A Level Mathematics 9709 Paper 52, Oct/Nov 2023

Using sum of probabilities = 1 to form an equation. Accept $p+r=0.45$ oe.

Use $\text{E}(X)=1.1$ to form an equation. Accept $p+2r=0.65$ oe. NB: These marks can be gained in either order; the second M may have an algebraic substitution.

$p=0.25,r=0.2$

Correct formula for variance method using their probability distribution table, $0<their\text{ P}(x)<1$.

$=1.19,1\frac{19}{100}$

$1-{0.8}^n,n=6,7$ or $0.2+0.2\times 0.8+0.2\times {0.8}^2+0.2\times {0.8}^3+0.2\times {0.8}^4+0.2\times {0.8}^5(+0.2\times {0.8}^6)$

$=0.738,\frac{11529}{15625}$

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

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

$=0.0327$

30 and 25.5, $25\frac{1}{2},\frac{51}{2}$ seen, allow unsimplified. May be seen in standardisation formula. $[\sigma =]5.049⩽\sigma ⩽5.05[0],\frac{\sqrt{102}}{2}$ implies correct variance. Correct notation is required.

Substituting their mean and their positive 5.04975 into $\pm$standardisation formula (any number for 40.5), not their ${\sigma }^2$ or $\sqrt{their\sigma }$.

Using continuity correction 39.5 or 40.5 in their standardisation formula.

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

$=0.0188$

Correct stem, ignore extra values (not in reverse, not split). If a split stem-and-leaf plot is used (i.e., stem values are repeated), the remaining B marks are available.

Correct Aces labelled on left, leaves in order from right to left and lined up vertically, no commas or other punctuation.

Correct Jets labelled on same diagram, leaves in order and lined up vertically, no commas or other punctuation.

Correct key for their diagram, need both teams labelled and ‘cm’ stated at least once here, or in leaf headings or title.

Median = 173 [cm]

$[\text{IQR}=]181-168$

13 [cm]

Jets have a greater variety of heights. Jets have a wider range of height. Jets have a greater/larger/bigger/wider/’more’ spread of heights. Aces have a smaller variety of height etc…

Use of $\pm$ standardisation formula with 170, 166 and 10 substituted appropriately, condone ${10}^2,\sqrt{10}$, condone continuity correction $\pm 0.5$.

$[=\text{P}(Z<0.4)=]0.655$

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

Use of the $\pm$standardisation formula with $h$, 166, 10 and a $z$-value (not $1-z$-value), not ${10}^2,\sqrt{10}$, no continuity correction.

$h=168.53$

Use of the $\pm$standardisation formula with $0,\mu$ and $\frac{2}{3}\mu$ substituted for $\sigma$. Or use of the $\pm$standardisation formula with $0,\sigma$ and $\frac{3}{2}\sigma$ substituted for $\mu$.

$=\text{P}(Z>-1.5)$

$=0.933\text{ final answer}$

1st column, 2 branches identified X, Y with probabilities ½, ½ indicated.

2nd column (1st marble pick) of 4 branches identified R B R B (oe) and probabilities $\frac{7}{10},\frac{3}{10},\frac{4}{5},\frac{1}{5}$ indicated appropriately.

3rd column (2nd marble pick) of 8 branches identified R B R B R B R [B] (oe) and probabilities $\frac{7}{10},\frac{3}{10},\frac{8}{10},\frac{2}{10},\frac{4}{5},\frac{1}{5},1,[0]$. Condone omission of YBB branch if YBR branch is fully correct. Ignore any additional columns of branches.

$[\text{P}(\text{BB})=][\frac{1}{2}\times \frac{3}{10}\times \frac{2}{10}[+\frac{1}{2}\times \frac{1}{5}\times 0]=]\frac{6}{200}$ seen. Accept unsimplified. FT from 6(a) unsimplified only with 3 term probabilities.

Either $[\text{P}(\text{XRR})=]\frac{1}{2}\times \frac{7}{10}\times \frac{7}{10}$ or $[\text{P}(\text{YRR})=]\frac{1}{2}\times \frac{4}{5}\times \frac{4}{5}$ seen. FT from 6(a) unsimplified only with 3 term probabilities.

$[\text{P}(\text{BB})+\text{P}(\text{XRR})+\text{P}(\text{YRR})=]$ $their\frac{6}{200}+their\frac{49}{200}+their\frac{16}{50}$ Accept unsimplified, consistent with tree diagram if not clearly identified by notation.

$=\frac{119}{200},0.595$

FT from their 6(a) and their 6(b) with 3 term probabilities unsimplified only or correct. Accept $\frac{\frac{4}{50}+\frac{1}{10}}{\frac{81}{200}},\frac{\frac{2}{25}+\frac{1}{10}}{\frac{81}{200}},\frac{0.08+0.1}{0.405}$.

$=\left[\frac{\frac{9}{50}}{\frac{81}{200}}\right]=\frac{4}{9},0.\dot{4}44$ Accept $\frac{36}{81},0.\dot{4}$.

$\frac{5!\times 4!}{e}$, $e$ a positive integer, 1 can be implied. No other terms on numerator. No addition etc.

$\frac{f}{2!\times g!}$, $f$ a positive integer, $g=1,2$. No other terms on denominator.

720

$\frac{7!}{2!}-e$, ${}^7{\text{P}}_5-e$, $e$ a positive integer or 0. OR $5!\times s$, $s$ a positive integer, 1 may be implied. OR Correct outcome/value for 1 identified scenario, accept unsimplified, www.

$d-\frac{6!}{r!}$, $d>720,r=1,2$. OR $t\times \frac{6\times 5}{u}$, $t$ a positive integer $>1,u=1,2$. OR Add values of 5 correct scenarios, no incorrect/repeated scenarios.

$=1800$

At least one correct unsimplified expression for an identified scenario. OR Numerator for at least one correct unsimplified expression for an identified scenario, either $\frac{2\times 2\times 5\times 4\times 12}{a\times b\times c\times d}$ or $\frac{2\times 2\times 1\times 5\times 12}{a\times b\times c\times d}$ seen, $6⩽a,b,c,d⩽9$. OR $a\times {(}^7{\text{C}}_3{-}^5{\text{C}}_3),a=1,2$. OR Either 10 correct outcomes for ADD^ listed or 40 correct outcomes for AD^^ listed.

$[\text{Total}=]40+10$ or $50\text{ soi}$ www OR $\frac{2\times 2\times 5\times 4\times 12}{a\times b\times c\times d}+\frac{2\times 2\times 1\times 5\times 12}{a\times b\times c\times d},6⩽a,b,c,d⩽9$ OR $[\text{Total}=]50$ www OR 50 stated www

$[\text{Total number of selections}={]}^9{\text{C}}_4[=126]$ OR $\frac{p}{9}\times \frac{q}{8}\times \frac{r}{7}\times \frac{s}{6}$ present in all scenarios attempted, accept $\frac{t}{3024},t<3024$. OR 126 stated or correct outcomes listed.

$[\text{Probability}=]\frac{50}{126},\frac{25}{63}$ FT $\frac{their\text{ attempted }40+10}{126}$ or $\frac{their\text{ attempted }960+240}{3024}$.