| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Determine p from given mean or variance |
| Difficulty | Standard +0.3 This question tests standard geometric distribution formulas and basic algebraic manipulation. Part (i) requires recall of P(X≥k) = (1-p)^(k-1) and standard E(X) and Var(X) formulas. Part (ii) involves setting up and solving 1-(1-p)³ = 0.986, which is straightforward algebra. All techniques are routine for Further Maths students 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)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | (a) |
| Answer | Marks |
|---|---|
| (cid:172) (cid:188) | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 1.1 | 128 |
| Answer | Marks |
|---|---|
| 78125 | Allow M1 for 0.000655 or 0.48 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | (b) |
| 3 | B1 | |
| [1] | 1.1 | n |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | (c) |
| 9 | B1 | |
| [1] | 1.1 | e |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (ii) | P(Y (cid:31)4)(cid:32)1(cid:16)q3 (cid:32)0.986 |
| Answer | Marks |
|---|---|
| So p(cid:32)1(cid:16)q(cid:32)0.759to 3 sf | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| 1.1 | m | 1(cid:16)q4 [but not q3p] gets |
| Answer | Marks |
|---|---|
| 5(ii) | p |
Question 5:
5 | (i) | (a) | 0.47
(cid:32)0.00164 (cid:170)1.6384(cid:117)10(cid:16)3(cid:186)
(cid:172) (cid:188) | M1
A1
[2] | 1.1a
1.1 | 128
Or
78125 | Allow M1 for 0.000655 or 0.48
OR
1(cid:16)(0.6(cid:14)0.6(cid:117)0.4(cid:14)0.6(cid:117)0.42 (cid:14)
...(cid:14)0.6(cid:117)0.46)
5 | (i) | (b) | 5
3 | B1
[1] | 1.1 | n
Any equivalent exact fraction or
awrt 1.67
5 | (i) | (c) | 10
9 | B1
[1] | 1.1 | e
Any equivalent exact fraction or
awrt 1.11
5 | (ii) | P(Y (cid:31)4)(cid:32)1(cid:16)q3 (cid:32)0.986
q(cid:32)30.014 (cid:32)0.241...
So p(cid:32)1(cid:16)q(cid:32)0.759to 3 sf | M1
c
M1
A1
[3] | 3.1a
i
2.1
1.1 | m | 1(cid:16)q4 [but not q3p] gets
M1M1A0
0.75898577…
--- 5(ii) ---
5(ii) | p
S\begin{enumerate}[label=(\roman*)]
\item The random variable $X$ has the distribution $\mathrm{Geo}(0.6)$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm{P}(X \geq 8)$. [2]
\item Find the value of $\mathrm{E}(X)$. [1]
\item Find the value of $\mathrm{Var}(X)$. [1]
\end{enumerate}
\item The random variable $Y$ has the distribution $\mathrm{Geo}(p)$. It is given that $\mathrm{P}(Y < 4) = 0.986$ correct to 3 significant figures. Use an algebraic method to find the value of $p$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS Q5 [7]}}