

$\text{Var}(X)=\frac{6\times (-2{)}^2+3\times (-1{)}^2+6\times 2^2+11\times 3^2}{28}-their{\left(\frac{15}{14}\right)}^2$

$=4.21,4\frac{41}{196}$

At least 4 points plotted within tolerance at upper bounds. Linear cf scale $0⩽\text{cf}⩽200$ and linear time scale $0⩽t⩽70$, with at least 3 values identified on each. Minimum scale uses at least ½ the grid.

All points plotted correctly. Curve drawn and joined to $(0,0)$. Axes labelled cumulative frequency (cf), time ($t$) and minutes (min) – or a suitable title.

Median = 33

$[\text{IQR}=]42-26$

16

At least 5 correct midpoints or 5 correct frequencies seen.

$\text{Mean}=\frac{18\times 7.5+28\times 20+42\times 27.5+52\times 35+36\times 45+24\times 60}{200}$

$=33.65,33\frac{13}{20}$

2 clearly identified unsimplified probabilities from $\text{P}(\text{HRR})$, $\text{P}(\text{TRR})$, $\text{P}(\text{HBB})$ and $\text{P}(\text{TBB})$ correct.

$\frac{4}{36}+\frac{3}{28}+\frac{1}{36}+\frac{6}{28}$

$=\frac{29}{63}$ or 0.460

$\frac{3}{4}\times \frac{4}{7}\times \frac{3}{6}$ seen as numerator of a fraction

$their\frac{1}{36}+their\frac{6}{28}$ seen as denominator of a fraction

$\frac{54}{61}$ or 0.885

$\text{P}(83<X<95)=\text{P}(\frac{83-90}{8}<Z<\frac{95-90}{8})$

$=\text{P}(-0.875<Z<0.625)$

$[\mathrm{\Phi }(0.625)+\mathrm{\Phi }(0.875)-1]=0.7340+0.8092-1$

$=0.543$

$[\text{Mean}=160\times 0.6=]96$ $[\text{Var}=160\times 0.6\times 0.4=]38.4$

$\text{P}(X<105)=\text{P}(Z<\frac{104.5-96}{\sqrt{38.4}})$

Use continuity correction 104.5 or 105.5 in their standardisation formula.

$[\text{P}(Z<1.372)=\mathrm{\Phi }(1.372)]$

$=0.915[0]$

1.282 or – 1.282 seen, CAO (critical value).

$-0.2535<z⩽-0.253$ or $0.253⩽z<0.2535$ seen.

One standardisation formula, not ${\sigma }^2$, or $\sqrt{\sigma }$, with 182.7 or 162.5 substituted correctly equated to a $z$ value (not 0.9, 0.1, 0.8159, 0.5398, 0.4, 0.6, 0.6554, 0.7257, ...).

Solve, obtaining values for $\mu$ and $\sigma$

$\mu =165.8,\sigma =13.2$

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

Correct unsimplified expression.

= 0.833

$\frac{8!}{2!2!}$

= 10080

$\frac{7!}{2!2!}-r$ or $t\times \frac{6\times 5}{2}$ / $t\times {}^6{\text{C}}_2$

$s-\frac{6!}{2!}$ or $\frac{5!}{2!}\times u$

= 900

$\text{TE}\mathrm{\_}\mathrm{\_}={}^2{\text{C}}_1\times {}^2{\text{C}}_1\times {}^5{\text{C}}_2=40$ or $\text{T}\wedge \wedge \wedge ={}^2{\text{C}}_1\times {}^7{\text{C}}_3=70$

$\text{TEE}\mathrm{\_}={}^2{\text{C}}_1\times {}^2{\text{C}}_2\times {}^5{\text{C}}_1=10$ or $\text{T}\ast \ast \ast ={}^2{\text{C}}_1\times {}^5{\text{C}}_3=20$

Probability $\frac{(40+10)}{{}^9{\text{C}}_4}$ or Probability $\frac{(70-20)}{{}^9{\text{C}}_4}$

$[\text{Percentage}=\frac{50}{126}\times 100=]39.7\mathrm{\%}$