| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2021 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Finding binomial parameters from properties |
| Difficulty | Standard +0.8 This question requires students to work backwards from binomial properties to find parameters, involving algebraic manipulation of E(X)=mp, Var(X)=mp(1-p), and the given relationships. It's more conceptually demanding than routine binomial calculations, requiring insight to eliminate unknowns and solve for p, but remains a standard Further Maths exercise without requiring novel problem-solving approaches. |
| Spec | 5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| \(np = 2mp\) | B1 | Stated |
| \(npq = 1.2mp\) | B1 | Stated |
| \(q = 1.2/2\ [= 0.6]\) | M1 | Method for \(q\) or \(1-p\), e.g. \(2(1-p) = 1.2\) |
| \(p = 1 - q = 0.4\) | A1 | Exact, www |
## Question 4:
$np = 2mp$ | **B1** | Stated
$npq = 1.2mp$ | **B1** | Stated
$q = 1.2/2\ [= 0.6]$ | **M1** | Method for $q$ or $1-p$, e.g. $2(1-p) = 1.2$
$p = 1 - q = 0.4$ | **A1** | Exact, www
[4]
---4 Two random variables $X$ and $Y$ have the distributions $\mathrm { B } ( m , p )$ and $\mathrm { B } ( n , p )$ respectively, where $p > 0$. It is known that
\begin{itemize}
\item $\mathrm { E } ( Y ) = 2 \mathrm { E } ( X )$
\item $\operatorname { Var } ( Y ) = 1.2 \mathrm { E } ( X )$.
\end{itemize}
Determine the value of $p$.
\hfill \mbox{\textit{OCR Further Statistics AS 2021 Q4 [4]}}