| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Find minimum n for P(X ≤ n) > threshold |
| Difficulty | Standard +0.3 This is a straightforward geometric distribution question requiring standard formula application. Part (a)(i) uses basic probability calculations, (a)(ii) involves simple inequality solving with logarithms, and part (b) applies the variance formula to find the mean. All techniques are routine for Further Statistics students with no novel problem-solving required. |
| 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 |
|---|---|---|
| 3 | (a) | (i) |
| = 0.212 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 3.4 | Allow 1 term wrong at either end |
| awrt 0.212 | Or pq4 + … + pq9 |
| (ii) | 0.7n–1 < ⅓, or 0.103 > 0.1 > 0.072 |
| Answer | Marks |
|---|---|
| min | M1 |
| Answer | Marks |
|---|---|
| [2] | 2.1 |
| 1.1 | Solve 0.3 × 0.7n–1 = 0.1 or < 0.1, allow inequality error |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (b) | 1− p |
| Answer | Marks |
|---|---|
| E(X) = 7 | M1 |
| Answer | Marks |
|---|---|
| A1 | 3.1a |
| Answer | Marks |
|---|---|
| 2.1 | Equate correct variance formula to 42 |
Question 3:
3 | (a) | (i) | P(X ≥ 5) – P(X ≥ 11) = 0.74 – 0.710
= 0.212 | M1
A1
[2] | 3.1b
3.4 | Allow 1 term wrong at either end
awrt 0.212 | Or pq4 + … + pq9
(ii) | 0.7n–1 < ⅓, or 0.103 > 0.1 > 0.072
n = 5
min | M1
A1
[2] | 2.1
1.1 | Solve 0.3 × 0.7n–1 = 0.1 or < 0.1, allow inequality error
5 only
SC: 5 without sufficient justification: B1
3 | (b) | 1− p
=42 ⇒ 42p2 + p – 1 = 0
p2
p = 1
7
Explicitly reject p=−1
6
E(X) = 7 | M1
A1
A1
A1
A1 | 3.1a
1.1
2.2a
2.3
2.1 | Equate correct variance formula to 42
Correct simplified quadratic equation
SC: if – 1 and 1, allow A1 for explicitly rejecting –1
7 6 7
[5]3 In a large collection of coloured marbles of identical size, the proportion of green marbles is $p$. One marble is chosen randomly, its colour is noted, and it is then replaced. This process is repeated until a green marble is chosen.
The first green marble chosen is the $X$ th marble chosen.
\begin{enumerate}[label=(\alph*)]
\item You are given that $p = 0.3$.
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { P } ( 5 \leqslant X \leqslant 10 )$.
\item Determine the smallest value of $n$ for which $\mathrm { P } ( X = n ) < 0.1$.
\end{enumerate}\item You are given instead that $\operatorname { Var } ( X ) = 42$.
Determine the value of $\mathrm { E } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2021 Q3 [9]}}