2 In the manufacture of fibre optical cable (FOC), flaws occur randomly. Whether any point on a cable is flawed is independent of whether any other point is flawed. The number of flaws in 100 m of FOC of standard diameter is denoted by \(X\).
- State a further assumption needed for \(X\) to be well modelled by a Poisson distribution.
Assume now that \(X\) can be well modelled by the distribution \(\operatorname { Po } ( 0.7 )\).
- Find the probability that in 300 m of FOC of standard diameter there are exactly 3 flaws.
The number of flaws in 100 m of FOC of a larger diameter has the distribution \(\mathrm { Po } ( 1.6 )\).
- Find the probability that in 200 m of FOC of standard diameter and 100 m of FOC of the larger diameter the total number of flaws is at least 4 .
Judith believes that mathematical ability and chess-playing ability are related. She asks 20 randomly chosen chess players, with known British Chess Federation (BCF) ratings \(X\), to take a mathematics aptitude test, with scores \(Y\). The results are summarised as follows.
$$n = 20 , \Sigma x = 3600 , \Sigma x ^ { 2 } = 660500 , \Sigma y = 1440 , \Sigma y ^ { 2 } = 105280 , \Sigma x y = 260990$$
- Calculate the value of Pearson's product-moment correlation coefficient \(r\).
- State an assumption needed to be able to carry out a significance test on the value of \(r\).
- Assume now that the assumption in part (ii) is valid. Test at the \(5 \%\) significance level whether there is evidence that chess players with higher BCF ratings are better at mathematics.
- There are two different grading systems for chess players, the BCF system and the international ELO system. The two sets of ratings are related by
$$\text { ELO rating } = 8 \times \text { BCF rating } + 650$$
Magnus says that the experiment should have used ELO ratings instead of BCF ratings. Comment on Magnus's suggestion.
- Calculate the value of Pearson's product-moment correlation coefficient \(r\).
- State an assumption needed to be able to carry out a significance test on the value of \(r\).
- Assume now that the assumption in part (b) is valid. Test at the \(5 \%\) significance level whether there is evidence that chess players with higher BCF ratings are better at mathematics.
- There are two different grading systems for chess players, the BCF system and the international ELO system. The two sets of ratings are related by
$$\mathrm { ELO } \text { rating } = 8 \# \mathrm { BCF } \text { rating } + 650 .$$
Magnus says that the experiment should have used ELO ratings instead of BCF ratings. Comment on Magnus's suggestion.
An environmentalist measures the mean concentration, \(c\) milligrams per litre, of a particular chemical in a group of rivers, and the mean mass, \(m\) pounds, of fish of a certain species found in those rivers. The results are given in the table.
\end{table}
| Question | Answer | Marks | AO | Guidance |
| 1 | (a) | | \(\begin{aligned} | 0.25 + 0.36 + x + x ^ { 2 } = 1 |
| x ^ { 2 } + x - 0.39 = 0 |
| x = 0.3 \text { (or } - 1.3 \text { ) } |
| x \text { cannot be negative } |
| \mathrm { E } ( W ) = 2.23 |
| \mathrm { E } \left( W ^ { 2 } \right) = \Sigma w ^ { 2 } \mathrm { p } ( w ) \quad [ = 5.83 ] |
| \text { Subtract } [ \mathrm { E } ( W ) ] ^ { 2 } \text { to get } \mathbf { 0 . 8 5 7 1 } \end{aligned}\) | \(\begin{gathered} \text { M1 } |
| \text { A1 } |
| \text { A1 } |
| \text { B1ft } |
| \text { B1 } |
| \text { M1 } |
| \text { A1 } |
| { [ 7 ] } \end{gathered}\) | | 3.1a | | 1.1b | | 1.1b | | 2.3 | | 1.1b | | 1.1 | | 2.1 |
| | Equation using \(\Sigma p = 1\) | | Correct simplified quadratic Correctly obtain \(x = 0.3\) | | Explicitly reject other solution | | 2.23 or exact equivalent only Use \(\Sigma w ^ { 2 } \mathrm { p } ( w )\) | | Correctly obtain given answer, www |
| | Can be implied | | Method needed ft on their quadratic Allow for \(\mathrm { E } ( W ) ^ { 2 } = 4.9729\) | | Need 2.23 or 4.9729 and 5.83 or full numerical \(\Sigma w ^ { 2 } \mathrm { p } ( w )\) |
|
| 1 | (b) | | \(9 \times 0.8571 = 7.7139\) | | 1.1b | Allow 7.71 or 7.714 | |
| 2 | (a) | | Flaws must occur at constant average rate (uniform rate) | | 1.2 | | Context (e.g. "flaws") needed | | Extra answers, e.g. "singly": B0 |
| Not "constant rate" or "average constant rate". |
| 2 | (b) | | \(\operatorname { Po(2.1)~or~ } e ^ { - \lambda } \frac { \lambda ^ { 3 } } { 3 ! }\) | | | Po(2.1) stated or implied, or formula with \(\lambda = 2.1\) stated Awrt 0.189 | |
| 2 | (c) | | | Po(3) | | \(1 - \mathrm { P } ( \leq 3 )\) |
| | | | \(\operatorname { Po } ( 2 \times 0.7 + 1.6 )\) stated or implied | | Allow \(1 - \mathrm { P } ( \leq 4 ) = 0.1847\), or from wrong \(\lambda\) | | Awrt 0.353 |
| | Or all combinations \(\leq 3\) | | \(1 -\) above, not just \(= 3\) |
|
| Question | Answer | Marks | AO | Guidance |
| 3 | (a) | | 0.4(00) | | | SC: if B0, give SC B1 for two of \(S _ { x x } = 12500 , S _ { y y } = 1600 , S _ { x y } = 1790\) and \(S _ { x y } / \sqrt { } \left( S _ { x x } S _ { y y } \right)\) | Also allow SC B1 for equivalent methods using Covariance \ | SDs |
| 3 | (b) | | Data needs to have a bivariate normal distribution | | 1.2 | Needs "bivariate normal" or clear equivalent. Not just "both normally distributed" | Allow "scatter diagram forms ellipse" |
| 3 | (c) | | | \(\mathrm { H } _ { 0 }\) : higher maths scores are not associated with higher BCF grading; \(\mathrm { H } _ { 1 }\) : positively associated | | CV 0.3783 | | \(0.400 > 0.3783\) so reject \(\mathrm { H } _ { 0 }\) | | Significant evidence that higher maths scores are associated with higher BCF grading |
| | | | Needs context and clearly onetailed \(O R \rho\) used and defined Not "evidence that ..." | | Allow 0.378 | | Reject/do not reject \(\mathrm { H } _ { 0 }\) | | Contextualised, not too definite Needn't say "positive" if \(\mathrm { H } _ { 1 } \mathrm { OK }\) | | SC 2-tail: B0; 0.4438, or 0.3783 B1; then M1A0 |
| | \(\mathrm { H } _ { 0 } : \rho = 0 , \mathrm { H } _ { 1 } : \rho > 0\) where \(\rho\) is population pmcc (not \(r\) ) | | FT on their \(r\), but not CV | | Not "scores are associated | | ...". FT on their \(r\) only |
|
| 3 | (d) | | It makes no difference as this is a linear transformation | | 2.2a | Need both "unchanged" oe and reason, need "linear" or exact equivalent | "oe" includes "their 0.4" |
| 4 | (a) | | Neither | | 2.5 | OE | Not "neither is independent of the other" |
| 4 | (b) | | \(c = 2.848 - 0.1567 m \quad \mathbf { B C }\) | | | | Correct \(a\), awrt 2.85 | | Correct \(b\), awrt 0.157 | | Letters correct from correct method | | (If both wrongly rounded, e.g. \(c = 2.84 - 0.156 m\), give B2) |
| | \(\mathrm { SC } : m\) on \(c\) : | | \(m = 15.65 - 4.832 c\) : B2 | | \(y = 15.65 - 4.832 x\) : B1 | | \(c = 15.65 - 4.832 m : \mathrm { B } 1\) | | If B0B0, give B1 for correct letters from valid working |
|
| Question | Answer | Marks | AO | Guidance |
| 4 | (c) | \(a\) unchanged, \(b\) multiplied by 2.2 (allow " \(a\) unchanged, \(b\) increases", etc) | B1 [1] | 2.2a | oe, e.g. \(c = 2.848 - 0.345 m\); \(m = 7.114 - 2.196 c\) | SC: \(m\) on \(c\) in (b): Both divided by 2.2 B1 |
| 4 | (d) | | Draw approximate line of best fit | | Draw at least one vertical from line to point | | Say that "Best fit" line minimises the sum of squares of these distances |
| | | Needs M2 and "minimises" and "sums of squares" oe | | SC: Horizontal(s): | | full marks (indept of (b)) |
|