| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Determine p from given mean or variance |
| Difficulty | Standard +0.8 Part (a) is straightforward application of geometric distribution formulas (finding p from mean, then calculating tail probability). Part (b) requires setting up an equation with geometric probabilities, algebraic manipulation to get a quadratic in p, then using the constraint 0<p<1 to determine the range of a—this involves discriminant analysis and inequality reasoning beyond routine application. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Geo}(0.1)\) | M1 | \(\text{Geo}(0.1)\) stated or implied, e.g. by \(0.1 \times 0.9^{13}\) |
| \(P(R > 13) = (1-p)^{13}\) | M1 | Or sum. Allow \(0.9^{14}\) (= 0.229) or \(0.9^{12}\) (= 0.282) or 1 term omitted or extra in sum, but not \(1 - (0.9)^{\text{anything}}\) |
| \(= 0.254\) | A1 [3] | Awrt 0.254, needs one of first two lines ("Determine"). Assume that, e.g. \(1 - 0.746\) is from calculator. SC no working: 0.254 B2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(pq^2 - 0.4pq - ap = 0\) | M1 | Use formula for \(\text{Geo}(p)\) to get equation involving 3 terms |
| \(p^2 - 1.6p + 0.6 - a = 0\) or \(q^2 - 0.4q - a = 0\) | A1 | Correct quadratic stated or implied, e.g. by correct use of \(b^2 - 4ac\) |
| \(p = \dfrac{1.6 \pm \sqrt{0.16 + 4a}}{2}\) or \(q = \dfrac{0.4 \pm \sqrt{0.4^2 + 4a}}{2}\) | M1 | Obtain explicit formula for \(p\) or \(q\), e.g. from \((q - 0.2)^2 = a + 0.04\), needn't be simplified |
| \(p > 0 \Rightarrow 0.8 - \sqrt{0.04 + a} > 0\) *or* \(q < 1 \Rightarrow 0.2 + \sqrt{0.04 + a} < 1\) | M1 | Use \(p > 0\) from negative sign *or* \(q < 1\) from positive sign (ignore other combinations of inequality and sign). Allow \(p \geq 0\) or \(q \leq 1\) |
| \(\sqrt{0.04 + a} < 0.8 \Rightarrow a < 0.6\) | A1 | Obtain \(a < 0.6\), allow \(a \leq 0.6\) here |
| \(p = 0.8 - \sqrt{0.04 + a}\) decreases with \(a\) (so any small positive \(a\) gives a valid value of \(p\)) *(or similar for \(q\))* | B1 | Reason why the lower limit is (not greater than) 0, e.g. sketch of \(p\) against \(a\), or other valid justification other than just "\(a > 0\) is given" |
| \(0 < a < 0.6\) *(strict inequalities only)* | A1 [7] | Fully correct, allow just \(a < 0.6\), needs all previous marks apart from B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = p^2 - 1.6p + 0.6\) (or \(a = q^2 - 0.4q\)) | M1 | Write \(a\) in terms of \(p\) or \(q\) and draw graph |
| Intersections \((0, 0.6)\), \((0.6, 0)\) (and \((1, 0)\)) | A1 | Correct parabolic shape, and intersections at \((0.6, 0)\) and \((0, 0.6)\) clear |
| \(0 < p < 1 \Rightarrow -0.04 < a < 0.6\) | M1 | Identify range of \(a\) for which \(0 < p < 1\) |
| \(0 < a < 0.6\) | A2 | \(0 < a < 0.6\) (\(\leq\) used, or \(-0.04 < a < 0.6\): A1). Max 7/7 |
# Question 7:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Geo}(0.1)$ | **M1** | $\text{Geo}(0.1)$ stated or implied, e.g. by $0.1 \times 0.9^{13}$ |
| $P(R > 13) = (1-p)^{13}$ | **M1** | Or sum. Allow $0.9^{14}$ (= 0.229) or $0.9^{12}$ (= 0.282) or 1 term omitted or extra in sum, but not $1 - (0.9)^{\text{anything}}$ |
| $= 0.254$ | **A1 [3]** | Awrt 0.254, needs one of first two lines ("Determine"). Assume that, e.g. $1 - 0.746$ is from calculator. SC no working: 0.254 B2 |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $pq^2 - 0.4pq - ap = 0$ | **M1** | Use formula for $\text{Geo}(p)$ to get equation involving 3 terms |
| $p^2 - 1.6p + 0.6 - a = 0$ or $q^2 - 0.4q - a = 0$ | **A1** | Correct quadratic stated or implied, e.g. by correct use of $b^2 - 4ac$ |
| $p = \dfrac{1.6 \pm \sqrt{0.16 + 4a}}{2}$ or $q = \dfrac{0.4 \pm \sqrt{0.4^2 + 4a}}{2}$ | **M1** | Obtain explicit formula for $p$ or $q$, e.g. from $(q - 0.2)^2 = a + 0.04$, needn't be simplified |
| $p > 0 \Rightarrow 0.8 - \sqrt{0.04 + a} > 0$ *or* $q < 1 \Rightarrow 0.2 + \sqrt{0.04 + a} < 1$ | **M1** | Use $p > 0$ from negative sign *or* $q < 1$ from positive sign (ignore other combinations of inequality and sign). Allow $p \geq 0$ or $q \leq 1$ |
| $\sqrt{0.04 + a} < 0.8 \Rightarrow a < 0.6$ | **A1** | Obtain $a < 0.6$, allow $a \leq 0.6$ here |
| $p = 0.8 - \sqrt{0.04 + a}$ decreases with $a$ (so any small positive $a$ gives a valid value of $p$) *(or similar for $q$)* | **B1** | Reason why the lower limit is (not greater than) 0, e.g. sketch of $p$ against $a$, or other valid justification other than just "$a > 0$ is given" |
| $0 < a < 0.6$ *(strict inequalities only)* | **A1 [7]** | Fully correct, allow just $a < 0.6$, needs all previous marks apart from B1 |
**OR method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = p^2 - 1.6p + 0.6$ (or $a = q^2 - 0.4q$) | **M1** | Write $a$ in terms of $p$ or $q$ and draw graph |
| Intersections $(0, 0.6)$, $(0.6, 0)$ (and $(1, 0)$) | **A1** | Correct parabolic shape, and intersections at $(0.6, 0)$ and $(0, 0.6)$ clear |
| $0 < p < 1 \Rightarrow -0.04 < a < 0.6$ | **M1** | Identify range of $a$ for which $0 < p < 1$ |
| $0 < a < 0.6$ | **A2** | $0 < a < 0.6$ ($\leq$ used, or $-0.04 < a < 0.6$: A1). Max 7/7 |7 A town council is planning to introduce a new set of parking regulations. An interviewer contacts randomly chosen people in the town and asks them whether they are in favour of the proposal. The first person who is not in favour of the regulation is the $R$ th person interviewed. It can be assumed that the probability that any randomly chosen person is not in favour of the proposal is a constant $p$, and that $p$ does not equal 0 or 1 .
Assume first that $\mathrm { E } ( R ) = 10$.
\begin{enumerate}[label=(\alph*)]
\item Determine $\mathrm { P } ( R \geqslant 14 )$.
Now, without the assumption that $\mathrm { E } ( R ) = 10$, consider a general value of $p$.\\
It is given that $\mathrm { P } ( R = 3 ) - 0.4 \times \mathrm { P } ( R = 2 ) - a \times \mathrm { P } ( R = 1 ) = 0$, where $a$ is a positive constant.
\item Determine the range of possible values of $a$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2023 Q7 [10]}}