2 After a holiday organised for a group, the company organising the holiday obtained scores out of 10 for six different aspects of the holiday. The company obtained responses from 100 couples and 100 single travellers. The total scores for each of the aspects are given in the following table.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{Question} & Answer &  & Mark & AO & \multicolumn{2}{|l|}{Guidance} \\
\hline
1 & (a) & $\frac { 1 } { 0.2 } = 5$ &  & M1 A1 [2] & 3.3 1.1 & Geometric distribution soi 5 (or $5.00 \ldots$ ) only &  \\
\hline
1 & (b) & $0.8 ^ { 2 } - 0.8 ^ { 10 }$ \(= \mathbf { 0 . 5 3 3 } \quad ( 0.5326258 \ldots )\) &  & M1 A1 [2] & 1.1 3.4 & \begin{tabular}{l}
Allow for powers 2, 3, 4 and 9, 10, 11 . \\
Awrt 0.533, www. [5201424/976562] \\
\end{tabular} & Or $0.2 \left( 0.8 ^ { 2 } + \ldots . + 0.8 ^ { 9 } \right) , \pm 1$ term at either end [0.506, 0.378, 0.275, 0.405, 0.302, 0.554, 0.426, 0.324] \\
\hline
1 & (c) & \begin{tabular}{l}
\(\mathrm { P } ( \geq 10 ) = 0.8 ^ { 9 }\) \\
$= 0.1342 \ldots$ \\
B(30, 0.1342...) \\
Variance $= n p q$ = 3.486... \\
\end{tabular} &  & \begin{tabular}{l}
M1 \\
A1 \\
M1 \\
A1ft [4] \\
\end{tabular} & \begin{tabular}{l}
3.1b \\
1.1 \\
3.1b \\
1.1 \\
\end{tabular} & \begin{tabular}{l}
Or $0.8 ^ { 10 }$. Can be implied by correct $p$ \\[0pt]
[0.10737... is M1A0 here] \\
Stated or implied, their $0.8 ^ { 9 }$ or $0.8 ^ { 10 }$ \\
In range [3.48, 3.49] \\
\end{tabular} & \begin{tabular}{l}
SC: 0.134(2) oe not properly shown: B2 for correct final answer. \\
SC: 2.875 from $0.8 ^ { 10 }$ : M1A0M1A1ft \\
\end{tabular} \\
\hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{Question} & Answer & Mark & AO & Guidance \\
\hline
2 & (a) & Test is for rankings/rankings arbitrary/not bivariate normal etc & B1 [1] & 2.4 & OE \\
\hline
2 & (b) & \begin{tabular}{l}
$\mathrm { H } _ { 0 } : \rho _ { s } = 0 , \mathrm { H } _ { 1 } : \rho _ { s } > 0$, where $\rho _ { s }$ is the population rank correlation coefficient \\
Ranks 543612 \\
512643 \\
$\Sigma d ^ { 2 } = 20$ \\
$r _ { s } = 1 - \frac { 6 \times 20 } { 6 \times 35 }$ \\
$= 3 / 7$ or $0.42857 \ldots$ \\
<0.9429 \\
\end{tabular} & \begin{tabular}{l}
B1 \\
B1 \\
M1 \\
A1 \\
B1 \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
1.1 \\
1.1 \\
1.1 \\
1.1 \\
\end{tabular} & \begin{tabular}{l}
Allow $\rho _ { s }$ not defined; allow $\rho$. \\
Allow: $\mathrm { H } _ { 0 }$ : no association between rankings. \\
$\mathrm { H } _ { 1 }$ : positive association (but not $\mathrm { H } _ { 1 }$ : association) \\
\end{tabular} \\
\hline
 &  & \begin{tabular}{l}
Do not reject $\mathrm { H } _ { 0 }$ \\
Insufficient evidence of association between ranking given by the two categories \\
\end{tabular} & \begin{tabular}{l}
M1ft \\
A1ft \\[0pt]
[7] \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
2.2b \\
\end{tabular} & FT on their $\Sigma d ^ { 2 }$ only \\
\hline
2 & (c) & Not dependent on any distributional assumptions & \begin{tabular}{l}
B1 \\[0pt]
[1] \\
\end{tabular} & 1.2 & Oe (cf. Specification, 5.08f) \\
\hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
\multicolumn{3}{|r|}{Question} & Answer & Mark & AO & Guidance \\
\hline
3 & (a) &  & Failures occur to no fixed pattern/are not predictable & B1 [1] & 1.1 & OE. NOT "independent" \\
\hline
3 & (b) &  & Failures occur independently of one another and at constant average rate & \begin{tabular}{l}
B1 \\
B1 \\[0pt]
[2] \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
1.1 \\
\end{tabular} & \begin{tabular}{l}
Not recoverable from (a) if independence not restated here; must be contextualised \\
Ignore "singly". Allow "uniform" rate, not "constant rate" or "constant probability"; must be contextualised \\
\end{tabular} \\
\hline
3 & (c) &  & \begin{tabular}{l}
Variance (1.6384) $\approx$ mean \\
So suggests that it is likely to be well modelled \\
\end{tabular} & \begin{tabular}{l}
M1 \\
A1 \\[0pt]
[2] \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
3.5a \\
\end{tabular} & \begin{tabular}{l}
Compare variance (or SD). Allow square/square-root confusion \\
Correct comparison and conclusion, 1.64 or better seen \\
\end{tabular} \\
\hline
3 & (d) &  & $\mathrm { e } ^ { - 1.61 }$ & \begin{tabular}{l}
B1 \\[0pt]
[1] \\
\end{tabular} & 3.4 & Exact needed, allow even if $0 !$ or $1.61 ^ { 0 }$ or both left in \\
\hline
3 & (e) &  & \begin{tabular}{ | c | c | }
\hline
1 & $\geq 2$ \\
\hline
0.322 & 0.478 \\
\hline
\end{tabular} & \begin{tabular}{l}
B1 \\
B1 \\[0pt]
[2] \\
\end{tabular} & \begin{tabular}{l}
3.4 \\
1.1 \\
\end{tabular} & \begin{tabular}{l}
One correct e.g. 0.3218 \\
Other correct e.g. 0.4783 \\
\end{tabular} \\
\hline
3 & (f) &  & $\mathrm { P } ( F = 1 )$ will be smaller as single failures are less likely & \begin{tabular}{l}
B1* \\
depB1 \\[0pt]
[2] \\
\end{tabular} & \begin{tabular}{l}
3.5c \\
3.3 \\
\end{tabular} & OE. Partial answer: B1 \\
\hline
\end{tabular}
\end{center}

\hfill \mbox{\textit{OCR FS1 AS 2021 Q2 [9]}}