A Level Mathematics 9709 Paper 51, May/June 2024

[For all 45 values Mean =] $\frac{439+470}{45}+30$

$=50.2$

For all 45 values ${\text{Sd}}^2=\frac{12405+11346}{45}-{\left(\frac{909}{45}\right)}^2$

sd $[=\sqrt{119.76}]=10.9$

$\text{P}(23<X<35)=\text{P}(\frac{23-28}{3.3}<Z<\frac{35-28}{3.3})$

One fully correct $\pm$ standardisation formula.

$[=\mathrm{\Phi }(2.121)+\mathrm{\Phi }(1.5151)-1]=0.9830+0.9351-1$

$=0.918$

0.674 or $-0.674$ seen.

$\frac{7.6-8.5}{\sigma }=-0.674$

$\sigma =1.34$

At least four frequency densities calculated $\frac{f}{\text{cw}}$ $\left(\text{e.g. }\frac{16}{20}\right)$.

All bar heights correct on graph, not FT. Using their suitable linear scale with at least three values indicated, no lower than $1\text{ cm}=\text{fd }2$.

Bar ends at 150, 160, 170, 175, 195. Five bars drawn with a horizontal linear scale no lower than $1\text{ cm}=10\text{ cm}$, with at least three values indicated, $130⩽\text{horizontal scale}⩽195$.

Axes labelled frequency density (fd) height (h) and cm, OE, or an appropriate title. (Axes may be reversed)

$170⩽h<175-160⩽h<170$ UQ and LQ classes seen.

$175-160=15$

$1-(s{)}^3$, $s=\frac{3}{4}$ or $\frac{1}{4}$.

AG

${(1-\frac{37}{64})}^4=0.0317$

Correct probability for 1 identified scenario. Accept unsimplified.

Add values of 2 correct scenarios. Identification may be implied by correct unsimplified expressions (condone omission of $\times 3$). Values may not be probabilities.

Probability = 0.0914

One term ${}^7{\text{C}}_x(p{)}^x(1-p{)}^{7-x}$, with $0<p<1$, $x\mathrm{\ne }0$ or 7.

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

$=0.647$

$1-p^3$, $0<p<1$, $p=1-\text{their (a)}$, or correct.

0.956

48 and 9.6, $9\frac{3}{5}$, $\frac{48}{5}$ seen, allow unsimplified.

Substituting their $\mu$ and $\sigma$ into $\pm$ standardising formula (any number for 46.5), not their ${\sigma }^2$ or $\sqrt{\text{their }\sigma }$.

Use continuity correction 46.5 or 47.5 in *their* standardised formula.

Appropriate area $\mathrm{\Phi }$, from final process, must be a probability. Expect final answer < 0.5.

$=0.314$

Table with correct $X$ values and at least one probability. Values need not be in order, lines may not be drawn, may be vertical, $X$ and $\text{P}(X)$ may be omitted. Condone any additional $X$ values if probability stated as 0.

$\text{P}(X=1)$ or $\text{P}(X=2)$ correct and identified, need not be in table, accept unsimplified.

Two more correct and identified probabilities, need not be in table, accept unsimplified.

4 correct probabilities linked with correct outcomes, may not be in table. Decimals correct to at least 3sf.

Either $\frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times (1-p{)}^2$, not $\frac{2}{5}\times (1-p{)}^2$; or $\frac{1}{3}\times \frac{1}{4}\times \frac{1}{5}\times p^2$, not $\frac{1}{60}\times p^2$.

Equating and forming a 3 term quadratic equation. Their $\text{P}(Y=0)=6\times \text{their P}(Y=5)$.

$p=\frac{2}{3}$

$[\frac{8!}{2!3!}=]3360$

$\frac{6!}{3!}\times r-s$, $r=1,2$ and $s$ a positive integer (including 0).

$4!$ Seen either alone or in $t-4!$, $t$ an integer value > 24.

$\frac{6!}{3!}\times r-4!\times u$, $r=1,2$ and $u=1,2$.

$=96$

One correct calculation, unsimplified for an identified scenario containing 1, 2 and/or 1, 4.

Two correct outcomes evaluated, accept unsimplified.

Four correct scenarios added.

[Total number of selections = ] ${}^8{\text{C}}_3$ [= 56] Used as denominator of probability expression.

[Probability =] $[\frac{34}{{}^8{\text{C}}_3}=]\frac{17}{28}$, 0.607