| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Topic | Geometric Distribution |
| Type | Variance of geometric distribution |
| Difficulty | Standard +0.3 This is a straightforward application of geometric distribution properties. Part (a) uses basic variance scaling (1 mark). Part (b) requires finding p from the variance formula, then computing the expectation and applying linear transformation (6 marks total but routine). Part (c) needs calculating a probability using the geometric distribution formula. All steps are standard textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^25.04a Linear combinations: E(aX+bY), Var(aX+bY) |
The random variable $D$ has the distribution Geo$(p)$. It is given that Var$(D) = \frac{40}{9}$.
Determine
\begin{enumerate}[label=(\alph*)]
\item Var$(3D + 5)$. [1]
\item E$(3D + 5)$. [6]
\item $\text{P}(D > \text{E}(D))$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2021 Q4 [10]}}