S2 January 2013 Q3
3. A random variable \(X\) has the distribution B(12, \(p\)).
| Scheme | Marks |
|---|---|
| (i) \(\mathrm{P}(X \lt 5) = 0.8424\) awrt 0.842 | B1 |
| (ii) \(\mathrm{P}(X \geqslant 7) = 1 - \mathrm{P}(X \leqslant 6)\) | M1 |
| \(= 1 - 0.9857\) | |
| \(= 0.0143\) awrt 0.0143 | A1 |
| (3) |
Notes
(ii) M1 writing or using \(1 - \mathrm{P}(X \leqslant 6)\) Do not accept \(1 - \mathrm{P}(X \lt 7)\) unless \(1 - \mathrm{P}(X \leqslant 6)\) has been used
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X = 0) = (1 - p)^{12}\) | |
| \((1 - p)^{12} = 0.05\) | M1 |
| \((1 - p) = \sqrt[12]{0.05}\) | M1 |
| \(p = 0.221\) awrt 0.221 | A1 |
| (3) |
Notes
1st M1 \((1 - p)^n = 0.05\)
2nd M1 taking \(n\)th root. If they have used logs they need to get to a correct expression for \(1 - p\) for their equation.
| Scheme | Marks |
|---|---|
| Variance \(= 12p(1 - p)\) | |
| \(12p(1 - p) = 1.92\) | M1 |
| \(12p - 12p^2 = 1.92\) | |
| \(12p^2 - 12p + 1.92 = 0\) or \(p^2 - p + 0.16 = 0\), \(25p^2 - 25p + 4 = 0\) | |
| \(p = \dfrac{12 \pm \sqrt{12^2 - 4 \times 12 \times 1.92}}{24}\) or \((5p - 1)(5p - 4) = 0\) | M1 |
| \(p = 0.2\) or \(0.8\) | A1,A1 |
| (4) | |
| (10 marks) |
Notes
1st M1 \(12p(1 - p) = 1.92\) o.e.
2nd M1 solving a quadratic either by factorising / completing the square / or formula. Working must either be correct for their quadratic (they may use a quadratic from an incorrect rearrangement) or they must have written the appropriate formula down correctly and only made 1 error substituting into it. May be implied by a correct value of \(p\).
1st A1 for 0.2
2nd A1 for 0.8