| Relationship may not continue | B1 | Can't extrapolate. Any indication that pattern may not continue. Must state or imply referring to future | Allow without context. Examples: Can't predict future; Things can change. May be recession ahead; Economic situation may change. Cost of advertising may increase. If spend too much on ads, profit may be reduced as a result. Advertising may not be as successful in the future. Item may go out of fashion. NOT Spending on adverts may not bring high profits. NOT Spending more on adverts may not bring higher profits (Since these just restate the question). NOT More money spent on ads will not affect profit |
| Corr'n not imply causation | B1 2 | Increase in profit may not be due to increase in spend on advertising. Variables may be increasing separately | Both variables may be affected by a third. Other factors may affect profits. Advertising not the sole factor affecting profits. Two different categories of reason needed, as given above. Two reasons which both fall under the same category: only B1. NOT Because corr'n not equal to 1 |
| $b = \frac{7351.12-\frac{86.6×943.8}{12}}{658.76-\frac{86.6^2}{12}}$ = 15.9788 or 16.0. $y–\frac{943.8}{12}$ = "16.0"$(x–\frac{86.6}{12})$. $y$ = 16$x$ – 37 or better | M1, A1, M1, A1 4 | or $a = \frac{943.8}{12}$ – "16.0" × $\frac{86.6}{12}$ (y = 15.9788x – 36.664). Coeffs not nec'y rounded, but would round to 16 & 37. These marks can be earned in (v) if not contradicted in (iv). If $x$ on $y$ line found: M-marks only ($x$ = 2.71 + 0.0572y). "16" × 7400 – "37": M0A0. If their (iv) |
| "16" × 7.4 – "37", 81400 to 81750 | M1, A1f 2 | 81.4 thousand to 81.7 thousand: M1A1 but 81.4 to 81.7 alone: M1A0 | ft their (iv) |

## Question 4i

| 0.4 × 0.7, 0.6 + 0.4 × 0.7 = 0.88 | M1, M1, A1 3 | or 0.6 + prod of 2 probs. Condone 0.6 × 0.7 + 0.6×0.3 + 0.4×0.7 or 0.6 × 0.6 + 0.6×0.4 + 0.4×0.7 | 1– prod of 2 P's or 0.4 × 0.3. 1 – 0.4 × 0.3 |
| $p + (1–p) × p$ = 0.51 or $2p – p^2$ = 0.51. $p^2 – 2p + 0.51$ = 0. $(p–0.3)(p–1.7)$ = 0 or $p = \frac{2±\sqrt{4-4×0.51}}{2}$ oe. $p$ = 0.3 | M1, M1, A1, A1 4 | or $p^2 + p×(1–p) + (1–p) × p$ = 0.51. Correct QE = 0. Condone omission of "= 0". Correct method for their 3-term QE. Not $p$ = 0.3 or 1.7 | Condone $p + p × 1–p$ M1, but $p + qp$ = 0.51 M0. Correct ans from correct but reduced wking or T & I or verification or no wking: 4 mks. Ans $p$ = 0.3 or 1.7 from correct but reduced wking or T & I or no wking: M1M1M1A0. Ans $p$ = 0.3 following correct wking except other solution incorrect: BOD 4 mks. ($eg p = \frac{2±\sqrt{4-4×0.51}}{2}$ so $p$ = 0.3 or –1.3 so $p$ = 0.3; 4 mks)). $p$ = 0.3 from wrong wking alone: M0A0M0A0 |