A Level Mathematics 9709 Paper 53, May/June 2024

Table with correct $X$ values and at least one correct probability associated with the correct $X$ value.

Three other probabilities associated with correct $x$ values, need not be in table, accept unsimplified.

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

$\text{E}(X)=\frac{1\times 2+4\times 3+10\times 4+12\times 5+9\times 6}{36}=[\frac{2}{36}+\frac{12}{36}+\frac{40}{36}+\frac{60}{36}+\frac{54}{36}=\frac{14}{3}\text{ or }4.67]$

$\text{Var}(X)=\frac{1\times 2^2+4\times 3^2+10\times 4^2+12\times 5^2+9\times 6^2}{36}-{\left(\frac{14}{3}\right)}^2$

$[=\frac{824}{36}-\frac{196}{9}=22.89-21.78]=\frac{10}{9}$

$\text{P}(X\text{ even}\mid X>3)=\frac{\frac{10}{36}+\frac{9}{36}}{\frac{31}{36}}$

$=\frac{19}{31}$

$\text{P}(Z>\frac{1.93-1.64}{0.25})=\text{P}(Z>1.16)$

$1-0.8770$

$0.123$

$-0.441<z_1<-0.439$ or $0.439<z_1<0.441$ seen.

$z_2=0.674$ or $z_2=-0.674$ seen, CAO, critical value.

Use of the $\pm$standardisation formula once with $\mu ,\sigma$ equating to a $z$-value (not 0.33, 0.67, 0.25, 0.75, 0.6293, 0.5987, 0.7486, 0.7734, $(1-0.44)$, $(1-0.674)$). Condone continuity correct $\pm 0.005$, not ${\sigma }^2$, $\sqrt{\sigma }$.

Solve two equations in $\mu$ and $\sigma$ using the elimination method, substitution method or other appropriate approach to obtain values for both $\mu$ and $\sigma$.

$\mu =1.68,\sigma =0.269$

Correct structure and probabilities for Box A branches.

Completely correct structure and one correct probability for a Box B branch including label for G or Y.

Completely correct structure and second correct probability on a Box B branch including label G or Y.

Completely correct structure and remaining two probabilities correct on Box B branches, including labels for G or Y.

$\text{P}(\text{GG})+\text{P}(\text{YY})=\left(\frac{6}{9}\text{ or }\frac{3}{9}\right)\times \text{their}\left(\frac{5}{5+x}\right)+\left(\frac{3}{9}\text{ or }\frac{6}{9}\right)\times \text{their}\left(\frac{1+x}{5+x}\right)$

$\frac{6}{9}\times \frac{5}{5+x}+\frac{3}{9}\times \frac{1+x}{5+x}=\frac{8}{15}$ and arrange as a linear equation

Solve: $x=5$

Correct stem, ignore extra values (not in reverse, not split).

Correct Penguins labelled on left, leaves in order from right to left and lined up vertically (less than halfway to next column), no commas or other punctuation.

Correct Dolphins labelled on same diagram, leaves in order and lined up vertically (less than halfway to next column), no commas or other punctuation.

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

For Dolphins, median is 51

$\text{LQ}=48,\text{UQ}=60$

Correct end points for whiskers and diagram labelled Dolphins

Dolphins have more consistent times than Penguins or Penguins are faster (have faster times) than Dolphins

$[(0.35{)}^4\times 0.65=]0.00975$

$(0.35{)}^3\times (0.65{)}^2\times 4$

$=0.0725$

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

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

$=0.468$

$[\text{Mean}=]84\times 0.65=54.6$, $[\text{Var}=]84\times 0.65\times 0.35=19.11$

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

Use continuity correction 49.5 or 50.5 in their standardisation formula.

$\text{P}(Z>-0.9379)=\mathrm{\Phi }(0.9379)$

$0.826$

$[\frac{9!}{2!3!}=]30240$

$\frac{7!}{3!}-e,{}^7{\text{P}}_4-e,e$ a positive integer.

$f-\frac{5!}{r!},f>120,r=1,2$

$720$

Correct no of ways for two correct identified scenarios other than three Rs two Es.

No of ways for five correct identified scenarios added or correct.

${[}^9{\text{C}}_5(=126)$ seen as a denominator] Accept evaluated, accept ${}^9{\text{C}}_4$.

$\text{Probability}=\frac{21}{126}=\frac{1}{6}(0.167)$