A Level Further Mathematics 9231 Paper 43, May/June 2023

Integrate f(x), + c not required, at least one power correct.

$$\frac{1}{4}{x}^{\frac{2}{3}}-\frac{1}{2}{x}^{\frac{1}{3}}+\frac{1}{4}$$

Correct ranges including $1⩽x⩽27$ associated with their $\text{F}(x)$ $\text{F}(x)=0$ for $x<1$ and $\text{F}(x)=1$ for $x>27$

$$\text{G}(y)=\frac{1}{4}(y^2-2y+1)$$

$$\text{g}(y)=\frac{1}{2}(y-1)$$

For $1⩽y⩽3$

$$\frac{1}{4}(y^2-2y+1)=\frac{1}{2},(y-1{)}^2=2\text{ OE}$$

$$[y=]1+\sqrt{2}$$

$$\bar{x}+1.771\sqrt{\frac{s^2}{14}}=1.14\text{ or }\bar{x}-1.771\sqrt{\frac{s^2}{14}}=1.11$$

$$[\text{Add: }\bar{x}=\frac{1}{2}(1.14+1.11)=1.125]\mathrm{\Sigma }x=15.75$$

$$\text{Subtract or substitute: }\sqrt{\frac{s^2}{14}}=\frac{1}{2}\left(\frac{1.14-1.11}{1.771}\right)$$

$$s^2=0.00100[4]\text{ or }s=0.0316[9]$$

$$\mathrm{\Sigma }{x}^2=13s^2+\frac{(\mathrm{\Sigma }x{)}^2}{14}$$

$$\mathrm{\Sigma }{x}^2=17.7(3)$$

Differences: 5 7 -6 1 –4 2 8 26 –10

Ranks: 4 6 -5 1 –3 2 7 9 –8

$$[P=29]Q=16$$

$${\text{H}}_0\text{: population medians equal or }{m}_1=m_2$$ $${\text{H}}_1\text{: population median }X>\text{population median }Y\text{ or }{m}_1>m_2$$

Critical value = 8

16 > 8 and accept ${\text{H}}_0$ / not significant

Insufficient evidence to support teacher’s belief or insufficient evidence that the marks/median in Paper 1 are/is higher than the marks/median in Paper 2

The population differences are symmetrical (about the median difference)

$${\text{H}}_0\text{: }{\mu }_x={\mu }_y,{\text{H}}_1\text{: }{\mu }_x<{\mu }_y$$

$$s_x^2=\frac{1}{79}[338.1-\frac{{164}^2}{80}][=0.02405]\text{ and }{s}_y^2=\frac{1}{59}[261.1-\frac{{124.8}^2}{60}][=0.02569]$$

$$s^2=\frac{0.02405}{80}+\frac{0.02569}{60}$$

$$s^2=0.0007289\text{ or }s=0.026998\text{ or }0.0270$$

$$z=\frac{\frac{164.0}{80}-\frac{124.8}{60}}{s}$$

$$-1.11$$

‘1.11’ < 1.282 Accept ${\text{H}}_0$ / not significant

Insufficient evidence to support the inspector’s suspicion/ Insufficient evidence that the (mean) lengths of rods from machine $X$ are shorter than the (mean) lengths of rods from machine $Y$

Large sample sizes OR central limit theorem applies.

$$[{\text{G}}_X(t)=k(1+3t+4t^2)]k(1+3+4)=1,k=\frac{1}{8}$$

$${\text{G}}_X^{\mathrm{\prime }}(t)=k(3+8t),$$

$$\text{E}(X)={\text{G}}_X^{\mathrm{\prime }}(1)=\frac{11}{8}$$

$${\text{G}}_Z(t)=\frac{1}{8}(1+3t+4t^2)\times \frac{1}{3}{t}^2(1+2t)$$

$$\frac{1}{24}(t^2+5t^3+10t^4+8t^5)$$

$${\text{G}}_Z^{\mathrm{\prime }}(t)=\frac{1}{24}(2t+15t^2+40t^3+40t^4)$$ $${\text{G}}_Z^{\mathrm{\prime }\mathrm{\prime }}(t)=\frac{1}{24}(2+30t+120t^2+160t^3)$$

$$\text{Var}(X)={\text{G}}_Z^{\mathrm{\prime }\mathrm{\prime }}(1)+{\text{G}}_Z^{\mathrm{\prime }}(1)-({\text{G}}_Z^{\mathrm{\prime }}(1){)}^2=\frac{1}{24}(312)+\frac{97}{24}-{\left(\frac{97}{24}\right)}^2$$

$$0.707$$

$$(Z=)4$$

$${\text{H}}_0\text{: grade is independent of age}$$ $${\text{H}}_1\text{: grade is not independent of age}$$

Calculate expected values, shown in table in bold (at least 2 correct)

6 correct expected values or expressions: 21.2, 24.4, 14.4, 31.8, 36.6, 21.6

Calculate chi-squared values: $$0.6811+2.8918+1.4694+0.4541+1.9279+0.9796$$

$$8.40[4]$$

Tabular value, 2 degrees of freedom = 7.378 ‘8.404’ > 7.378 and reject ${\text{H}}_0$/significant

Sufficient evidence to suggest that grade is not independent of age.

Value, 4 degrees of freedom = 11.14 10.91 < 11.14 Accept ${\text{H}}_0$/not significant

Insufficient evidence to suggest that grade is not independent of age.

For example, result in part (b) because the table contains more information e.g. More degrees of freedom, more groups, more detail