| Exam Board | OCR |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2017 |
| Session | Specimen |
| Marks | 7 |
| Topic | Geometric Distribution |
| Type | Determine p from given mean or variance |
| Difficulty | Standard +0.3 This is a straightforward application of geometric distribution formulas from Further Statistics 1. Part (i) requires standard formula recall for P(X≥8), E(X), and Var(X). Part (ii) involves solving a simple equation P(Y<4)=1-(1-p)³=0.986 for p, which is algebraically routine. While this is Further Maths content, the question demands only direct formula application with minimal problem-solving, making it slightly easier than an average A-level question overall. |
| 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 |
\begin{enumerate}[label=(\roman*)]
\item The random variable $X$ has the distribution $\text{Geo}(0.6)$.
\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 $\text{Var}(X)$. [1]
\end{enumerate}
\item The random variable $Y$ has the distribution $\text{Geo}(p)$. 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 FS1 AS 2017 Q5 [7]}}