A Level Further Mathematics 9231 Paper 42, Oct/Nov 2023

$$s_x^2=\frac{243.8}{59}[=\frac{1219}{295}=4.132],s_y^2=\frac{384.9}{74}[=\frac{3849}{740}=5.201]$$

$$s^2=\frac{4.132}{60}+\frac{5.201}{75}[=0.1382]\text{ or }s=0.3718$$

$$s=0.3718\text{ or }{s}^2=0.1382$$

$$\text{CI: }\frac{6345}{60}-\frac{7614}{75}\pm z\times s$$

$$\text{CI: }\frac{6345}{60}-\frac{7614}{75}\pm 1.96\times \text{’0.3718’}$$

$$[3.5[0],4.96]\text{ or }(3.5,4.96)$$

Correct expected frequencies for first group of cells

Correct expected frequencies for remaining cells

${\text{H}}_0$: $\text{Po}(3.5)$ is a good fit to the data ${\text{H}}_1$: $\text{Po}(3.5)$ is not a good fit to the data

Combine first 2 columns: 5, 12.23 And last 2 columns: 9, 5.876

Chi-squared values: $4.274+0.799+0.302+0.945+1.417+0.612+1.661$

10.0

Tabular value: 10.64 '10.0' < 10.64, accept ${\text{H}}_0$/not significant

Insufficient evidence to suggest that $\text{Po}(3.5)$ is not a good fit to the data

$$\text{P}(3\text{R})=\frac{120}{504}\text{P}(2\text{R})=\frac{270}{504}\text{P}(1\text{R})=\frac{108}{504}\text{P}(0\text{R})=\frac{6}{504}$$

$${\text{G}}_X(t)=\frac{1}{504}(6+108t+270t^2+120t^3)=\frac{1}{84}(1+18t+45t^2+20t^3)$$

$$\frac{1}{84}+\frac{3}{14}t+\frac{15}{28}{t}^2+\frac{5}{21}{t}^3$$

$${\text{G}}_Z(t)=\frac{1}{84}(1+18t+45t^2+20t^3)\times \frac{1}{12}(1+6t+5t^2)$$

$$\frac{1}{1008}(1+24t+158t^2+380t^3+345t^4+100t^5)$$

$$\frac{1}{1008}+\frac{1}{42}t+\frac{79}{504}{t}^2+\frac{95}{252}{t}^3+\frac{115}{336}{t}^4+\frac{25}{252}{t}^5$$

$${\text{G}}^{\mathrm{\prime }}=\frac{1}{1008}(24+316t+1140t^2+1380t^3+500t^4)$$ $${\text{G}}^{\mathrm{\prime }\mathrm{\prime }}=\frac{1}{1008}(316+2280t+4140t^2+2000t^3)$$

$$[\text{E}(Z)=\frac{3360}{1008}]=\frac{10}{3}$$

$$\text{Var}(Z)=\frac{8736}{1008}+\frac{10}{3}-{\left(\frac{10}{3}\right)}^2[=\frac{26}{3}+\frac{10}{3}-\frac{100}{9}]$$

$$\frac{8}{9}$$

$$\text{UQ}=4\text{ so }{\int }_4^{6}c\text{d}x=\frac{1}{4}:c[6-4]=\frac{1}{4},c=\frac{1}{8}$$

$${\int }_0^4\frac{1}{128}(4ax-b{x}^3)\text{d}x=\frac{3}{4},\frac{1}{128}{[2a{x}^2-\frac{b}{4}{x}^4]}_0^4=\frac{3}{4},32a-64b=96$$

$$\text{Equate at }x=4:a-4b=1$$

$$\text{Solve: }a=5,b=1$$

$${\int }_0^m\frac{1}{128}(20x-x^3)\text{d}x=\frac{1}{2}$$ OR $$\text{F}(x)=\frac{1}{128}(2a{x}^2-\frac{b}{4}{x}^4)=\frac{5}{64}{x}^2-\frac{1}{512}{x}^4$$

$$m^4-40m^2+256=0$$

$$m=2\sqrt{2}$$

$$\text{E}(\sqrt{X})=\frac{1}{128}{\int }_0^{4}20x^{\frac{3}{2}}-x^{\frac{7}{2}}\text{d}x+{\int }_4^6\frac{1}{8}{x}^{\frac{1}{2}}\text{d}x$$

$$\frac{1}{128}{[8x^{\frac{5}{2}}-\frac{2}{9}{x}^{\frac{9}{2}}]}_0^4+\frac{1}{12}{\left[x^{\frac{3}{2}}\right]}_4^6$$

$$\frac{1}{128}(2^8-\frac{1}{9}{2}^{10})+\frac{1}{12}(6\sqrt{6}-8)=\frac{4}{9}+\frac{1}{2}\sqrt{6}=1.67$$

${\text{H}}_0$: population medians are equal or $m_x=m_y$ ${\text{H}}_1$: population median for $X<$ population median for $Y$ or $m_x<m_y$

Rankings table with sums: 55, 98

Test statistic = 55

Tabular value for $m=8,n=9$ is 54

'55' > '54', accept ${\text{H}}_0$/not significant

Insufficient evidence to support manager's claim. Insufficient evidence to suggest that the median time of machine $X$ is less than the median time of machine $Y$.

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

$$\sum x=39.7\sum x^2=199.33\sum y=49.5\sum y^2=275.47$$ $$s_x^2=\frac{1}{7}(199.33-\frac{{39.7}^2}{8})=0.33125$$

$$s_y^2=\frac{1}{8}(275.47-\frac{{49.5}^2}{9})=0.4025$$

$$\text{Pooled variance }{s}^2=\frac{7\times 0.33125+8\times 0.4025}{8+9-2}$$

0.36925

$$t=\frac{\frac{39.7}{8}-\frac{49.5}{9}}{s\sqrt{\frac{1}{8}+\frac{1}{9}}}$$

$$t=-1.82$$

$$\text{Tabular value}=1.753:1.82>1.753$$

Reject ${\text{H}}_0$, sufficient evidence that mean for machine $X$ is less than mean for machine $Y$.

$t$-test is assuming a normal distribution, and with equal variances. This may not be true. So, no reason to expect results to be the same. Outliers affect part (b) but not part (a).