| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | State assumptions for geometric model |
| Difficulty | Standard +0.3 This is a straightforward geometric distribution question testing standard knowledge. Part (a) requires recall of standard assumptions, (b) is a direct probability calculation, (c) involves standard formulas for E(Y) and Var(Y) with simple algebraic manipulation, and (d) requires comparing two geometric probabilities to find a range for k. All parts are routine applications of geometric distribution theory with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Whether or not one car is white is independent of whether or not any other is. | B1 | Independence *or* "selected randomly", stated, context used. Allow BOD in application of context |
| For any car, the probability that it is white is constant. | B1 [2] | \(p\) constant, oe, stated, in context. *Not* "singly". More than 2 conditions: max B1 |
| *OR*: The sample is a random sample of residents | B2 | "Selected randomly": B1 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.9^6\) | M1 | \(0.9^7 = 0.478\ldots\) or \(0.9^5 = 0.590\): M1A0. *Not* \(1 - 0.9^6\) |
| \(= 0.531(441)\) | A1 [2] | Clearly correct numerical answer, in range [0.532, 0.532], www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(Y) = \frac{1}{p} - 1\) and \(\text{Var}(Y) = \frac{1-p}{p^2}\) | M1 | \(Y = X - 1\) stated or implied |
| B1 | Correct variance formula, allow if unclear whether it applies to \(X\) or \(Y\) | |
| Hence \(E(Y) \div \text{Var}(Y) = p\) | A1 [3] | CWO, needs to be simplified to \(p\) only. Typically \(\frac{1}{p} \div \frac{1-p}{p^2} = \frac{p}{1-p}\): M0B1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}p(1 - \frac{1}{2}p)^2 = kp(1-p)^2\) | M1 | Equation with at least one geometric formula correct in terms of \(p\) or \(\frac{1}{2}p\) |
| \(2k = \left(\frac{1-\frac{1}{2}p}{1-p}\right)^2\) or \(8k = \left(\frac{2-p}{1-p}\right)^2\) oe | A1 | Correct equation with \(p\) cancelled, aef, can be implied by correct \(k\) etc |
| \(\frac{1}{2}p < p \Rightarrow 1 - \frac{1}{2}p > 1 - p\) | B1 | Reason for linear inequality for \(p\), e.g. LHS \(> 1\) |
| \(\Rightarrow\) LHS \(> 1\) | B1 | Independent but needs justification |
| \(\Rightarrow k > \frac{1}{2}\) | B1 [5] | Final answer \(k > \frac{1}{2}\) or \(k > 0.5\) only. *Not* \(k \geq 0.5\) |
| *OR*: \(p = 0, k = 0.5\); \(p > 0, k > 0.5\); (and \(k\) increases with \(p\) because \(1 - \frac{1}{2}p > 1-p\)); \(k > 0.5\) | M1, B1, B1 | Use one value of \(p\) in \([0,1)\) to find \(k\); assert \(k\) increasing with \(p\); final answer |
| *OR*: \(p = \frac{8k - 2 \pm \sqrt{8k}}{8k-1}\) and \(0 < p < 1\) | M1 | Solve quadratic for \(p\) and use at least one correct inequality |
| Ignore negative sign (inequalities satisfied for all \(k\)); \(k > 0.5\) | B1, A1 | Correct choice of signs; final answer \(k > \frac{1}{2}\) or \(k > 0.5\) only. *Not* \(k \geq 0.5\) |
# Question 6:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Whether or not one car is white is independent of whether or not any other is. | **B1** | Independence *or* "selected randomly", stated, context used. Allow BOD in application of context |
| For any car, the probability that it is white is constant. | **B1 [2]** | $p$ constant, oe, stated, in context. *Not* "singly". More than 2 conditions: max B1 |
| *OR*: The sample is a random sample of residents | **B2** | "Selected randomly": B1 only |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.9^6$ | **M1** | $0.9^7 = 0.478\ldots$ or $0.9^5 = 0.590$: M1A0. *Not* $1 - 0.9^6$ |
| $= 0.531(441)$ | **A1 [2]** | Clearly correct numerical answer, in range [0.532, 0.532], www |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(Y) = \frac{1}{p} - 1$ and $\text{Var}(Y) = \frac{1-p}{p^2}$ | **M1** | $Y = X - 1$ stated or implied |
| | **B1** | Correct variance formula, allow if unclear whether it applies to $X$ or $Y$ |
| Hence $E(Y) \div \text{Var}(Y) = p$ | **A1 [3]** | CWO, needs to be simplified to $p$ only. Typically $\frac{1}{p} \div \frac{1-p}{p^2} = \frac{p}{1-p}$: M0B1A0 |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}p(1 - \frac{1}{2}p)^2 = kp(1-p)^2$ | **M1** | Equation with at least one geometric formula correct in terms of $p$ or $\frac{1}{2}p$ |
| $2k = \left(\frac{1-\frac{1}{2}p}{1-p}\right)^2$ or $8k = \left(\frac{2-p}{1-p}\right)^2$ oe | **A1** | Correct equation with $p$ cancelled, aef, can be implied by correct $k$ etc |
| $\frac{1}{2}p < p \Rightarrow 1 - \frac{1}{2}p > 1 - p$ | **B1** | Reason for linear inequality for $p$, e.g. LHS $> 1$ |
| $\Rightarrow$ LHS $> 1$ | **B1** | Independent but needs justification |
| $\Rightarrow k > \frac{1}{2}$ | **B1 [5]** | Final answer $k > \frac{1}{2}$ or $k > 0.5$ only. *Not* $k \geq 0.5$ |
| *OR*: $p = 0, k = 0.5$; $p > 0, k > 0.5$; (and $k$ increases with $p$ because $1 - \frac{1}{2}p > 1-p$); $k > 0.5$ | **M1, B1, B1** | Use one value of $p$ in $[0,1)$ to find $k$; assert $k$ increasing with $p$; final answer |
| *OR*: $p = \frac{8k - 2 \pm \sqrt{8k}}{8k-1}$ and $0 < p < 1$ | **M1** | Solve quadratic for $p$ and use at least one correct inequality |
| Ignore negative sign (inequalities satisfied for all $k$); $k > 0.5$ | **B1, A1** | Correct choice of signs; final answer $k > \frac{1}{2}$ or $k > 0.5$ only. *Not* $k \geq 0.5$ |6 Anika walks along a street that contains parked cars. The number of cars that Anika passes, up to and including the first car that is white, is denoted by $X$.
\begin{enumerate}[label=(\alph*)]
\item State two assumptions needed for $X$ to be well modelled by a geometric distribution.
Assume now that $X$ can be well modelled by the distribution $\operatorname { Geo } ( p )$, where $0 < p < 1$.
\item For $p = 0.1$, find $\mathrm { P } ( X > 6 )$.
The number of cars that Anika passes, up to but not including the first car that is white, is denoted by $Y$.
\item For a general value of $p$, determine a simplified expression for $\mathrm { E } ( Y ) \div \operatorname { Var } ( Y )$, in terms of $p$.
Ben walks along a different street that also contains parked cars. The number of cars that Ben passes, up to and including the first white car on which the last digit of the number plate is even is denoted by $Z$.
It may be assumed that $Z$ can be well modelled by the distribution $\operatorname { Geo } \left( \frac { 1 } { 2 } p \right)$, where $p$ is the parameter of the distribution of $X$.
It is given that $\mathrm { P } ( \mathrm { Z } = 3 ) = \mathrm { kP } ( \mathrm { X } = 3 )$, where $k$ is a positive constant.
\item Determine the range of possible values of $k$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2024 Q6 [12]}}