S1 June 2013 (R) Q7
7. The score \(S\) when a spinner is spun has the following probability distribution.
| \(s\) | 0 | 1 | 2 | 4 | 5 |
|---|---|---|---|---|---|
| \(\mathrm{P}(S = s)\) | 0.2 | 0.2 | 0.1 | 0.3 | 0.2 |
The spinner is spun twice.
The score from the first spin is \(S_1\) and the score from the second spin is \(S_2\)
The random variables \(S_1\) and \(S_2\) are independent and the random variable \(X = S_1\times S_2\)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(S) = 0 + 1\times 0.2 + 2\times 0.1 + 4\times 0.3 + 5\times 0.2 = [0.2 + 0.2 + 1.2 + 1.0]\) | M1 |
| \(\underline{\mathbf{2.6}}\) | A1 |
| (2) |
Notes
M1 for an attempt at \(\sum x\mathrm{P}(X = x)\), at least 2 non-zero terms seen. Correct answer 2/2
A1 for 2.6 or any exact equivalent
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(S^2) = 0 + 1\times 0.2 + 2^2\times 0.1 + 4^2\times 0.3 + 5^2\times 0.2\) or \(0.2 + 0.4 + 4.8 + 5\) | M1 |
| \(\underline{\mathbf{10.4}}\) (*) | A1cso |
| (2) |
Notes
M1 for a correct attempt, at least 3 non-zero terms seen
A1cso for 10.4 provided M1 is scored and no incorrect working seen
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(S) = 10.4 - (\text{"}2.6\text{"})^2\) | M1 |
| \(\underline{\mathbf{3.64}}\) or \(\frac{91}{25}\) (o.e.) | A1 |
| (2) |
Notes
M1 for \(10.4 - \mu^2\), ft their \(\mu\). Must see their value of \(\mu\) squared (A1 for 3.64 or any exact equiv.)
| Scheme | Marks |
|---|---|
| (i) \(5\mathrm{E}(S) - 3 = 5\times\text{"}2.6\text{"} - 3\), \(= \underline{\mathbf{10}}\) | M1, A1 |
| (ii) \(5^2\,\mathrm{Var}(S) = 25\times 3.64\), \(= \underline{\mathbf{91}}\) | M1, A1 |
| (4) |
Notes
(i) M1 for a correct expression using their 2.6 (A1 for 10)
(ii) M1 for \(25\times\mathrm{Var}(S)\)- ft their \(\mathrm{Var}(S)\) (A1 for 91)
| Scheme | Marks |
|---|---|
| \(5S - 3 \gt S + 3 \;\Rightarrow\; 4S \gt 6\) or \(S \gt 1.5\), so need \(\mathrm{P}(S \geqslant 2)\) | M1, A1 |
| \(\mathrm{P}(S \geqslant 2) = \underline{\mathbf{0.6}}\) | A1 |
| (3) |
Notes
M1 for solving the inequality as far as \(pS \gt q\) where one of \(p\) or \(q\) are correct
1st A1 for \(\mathrm{P}(S \geqslant 2)\)
2nd A1 for 0.6 (provided \(S \gt 1.5\) was obtained). Ans only of 0.6 scores 3/3
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(S_1 = 1)\times\mathrm{P}(S_2 \leqslant 4), = 0.2\times 0.8 = 0.16\) (*) | M1,A1cso |
| (2) |
Notes
A table showing all 25 cases can only score M1 in (g) if the correct cases are indicated.
M1 for using independence (so multiplying) and attempting \(\mathrm{P}(S_2 \leqslant 4)\)
e.g. \(0.2\times(0.2 + 0.2 + 0.1 + 0.3)\) or \(0.04 + 0.04 + 0.02 + 0.06\) score M1 BUT \(\frac{4}{25}\) (not from \(0.2\times 0.8\)) is M0A0
A1cso for a fully correct explanation leading to 0.16. Must come from \(0.2\times 0.8\) not \(\frac{4}{25}\)
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(S_1 = 2)\times\mathrm{P}(S_2 \leqslant 2) = 0.1\times 0.5 = 0.05\) \(\mathrm{P}(S_1 = 4)\times\mathrm{P}(S_2 \leqslant 1) = 0.3\times 0.4 = 0.12\) \(\mathrm{P}(S_1 = 5)\times\mathrm{P}(S_2 = 0) = 0.2\times 0.2 = 0.04\) \(\mathrm{P}(S_1 = 0)\times\mathrm{P}(S_2 = \text{any value}) = 0.2\times 1 = 0.20\) Full method – all cases listed | M1 |
| all correct products | A1 |
| \(= \underline{\mathbf{0.57}}\) | A1 |
| (3) | |
| (18 marks) |
Notes
M1 for all cases for \(S_1\) or all 15 cases for \(X\)
1st A1 for all correct probability products for \(S_1\) or \(X\)
2nd A1 for 0.57 Correct answer scores 3/3. Probabilities out of 25 score A0A0