| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Mean/expectation of geometric distribution |
| Difficulty | Moderate -0.8 This is a straightforward application of standard geometric distribution formulas. Part (a) requires recalling that mean = 1/p, part (b) uses the CDF formula P(X≤k) = 1-(1-p)^k, and part (c) applies binomial variance to a probability calculated from geometric distribution. All parts are direct formula application with minimal problem-solving or conceptual challenge, making this easier than average even for Further Maths. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{0.2} = 5\) | M1, A1 [2] | Geometric distribution soi; \(5\) (or \(5.00\ldots\)) only |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.8^2 - 0.8^{10}\) | M1 | Allow for powers \(2, 3, 4\) and \(9, 10, 11\). Values: \(0.506, 0.378, 0.275, 0.405, 0.302, 0.554, 0.426, 0.324\) |
| \(= 0.533\) \((0.5326258\ldots)\) | A1 [2] | Awrt \(0.533\), www. \([5201424/9765625]\); Or \(0.2(0.8^2 + \ldots + 0.8^9)\), \(\pm 1\) term at either end |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\geq 10) = 0.8^9\) | M1 | Or \(0.8^{10}\). Can be implied by correct \(p\); SC: \(0.134(2)\) oe not properly shown: B2 for correct final answer |
| \(= 0.1342\ldots\) | A1 | \([0.10737\ldots\) is M1A0 here\(]\) |
| \(B(30, 0.1342\ldots)\) | M1 | Stated or implied, their \(0.8^9\) or \(0.8^{10}\) |
| Variance \(= npq\) \(= 3.486\ldots\) | A1ft [4] | In range \([3.48, 3.49]\); \(2.875\) from \(0.8^{10}\): M1A0M1A1ft |
# Question 2:
## Part (a)
$\frac{1}{0.2} = 5$ | **M1, A1** [2] | Geometric distribution soi; $5$ (or $5.00\ldots$) only
## Part (b)
$0.8^2 - 0.8^{10}$ | **M1** | Allow for powers $2, 3, 4$ and $9, 10, 11$. Values: $0.506, 0.378, 0.275, 0.405, 0.302, 0.554, 0.426, 0.324$
$= 0.533$ $(0.5326258\ldots)$ | **A1** [2] | Awrt $0.533$, www. $[5201424/9765625]$; Or $0.2(0.8^2 + \ldots + 0.8^9)$, $\pm 1$ term at either end
## Part (c)
$P(\geq 10) = 0.8^9$ | **M1** | Or $0.8^{10}$. Can be implied by correct $p$; SC: $0.134(2)$ oe not properly shown: B2 for correct final answer
$= 0.1342\ldots$ | **A1** | $[0.10737\ldots$ is M1A0 here$]$
$B(30, 0.1342\ldots)$ | **M1** | Stated or implied, their $0.8^9$ or $0.8^{10}$
Variance $= npq$ $= 3.486\ldots$ | **A1ft** [4] | In range $[3.48, 3.49]$; $2.875$ from $0.8^{10}$: M1A0M1A1ft
---2 Every time a spinner is spun, the probability that it shows the number 4 is 0.2 , independently of all other spins.
\begin{enumerate}[label=(\alph*)]
\item A pupil spins the spinner repeatedly until it shows the number 4.
Find the mean of the number of spins required.
\item Calculate the probability that the number of spins required is between 3 and 10 inclusive.
\item Each pupil in a class of 30 spins the spinner until it shows the number 4. Out of the 30 pupils, the number of pupils who require at least 10 spins is denoted by $X$.
Determine the variance of $X$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2020 Q2 [8]}}