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.
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| Question | Answer | Mark | AO | Guidance |
| \multirow[t]{3}{*}{1} | \multirow[t]{3}{*}{(a)} | | \(A + B + C \sim \mathrm {~N} ( 701 , \ldots\) | | .. 419) |
| M1 | 1.1a | Normal, mean \(\mu _ { A } + \mu _ { B } + \mu _ { C }\) | \multirow{3}{*}{} |
| | | A1 | 1.1 | Variance 419 | |
| | \(\mathrm { P } ( > 720 ) = 0.176649\) | A1 | 1.1 | Answer, 0.177 or better, www | |
| \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}{*}{} |
| | \(\mathrm { P } ( > 720 ) = 0.244919\) | A1 [2] | 1.1 | Answer, art 0.245 | |
| \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 \(\frac { 8 } { 28 } \times \frac { 7 } { 27 } \times \frac { 6 } { 26 } \times \frac { 20 } { 25 } \times \ldots \times \frac { 16 } { 21 } \times { } ^ { 8 } C _ { 3 } = 0.27934 \ldots\)} |
| | \(\frac { 56 \times 15504 } { 3108105 } = 0.27934 \ldots\) | A1 [3] | 1.1 | Any exact form or awrt 0.279 | |
| 2 | (b) | | × B × B × B × B × B × B × B × B x | | GGG in one \(\mathrm { x } , \mathrm { G }\) in another: \(9 \times 8\) \(\div \frac { 12 ! } { 8 ! \times 4 ! }\) \(= \frac { 72 } { 495 } = \frac { 8 } { 55 } \text { or } 0.145 \ldots\) |
| M1 A1 | 3.1b 2.1 | | Or e.g. \(\frac { 10 ! } { 8 ! } - 2 \times 9\) | | Divide by \({ } _ { 12 } \mathrm { C } _ { 4 }\) oe |
| Or, e.g. find \({ } _ { 12 } \mathrm { C } _ { 4 }\) - (\# (all separate) +\#(all together) \(+ \# ( 2,1,1 ) \times 3 +\) \#(2,2)) |
| | | М1 | 1.1 | | |
| | | A1 | 1.1 | | |
| | | [4] | | | |
| Question | Answer | Mark | AO | Guidance |
| \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 \(\mu\) |
| | | \(\mathrm { H } _ { 1 } : \mu < 700\) where \(\mu\) is the mean reaction | | 1.1 | letter, \(\neq\), etc : B1 | here |
| | | \(\bar { x } = 607\) | М1 | 3.3 | Find sample mean | |
| | | \(z = - 1.822\) or \(p = 0.0342\) or \(\mathrm { CV } = 616.05 \ldots\) | A1 | 3.4 | Correct \(z , p\) or CV | |
| | | \(z < - 1.645\) or \(p < 0.05\) or \(607 < \mathrm { CV }\) | A1 | 1.1 | Correct comparison | |
| | | Reject \(\mathrm { H } _ { 0 }\) | M1ft | 1.1 | Correct first conclusion | Needs correct method, like- |
| | | 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 |
| 3 | (b) | (i) | Uses more information (e.g. magnitudes of differences) | B1 [1] | 2.4 | | |
| \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 | |
| | | | \(W _ { - } = 18\) | | \(W _ { + } = 3\) so \(T = 3\) |
| | | | |
| | | | | | For both, and \(T\) correct | |
| | | \(n = 6 , \mathrm { CV } = 2\) | A1 | 1.1 | Correct CV | |
| | | 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 \(T\) |
| 3 | (c) | | They use different assumptions | B1 [1] | 2.3 | | Not "one is more accurate" |
| Question | Answer | Mark | AO | Guidance |
| 4 | (a) | | \(\begin{aligned} | \int _ { 0 } ^ { a } x \frac { 2 x } { a ^ { 2 } } d x = 4 |
| { \left[ \frac { 2 x ^ { 3 } } { 3 a ^ { 2 } } \right] = 4 } |
| \frac { 2 } { 3 } a = 4 \Rightarrow a = 6 \end{aligned}\) | | | | |
| 4 | (b) | | | \(\mathrm { F } ( x ) = \frac { x ^ { 2 } } { 36 }\) | | Let the CDF of \(M\) be \(\mathrm { H } ( m )\). Then \(\mathrm { H } ( m ) = \mathrm { P } (\) all observations less than \(m )\) \(= [ \mathrm { P } ( X \leqslant m ) ] ^ { 5 }\) \(= \left[ \frac { m ^ { 2 } } { 36 } \right] ^ { 5 }\) | | \(\mathrm { H } ( m ) = \begin{cases} 0 | m < 0 , | | \frac { m ^ { 10 } } { 60466176 } | 0 \leq m \leq 6 , | | 1 | m > 6 . \end{cases}\) |
| | | 1.1 | | 1.1 | | 2.1 | | 3.1a | | 2.2a | | 2.1 | | 2.1 | | 1.2 |
| | 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 |
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