| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Variance of geometric distribution |
| Difficulty | Moderate -0.3 This is a straightforward application of standard distribution formulas (binomial and geometric). Part (a) requires recalling the binomial variance formula, part (b) involves basic geometric distribution calculations using standard formulas, and part (c) requires setting up and solving a simple equation from the geometric PMF. All parts are routine textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02d Binomial: mean np and variance np(1-p)5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(10p(1-p)\) | B1 [1] | Allow \(10pq\) e.g. \(10p - 10p^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.7^4\) | M1 | \(0.7^5 = 0.168\) or \(0.7^6 = 0.118\): M1 |
| \(= 0.240(1)\) | A1 [2] | Allow 0.24 |
| Answer | Marks | Guidance |
|---|---|---|
| \(q/p^2 = \dfrac{70}{9}\) or \(7.777\ldots\) | B1 [1] | Allow 7.78, 7.778, etc |
| Answer | Marks | Guidance |
|---|---|---|
| \((1-p)^2 p = \dfrac{4}{25}p\) | B1 | Correct equation |
| \(p = 0\) or \((1-p)^2 = \dfrac{4}{25}\) \((p\neq 0)\) | M1 | Reduce to quadratic/cubic and solve |
| \((1-p) = \pm\dfrac{2}{5}\) | M1 | Obtain two non-zero solutions |
| \(p \neq \dfrac{7}{5}\) | B1ft | Explicitly discard one solution, *either* here *or* in line 2 |
| \(p = \dfrac{3}{5}\) | A1 [5] | Exact final answer exact (0.6) no others left, allow from \(\pm\) omitted |
# Question 6:
## Part (a)
$10p(1-p)$ | **B1 [1]** | Allow $10pq$ e.g. $10p - 10p^2$ | Not just $np(1-p)$
## Part (b)(i)
$0.7^4$ | **M1** | $0.7^5 = 0.168$ or $0.7^6 = 0.118$: M1 | Or $1 - 0.3(1+0.7+0.7^2+0.7^3)$; Allow M1 if also $0.3\times 0.7^4$
$= 0.240(1)$ | **A1 [2]** | Allow 0.24 | $[0.15$ is from binomial$]$
## Part (b)(ii)
$q/p^2 = \dfrac{70}{9}$ or $7.777\ldots$ | **B1 [1]** | Allow 7.78, 7.778, etc | Allow 8 only if evidence, e.g. $(1-0.3)/0.3^2$
## Part (c)
$(1-p)^2 p = \dfrac{4}{25}p$ | **B1** | Correct equation
$p = 0$ or $(1-p)^2 = \dfrac{4}{25}$ $(p\neq 0)$ | **M1** | Reduce to quadratic/cubic and solve | e.g. $p(p^2 - 2p + \frac{21}{25}) = 0$
$(1-p) = \pm\dfrac{2}{5}$ | **M1** | Obtain two non-zero solutions | $\pm$ omitted: M0B0A1
$p \neq \dfrac{7}{5}$ | **B1ft** | Explicitly discard one solution, *either* here *or* in line 2 | Allow "$p = 0, \frac{3}{5}, \frac{7}{5}$ but $p\leq 1$"
$p = \dfrac{3}{5}$ | **A1 [5]** | Exact final answer exact (0.6) no others left, allow from $\pm$ omitted | SC binomial: B0 then $75p^2=(1-p)^2$ & solve M1; $0.104\ [0.1035]$ A1; Explicitly reject 0 or $-0.13$ B1; SC Poisson: 0
---6 A bag contains a mixture of blue and green beads, in unknown proportions. The proportion of green beads in the bag is denoted by $p$.
\begin{enumerate}[label=(\alph*)]
\item Sasha selects 10 beads at random, with replacement.
Write down an expression, in terms of $p$, for the variance of the number of green beads Sasha selects.
Freda selects one bead at random from the bag, notes its colour, and replaces it in the bag. She continues to select beads in this way until a green bead is selected. The first green bead is the $X$ th bead that Freda selects.
\item Assume that $p = 0.3$.
Find
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X \geqslant 5 )$,
\item $\operatorname { Var } ( X )$.
\end{enumerate}\item In fact, on the basis of a large number of observations of $X$, it is found that $\mathrm { P } ( X = 3 ) = \frac { 4 } { 25 } \times \mathrm { P } ( X = 1 )$.
Estimate the value of $p$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2019 Q6 [9]}}