| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Minor (Further Statistics Minor) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Name geometric distribution and parameter |
| Difficulty | Moderate -0.8 This is a straightforward application of the geometric distribution with clearly stated parameters. Parts (a)-(d) require only direct recall of the definition, standard formulas for E(X) and Var(X), and basic probability calculations using p=0.15. No problem-solving insight or multi-step reasoning is needed—just routine application of memorized formulas. |
| 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 |
|---|---|---|
| 1 | (a) | The probability of success (and/or failure) on each |
| kick is constant (0.15). | B1 | |
| [1] | 1.2 | Does not need to be in context. |
| Answer | Marks |
|---|---|
| confused with “scoring a goal” | Probability of success is ‘fixed’, or |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (b) | E(X) = 1/0.15 = 20/3 or awrt 6.67 |
| Var(X) = 0.85/0.152 = 340/9 or awrt 37.8 | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Accept 6.7 |
| Accept 38 | If B0B0 then SC1 for 20/17 (or awrt |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (c) | P(X = 3) = P(Scores, Scores, Doesn’t) |
| Answer | Marks |
|---|---|
| , | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 1.1 | Understanding of what is required |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (d) | P(X ≥ 5) = P(X > 4) = 0.854 |
| = awrt 0.522 | M1 | |
| A1 | 1.1 | |
| 1.1 | Allow M1 for 0.154 implied by |
| Answer | Marks |
|---|---|
| Alternative method 1 | M1 |
| A1 | Accept 0.52 |
Question 1:
1 | (a) | The probability of success (and/or failure) on each
kick is constant (0.15). | B1
[1] | 1.2 | Does not need to be in context.
Does not matter if “success” is
confused with “scoring a goal” | Probability of success is ‘fixed’, or
‘remains the same’, or equivalent term
1 | (b) | E(X) = 1/0.15 = 20/3 or awrt 6.67
Var(X) = 0.85/0.152 = 340/9 or awrt 37.8 | B1
B1
[2] | 1.1
1.1 | Accept 6.7
Accept 38 | If B0B0 then SC1 for 20/17 (or awrt
1.18) and 60/289 (or awrt 0.208)
1 | (c) | P(X = 3) = P(Scores, Scores, Doesn’t)
= 0.8520.15 = awrt 0.108
, | M1
A1
[2] | 1.1a
1.1 | Understanding of what is required
in a geometric scenario to obtain
X = 3
Allow M1 for 0.1520.85 implied
by awrt 0.0191
Accept 0.11
1 | (d) | P(X ≥ 5) = P(X > 4) = 0.854
= awrt 0.522 | M1
A1 | 1.1
1.1 | Allow M1 for 0.154 implied by
awrt 0.000506
Accept 0.52
Alternative method 1 | M1
A1 | Accept 0.52
P(X ≥ 5) = 1 – P(X ≤ 4) =
1−0.15(1+0.85+0.852+0.853) oe
= awrt 0.522
[2]
M1
A1
Accept 0.521 When a footballer takes a penalty kick the result is that either a goal is scored or a goal is not scored.
It is known that, on average, a certain footballer scores a goal on $85 \%$ of penalty kicks. During one practice session, the footballer decides to take penalty kicks until a goal is not scored. You may assume that the outcome of each penalty kick that the footballer takes is independent of the outcome of each other penalty kick.
The random variable representing the number of penalty kicks up to and including the first penalty kick that does not result in a goal is denoted by $X$.
\begin{enumerate}[label=(\alph*)]
\item State one further assumption that is necessary for $X$ to be modelled by a Geometric distribution.
For the remainder of this question you may assume that this assumption is valid.
\item Find each of the following.
\begin{itemize}
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
\item Find the probability that the footballer takes exactly 3 penalty kicks.
\item Find the probability that the footballer takes at least 5 penalty kicks.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2024 Q1 [7]}}