A Level Mathematics 9709 Paper 51, Oct/Nov 2023

[IQR =] 31 – 23.7

7.3

[65% of 120 = ]78

28.5

$${\left(\frac{21}{36}\right)}^4\left(\frac{15}{36}\right)$$

$$=\frac{12005}{248832},0.0482$$

$$[\text{P}(X⩽4)=]1-{\left(\frac{21}{36}\right)}^4$$ or $$[\text{P}(X⩽4)=]\frac{15}{36}+\frac{15}{36}\times \frac{21}{36}+\frac{15}{36}\times {\left(\frac{21}{36}\right)}^2+\frac{15}{36}\times {\left(\frac{21}{36}\right)}^3$$

$$=\frac{18335}{20736},0.884$$

One term $${}^8{\text{C}}_x(q{)}^x(1-q{)}^{8-x},0<q<1,x\mathrm{\ne }0,8.$$

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

= 0.282

$$[\text{P}(X<76)=]\text{P}(Z<\frac{76-80.5}{6.6})$$

$$[=\mathrm{\Phi }(-0.6818)=1-\mathrm{\Phi }(0.6818)=]$$ $$1-0.7524=0.2476$$

24.8%

$$0.378⩽z<0.3791$$ or $$-0.3791<z⩽-0.378$$ seen.

$$\frac{x-80.5}{6.6}=0.378$$

$$x=83[.0]$$

$$\text{Mean}=150\times 0.4=60$$ $$\text{Var}=150\times 0.4\times 0.6=36$$

$$\text{P}(X>68)=\text{P}(Z>\frac{68.5-60}{\sqrt{36}})$$

Using continuity correction 67.5 or 68.5 in their standardisation formula.

$$\text{P}(Z>1.417)=1-\mathrm{\Phi }(1.417)$$ $$[=1-0.9217]$$

0.0783

At least 4 frequency densities calculated by $$\frac{f}{\text{cw}}$$, e.g. $$\frac{18}{20}$$ (condone $$\frac{f}{\text{cw}\pm 0.5}$$ if unsimplified). Accept unsimplified, may be read from graph using their scale, no lower than 1cm = 1 fd.

All bar heights correct on graph (no FT), using their suitable linear scale with at least 3 values indicated, no lower than 1cm = 1 fd.

Bar ends at 120.5, 130.5, 135.5, 145.5, 160.5. 5 bars drawn with a horizontal linear scale, no lower than 1 cm = 10min, with at least 3 values indicated. 100 ⩽ horizontal scale ⩽ 160.

Axes labelled frequency density (fd), time (t) and minutes (min, m) oe, or an appropriate title. (Axes may be reversed).

[Midpoints ] 110.5 125.5 133 140.5 153

$$\text{Mean}=\frac{18\times 110.5+48\times 125.5+34\times 133+32\times 140.5+18\times 153}{150}=\frac{1989+6024+4522+4496+2754}{150}$$

= 131.9

$$\text{Variance}=\frac{18\times {110.5}^2+48\times {125.5}^2+34\times {133}^2+32\times {140.5}^2+18\times {153}^2}{150}-(\text{their }131.9{)}^2$$

[Standard deviation =] 11.7

$$0.28+6p=1,p=0.12$$

Correct probability for 1 identified scenario, accept unsimplified, www.

Add values of 4 correct scenarios, may be implied by correct unsimplified expressions. No incorrect/repeated scenarios.

0.06

[P(X is odd) = 0.28 + 2×0.12 or 0.24 ]= 0.52[0]

Values of at least 5 identified correct scenarios added, accept unsimplified, condone incorrect scenarios in calculation.

[P(product of 3 scores ⩽ 4 ∩ X is odd) = ] $$0.28\times (0.25{)}^2\times 8+0.24\times (0.25{)}^2$$

$$[\text{P}(\text{product of 3 scores }⩽4\mid X\text{ is odd})=\frac{\text{P}(\text{product of 3 scores }⩽4\cap X\text{ is odd})}{\text{P}(X\text{ is odd})}=]\frac{0.155}{0.52}$$

$$=0.298,\frac{155}{520},\frac{31}{104}$$

$${}^5{\text{C}}_2\times r,r=\text{positive integer, 1 implied, no addition.}$$

$$s\times 2,s={}^5{\text{C}}_2\text{ or }{}^5{\text{P}}_2$$ or if $${}^5{\text{C}}_2\text{ or }{}^5{\text{P}}_2$$ not present, $$s=\text{a single integer}>1$$ or $$t!\times 2,2⩽t⩽8$$, no other terms.

20

Method 1: $${}^6{\text{C}}_2\times 2\times 2\times 2\times t,t=\text{positive integer}⩾1$$ $${}^6{\text{P}}_2\times 2\times 2\times t,t=\text{positive integer}⩾1$$ Method 2: $$6!\times v,v=\text{positive integer}⩾1$$

Method 1: $$u\times 4!,u=\text{positive integer}>1$$ Method 2: $$w\times 2\times 2,w=\text{positive integer}>1$$ condone $$w\times 4,w=\text{positive integer}>1$$

2880

Method 1: 7! × 2 – a, a = positive integer > 1. Method 2: 6! × e × f, e, f = positive integers ⩾ 1. Method 3: ⁶P₂ × e × f, e, f = positive integers ⩾ 1. Method 4: ⁶Pₙ × a × (6−n)!, a = positive integer, 1 ⩽ n ⩽ 5 seen once.

Method 1: b – 6! × 4, b = positive integer > 2880. Method 2: 6! × 2 × f, f = positive integer ⩾ 1. If 5! Used, SC B1 5! × 2 × f, f = positive integer > 1. Method 3: ⁶P₂ × 5! × f, f = positive integer ⩾ 1. Condone ⁶C₂ × 5! × f, f = positive integer ⩾ 1. Method 4: ⁶Pₙ × 2 × (6−n)!, a = positive integer, 1 ⩽ n ⩽ 5 seen at least 3 times in identified scenarios.

Method 1: 7! × c – 6! × d, c = 1,2 and d = 1, 4. Method 2: 6! × e × 5, e = positive integer ⩾ 1. Method 3: ⁶P₂ × e × 2, e = positive integer ⩾ 1. Condone ⁶C₂ × e × 2, e = positive integer ⩾ 1. Method 4: Add 5 values of appropriate scenarios only. No additional, incorrect or repeated scenarios. Accept unsimplified.

7200