| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | State assumptions for geometric model |
| Difficulty | Moderate -0.8 This is a straightforward question testing basic knowledge of the geometric distribution. Parts (i) and (v) require stating standard assumptions (independence/constant probability), while parts (ii)-(iv) involve direct application of standard formulas with no problem-solving required. The context is simple and all parts are routine textbook exercises. |
| Spec | 5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Receipt of books is independent OR receipt of one delivery does not affect receipt of another | B1 [1] | Contextualised (not "events"), properly expressed, no extras. Allow "probabilities independent" |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.1 \times 0.9^{10} = \mathbf{0.03486}\) | M1 A1 [2] | M1: Allow \(0.1 \times 0.9^{11}\). A1: Answer awrt 0.0349. Allow 0.03138 |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - 0.9^8 = \mathbf{0.5695\ldots}\) | M1 A1 [2] | M1: Allow \(1 - 0.9^7\) or \(1 - 0.9^9\), or add up all terms (\(\pm 1\) term). A1: Answer, allow 0.570 or 0.57. Allow 0.5217 or 0.6126 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{0.9}{0.1^2} = \mathbf{90}\) | M1 A1 [2] | M1: Correct formula used. A1: Exact answer only |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. probabilities different for first few days of year because of holidays | B1 [1] | Give contextualised reason for different probability within first few days, or not independent, or not random. Needn't state "different probs" or "not independent" explicitly. Not for reason that denies geometric for whole year |
## Question 1:
### Part (i)
Receipt of books is independent OR receipt of one delivery does not affect receipt of another | **B1** [1] | Contextualised (not "events"), properly expressed, no extras. Allow "probabilities independent"
### Part (ii)
$0.1 \times 0.9^{10} = \mathbf{0.03486}$ | **M1** A1 [2] | M1: Allow $0.1 \times 0.9^{11}$. A1: Answer awrt 0.0349. Allow 0.03138
### Part (iii)
$1 - 0.9^8 = \mathbf{0.5695\ldots}$ | **M1** A1 [2] | M1: Allow $1 - 0.9^7$ or $1 - 0.9^9$, or add up all terms ($\pm 1$ term). A1: Answer, allow 0.570 or 0.57. Allow 0.5217 or 0.6126
### Part (iv)
$\dfrac{0.9}{0.1^2} = \mathbf{90}$ | **M1** A1 [2] | M1: Correct formula used. A1: Exact answer only
### Part (v)
e.g. probabilities different for first few days of year because of holidays | **B1** [1] | Give contextualised reason for different probability within first few days, or not independent, or not random. Needn't state "different probs" or "not independent" explicitly. Not for reason that denies geometric for whole year
---1 A book reviewer estimates that the probability that he receives a delivery of books to review on any one weekday is 0.1 . The first weekday in September on which he receives a delivery of books to review is the Xth weekday of September.\\
(i) State an assumption needed for $X$ to be well modelled by a geometric distribution.\\
(ii) Find $\mathrm { P } ( X = 11 )$.\\
(iii) Find $\mathrm { P } ( X \leqslant 8 )$.\\
(iv) Find $\operatorname { Var } ( X )$.\\
(v) Give a reason why a geometric distribution might not be an appropriate model for the first weekday in a calendar year on which the reviewer receives a delivery of books to review.
\hfill \mbox{\textit{OCR Further Statistics AS 2018 Q1 [8]}}