A Level Further Mathematics 9231 Paper 43, May/June 2024

Attempt at ranking from 1 to 19.

all correct.

Test statistic: 70

${\text{H}}_0$: population median of $X=$ population median of $Y$ ${\text{H}}_1$: population median of $X<$ population median of $Y$

Critical value for $m=9,n=10$ is 61

‘70’ > ‘61’ accept ${\text{H}}_0$

Insufficient evidence to suggest that population medians of $X$ and $Y$ are not equal or Insufficient evidence to suggest that population median of $X$ is less than median of $Y$ or Insufficient evidence to support the interviewer’s belief

$\bar{x}=\frac{1}{2}(68.66+64.22)=66.44$

$\sum x=12\times 66.44=797.28$

$\frac{ts}{\sqrt{12}}=\frac{1}{2}(68.66-64.22)$

With $t=2.718$ gives $s^2=8.0055$

Use $s^2=\frac{1}{11}(\sum x^2-\frac{(\sum x{)}^2}{12})$ to find an expression for $\sum x^2$

$\sum x^2=53059$

Population is normally distributed

Calculate expected values, at least 4 correct to 3sf. These must be seen.

all correct to at least 3sf.

At least 4 correct values (to 2sf) or expressions.

Test statistic = 8.61

${\text{H}}_0$: Reliability of buses is independent of bus company ${\text{H}}_1$: Reliability of buses is not independent of/dependent on bus company

‘8.61’ < 9.488 Accept ${\text{H}}_0$

There is insufficient evidence to support/ suggest that reliability of buses is dependent/not independent of bus company

$c=\frac{1}{32}$

$G_X(t)=ct(1+t{)}^5$, $G_X^{\mathrm{\prime }}(t)=c(5t(1+t{)}^4+(1+t{)}^5)$

$\text{E}(X)=G_X^{\mathrm{\prime }}(1)=c(80+32)=3.5$

$G_Y(t)=c^2{t}^2(1+t{)}^{10}$

$G_Y^{\mathrm{\prime }}(t)=c^2(10t^2(1+t{)}^9+2t(1+t{)}^{10})$ $G_Y^{\mathrm{\prime }\mathrm{\prime }}(t)=c^2(90t^2(1+t{)}^8+40t(1+t{)}^9+2(1+t{)}^{10})$

Use correct formula and substitute values for $t=1$ (may be in terms of $c$).

$G_Y^{\mathrm{\prime }}(1)=7,G_Y^{\mathrm{\prime }\mathrm{\prime }}(1)=44.5$ $\text{Var}(Y)=G_Y^{\mathrm{\prime }\mathrm{\prime }}(1)+G_Y^{\mathrm{\prime }}(1)-(G_Y^{\mathrm{\prime }}(1){)}^2=2.5$

$\text{P}(Y=5)=\text{coefficient of }{t}^5\text{ in expansion of their }{c}^2{t}^2(1+t{)}^{10}$ Required term is $c^2\times 120$ Expansion = $c^2{t}^2(1+10t+45t^2+120t^3+\dots )$

$\frac{15}{128}$

Attempt at differentiating both parts of CDF to obtain linear expressions.

Correct shape (Two line segments touch the $x$-axis and meet at a point above the $x$-axis).

Correct shape with 2 and 8 and at least one of 4 or 1/3 correctly labelled.

${\int }_2^4\frac{1}{6}x(x-2)\text{d}x+{\int }_4^8\frac{1}{12}x(8-x)\text{d}x$

$\frac{1}{6}[\frac{1}{3}{x}^3-x^2]+\frac{1}{12}[4x^2-\frac{1}{3}{x}^3]$

$\frac{14}{3}$

LQ: $\frac{1}{12}(x-2{)}^2=\frac{1}{4},x=2+\sqrt{3}$

UQ: $1-\frac{(8-x{)}^2}{24}=\frac{3}{4},x=8-\sqrt{6}$

IQR = UQ – LQ, with values found for both quartiles

$6-\sqrt{6}-\sqrt{3}$

$s_x^2=\frac{1}{49}(23480-\frac{{1080}^2}{50})(=3.102)$ $s_y^2=\frac{1}{39}(22220-\frac{{940}^2}{40})(=3.333)$

$s^2=\frac{3.102}{50}+\frac{3.333}{40}(=0.1454)$

$s^2=\frac{2137}{14700}=0.1454$

CI: $\frac{1080}{50}-\frac{940}{40}\pm 1.645\sqrt{s^2}$ or $\frac{940}{40}-\frac{1080}{50}\pm 1.645\sqrt{s^2}$

1.645 (accept 1.64 or 1.65) used in CI formula.

$[1.27,2.53]$ or $[-2.53,-1.27]$

$z=\frac{1.9-1.1}{s}$

2.098

‘2.098’ > 1.28(2) Reject ${\text{H}}_0$ or $\mathrm{\Phi }(2.098)=0.9821>0.9$ Reject ${\text{H}}_0$ or $1-\mathrm{\Phi }(2.098)=0.0179<0.1$ Reject ${\text{H}}_0$

Sufficient evidence to suggest that the lengths of tails of wallabies in region $Y$ are, on average, more than 1.1 cm longer than the tails of wallabies in region $X$