1 The performance of a piece of music is being recorded. The piece consists of three sections, $A , B$ and $C$. The times, in seconds, taken to perform the three sections are normally distributed random variables with the following means and standard deviations.

\end{table}

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{Question} & Answer & Mark & AO & \multicolumn{2}{|l|}{Guidance} \\
\hline
\multirow[t]{3}{*}{1} & \multirow[t]{3}{*}{(a)} & \begin{tabular}{l}
\(A + B + C \sim \mathrm {~N} ( 701 , \ldots\) \\
.. 419) \\
\end{tabular} & M1 & 1.1a & Normal, mean $\mu _ { A } + \mu _ { B } + \mu _ { C }$ & \multirow{3}{*}{} \\
\hline
 &  &  & A1 & 1.1 & Variance 419 &  \\
\hline
 &  & $\mathrm { P } ( > 720 ) = 0.176649$ & A1 & 1.1 & Answer, 0.177 or better, www &  \\
\hline
\multirow[t]{2}{*}{1} & \multirow[t]{2}{*}{(b)} & $2 A + B \sim \mathrm {~N} ( 701,757 )$ & M1 & 1.1a & Normal, same mean, $4 \sigma _ { A } { } ^ { 2 } + \sigma _ { B } { } ^ { 2 }$ & \multirow{2}{*}{} \\
\hline
 &  & $\mathrm { P } ( > 720 ) = 0.244919$ & A1 [2] & 1.1 & Answer, art 0.245 &  \\
\hline
\multirow{2}{*}{2} & \multirow{2}{*}{(a)} & $\frac { { } ^ { 8 } C _ { 3 } \times { } ^ { 20 } C _ { 5 } } { { } ^ { 28 } C _ { 8 } }$ & M1 A1 & 3.1b 1.1 & (Product of two ${ } ^ { n } C _ { r }$ ) ÷ ${ } ^ { n } C _ { r }$ At least two ${ } ^ { n } C _ { r }$ correct & \multirow[t]{2}{*}{Or $\frac { 8 } { 28 } \times \frac { 7 } { 27 } \times \frac { 6 } { 26 } \times \frac { 20 } { 25 } \times \ldots \times \frac { 16 } { 21 } \times { } ^ { 8 } C _ { 3 } = 0.27934 \ldots$} \\
\hline
 &  & $\frac { 56 \times 15504 } { 3108105 } = 0.27934 \ldots$ & A1 [3] & 1.1 & Any exact form or awrt 0.279 &  \\
\hline
2 & (b) & \begin{tabular}{l}
× B × B × B × B × B × B × B × B x \\
GGG in one $\mathrm { x } , \mathrm { G }$ in another: $9 \times 8$ \(\div \frac { 12 ! } { 8 ! \times 4 ! }\) \(= \frac { 72 } { 495 } = \frac { 8 } { 55 } \text { or } 0.145 \ldots\) \\
\end{tabular} & M1 A1 & 3.1b 2.1 & \begin{tabular}{l}
Or e.g. $\frac { 10 ! } { 8 ! } - 2 \times 9$ \\
Divide by ${ } _ { 12 } \mathrm { C } _ { 4 }$ oe \\
\end{tabular} & Or, e.g. find ${ } _ { 12 } \mathrm { C } _ { 4 }$ - (\# (all separate) +\#(all together) $+ \# ( 2,1,1 ) \times 3 +$ \#(2,2)) \\
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 &  &  & M1 & 1.1 &  &  \\
\hline
 &  &  & A1 & 1.1 &  &  \\
\hline
 &  &  & [4] &  &  &  \\
\hline
\end{tabular}
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\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
\multicolumn{3}{|c|}{Question} & Answer & Mark & AO & \multicolumn{2}{|l|}{Guidance} \\
\hline
\multirow{7}{*}{3} & \multirow{7}{*}{(a)} &  & $\mathrm { H } _ { 0 } : \mu = 700$ & B2 & 1.1 & One error, e.g. no or wrong & Ignore failure to define $\mu$ \\
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 &  &  & $\mathrm { H } _ { 1 } : \mu < 700$ where $\mu$ is the mean reaction &  & 1.1 & letter, $\neq$, etc : B1 & here \\
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 &  &  & $\bar { x } = 607$ & M1 & 3.3 & Find sample mean &  \\
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 &  &  & $z = - 1.822$ or $p = 0.0342$ or $\mathrm { CV } = 616.05 \ldots$ & A1 & 3.4 & Correct $z , p$ or CV &  \\
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 &  &  & $z < - 1.645$ or $p < 0.05$ or $607 < \mathrm { CV }$ & A1 & 1.1 & Correct comparison &  \\
\hline
 &  &  & Reject $\mathrm { H } _ { 0 }$ & M1ft & 1.1 & Correct first conclusion & Needs correct method, like- \\
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 &  &  & Significant evidence that mean reaction times & A1ft & 2.2b & Context, not too definite (e.g. not "international athletes' reaction times are shorter" & ft on their $z , p$ or CV \\
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3 & (b) & (i) & Uses more information (e.g. magnitudes of differences) & B1 [1] & 2.4 &  &  \\
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\multirow{5}{*}{3} & \multirow{5}{*}{(b)} & \multirow{5}{*}{(ii)} & $\mathrm { H } _ { 0 } : m = 700 , \mathrm { H } _ { 1 } : m < 700$ where $m$ is the median reaction time for all international athletes & B1 & 2.5 & Same as in (i) but different letter or "median" stated &  \\
\hline
 &  &  & \begin{tabular}{l}
$W _ { - } = 18$ \\
$W _ { + } = 3$ so $T = 3$ \\
\end{tabular} &  &  &  &  \\
\hline
 &  &  &  &  &  & For both, and $T$ correct &  \\
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 &  &  & $n = 6 , \mathrm { CV } = 2$ & A1 & 1.1 & Correct CV &  \\
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 &  &  & Do not reject $\mathrm { H } _ { 0 }$. Insufficient evidence that median reaction times of international athletes are shorter & A1ft [6] & 2.2b & In context, not too definite & FT on their $T$ \\
\hline
3 & (c) &  & They use different assumptions & B1 [1] & 2.3 &  & Not "one is more accurate" \\
\hline
\end{tabular}
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\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
\multicolumn{3}{|c|}{Question} & Answer & Mark & AO & \multicolumn{2}{|l|}{Guidance} \\
\hline
4 & (a) &  & \(\begin{aligned} & \int _ { 0 } ^ { a } x \frac { 2 x } { a ^ { 2 } } d x = 4 \\ & { \left[ \frac { 2 x ^ { 3 } } { 3 a ^ { 2 } } \right] = 4 } \\ & \frac { 2 } { 3 } a = 4 \Rightarrow a = 6 \end{aligned}\) & \begin{tabular}{l}
M1 \\
B1 \\
A1 [3] \\
\end{tabular} & \begin{tabular}{l}
3.1a \\
1.1 \\
2.2a \\
\end{tabular} &  &  \\
\hline
4 & (b) &  & \begin{tabular}{l}
\(\mathrm { F } ( x ) = \frac { x ^ { 2 } } { 36 }\) \\
Let the CDF of $M$ be $\mathrm { H } ( m )$. Then $\mathrm { H } ( m ) = \mathrm { P } ($ all observations less than $m )$ \(= [ \mathrm { P } ( X \leqslant m ) ] ^ { 5 }\) \(= \left[ \frac { m ^ { 2 } } { 36 } \right] ^ { 5 }\) \\
$\mathrm { H } ( m ) = \begin{cases} 0 & m < 0 , \\ \frac { m ^ { 10 } } { 60466176 } & 0 \leq m \leq 6 , \\ 1 & m > 6 . \end{cases}$ \\
\end{tabular} & \begin{tabular}{l}
M1 A1ft \\
M1 \\
M1 \\
A1 \\
A1 \\
A1 \\
A1 \\[0pt]
[8] \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
1.1 \\
2.1 \\
3.1a \\
2.2a \\
2.1 \\
2.1 \\
1.2 \\
\end{tabular} & \begin{tabular}{l}
Find $\mathrm { F } ( x ) ; = \frac { x ^ { 2 } } { a ^ { 2 } }$ \\
Correct basis for CDF of $m$ \\
Correct function, any letter Range $0 \leq m \leq 6$ \\
Letter not $x$, and 0, 1 present \\
\end{tabular} & \begin{tabular}{l}
ft on their $a$ \\
Allow </ s throughout \\
\end{tabular} \\
\hline
\end{tabular}
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\hfill \mbox{\textit{OCR Further Statistics 2021 Q1 [5]}}