\(\mathbf { 4 }\) & \(\mathbf { ( a ) }\) & Geometric & M1 & 1.1 & Stated explicitly
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| Question | Answer | Mark | AO | Guidance |
| | | Mean \(= 400 \div 100 ( = 4 )\) and \(p = 1 /\) mean | M1 | 2.4 | Use mean (or P(1) etc) to deduce \(p\) ("Determine", so justification needed for 0.25) | Needs to deduce \(p\) in part (a), not defer it to (b) |
| \multirow{3}{*}{4} | \multirow[t]{3}{*}{(b)} | | Probability is \(0.75 ^ { 6 } ( = 0.1779785 \ldots )\) | M1 | 3.3 | | SC Geo(0.2): \(0.8 { } ^ { 6 }\) M1A0 |
| | | Or: 0.177978 or 0.177979 or better seen, or \(1 - [ \mathrm { P } ( 1 ) + \ldots + \mathrm { P } ( 6 ) ]\) with evidence, e.g. formula | M1 | | Allow ± 1 term | |
| | | Expected frequency \(=\) probability \(\times 100 = 17.798\) | A1 [2] | 2.1 | 17.798 correctly obtained, with sufficient evidence, www | \(100 - \Sigma\) (other frequencies): SC B1 |
| \multirow{6}{*}{4} | \multirow{6}{*}{(c)} | | Ho: data consistent with (geometric) | B1 | 1.1 | | Both, allow equivalents, but not "evidence that ...". 9.005 or 9.01 | | Compare their \(\Sigma X ^ { 2 }\) with 11.07 |
| E.g. \(\mathrm { H } _ { 0 } : X \sim \operatorname { Geo } ( p )\) Allow Geo(0.25) |
| | | \(\Sigma X ^ { 2 } = 9.005\) | B1 | 1.1 | | |
| | | \(9.005 < 11.07 ( v = 5 )\) | B1 | 1.1 | | |
| | | Do not reject \(\mathrm { H } _ { 0 }\). | M1ft | 1.1 | Correct first conclusion, ft on their 9.005 or on 12.59, needs like-with-like | Allow from comparison with 12.59 but nothing else |
| | | Insufficient evidence that a geometric distribution is not a good fit. | A1ft [5] | 2.2b | Contextualised, not too definite (needs double negative) Don't penalise "Geo(0.25)" | Allow addition slip in \(\Sigma X ^ { 2 }\) SC Geo(0.2): can get full marks if given data used, \(\Sigma X ^ { 2 } = 4.54\) used gets B1B1B0M1A1 |
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