| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Name geometric distribution and parameter |
| Difficulty | Moderate -0.3 Part (a) requires straightforward identification of a geometric distribution with p=0.01 and a routine probability calculation using tables or calculator. Part (b) involves algebraic verification that a specific (incorrect) formula happens to work for geometric distributions, requiring knowledge of E(X) and Var(X) formulas but only basic algebraic manipulation. This is slightly easier than average as it's mostly recall and standard application with minimal problem-solving. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | (i) |
| (a) | (ii) | qm – qn where m = 50, 49, 51, n = 150, 149, 151 |
| q50 – q150 = 0.384 (0.38355..) | M1 | |
| A1 | 2.1 | |
| 1.1 | e.g. 0.378, 0.386, 0.387, 0.381, 0.390, 0.392. Their q | |
| In range [0.383, 0.384], cao | Or pq50 + … + pq149 = pq50(1 – q100)/(1 – q) |
| Answer | Marks |
|---|---|
| (b) | E(R) = 1/p and correct Var(R) = (1 – p)/p2 |
| = (1/p)2 – (1/p) = [E(R)]2 – E(R) | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 2.1 | Art least one correct general formulae stated |
Question 1:
1 | (a) | (i) | Geo(0.01) | B1 | 1.1 | Allow if 0.01 not stated here but used in (ii)
(a) | (ii) | qm – qn where m = 50, 49, 51, n = 150, 149, 151
q50 – q150 = 0.384 (0.38355..) | M1
A1 | 2.1
1.1 | e.g. 0.378, 0.386, 0.387, 0.381, 0.390, 0.392. Their q
In range [0.383, 0.384], cao | Or pq50 + … + pq149 = pq50(1 – q100)/(1 – q)
1 term at either end for M1
M1 needs method for summing GP
(b) | E(R) = 1/p and correct Var(R) = (1 – p)/p2
= (1/p)2 – (1/p) = [E(R)]2 – E(R) | M1
A1
[2] | 1.1
2.1 | Art least one correct general formulae stated
Correctly show required relationship, either way
round, don’t need conclusion, not numerical1 A researcher wishes to find people who say that they support a specific plan. Each day the researcher interviews people at random, one after the other, until they find one person who says that they support this plan. The researcher does not then interview any more people that day. The total number of people interviewed on any one day is denoted by $R$.
\begin{enumerate}[label=(\alph*)]
\item Assume that in fact $1 \%$ of the population would say that they support the plan.
\begin{enumerate}[label=(\roman*)]
\item State an appropriate distribution with which to model $R$, giving the value(s) of any parameter(s).
\item Find $\mathrm { P } ( 50 < R \leqslant 150 )$.
The researcher incorrectly believes that the variance of a random variable $X$ with any discrete probability distribution is given by the formula $[ \mathrm { E } ( X ) ] ^ { 2 } - \mathrm { E } ( X )$.
\end{enumerate}\item Show that, for the type of distribution stated in part (a), they will obtain the correct value of the variance, regardless of the value(s) of the parameter(s).
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2022 Q1 [5]}}