OCR FS1 AS 2018 March — Question 1 7 marks

Exam BoardOCR
ModuleFS1 AS (Further Statistics 1 AS)
Year2018
SessionMarch
Marks7
TopicGeometric Distribution
TypeState assumptions for geometric model
DifficultyEasy -1.2 This is a straightforward recall question testing basic knowledge of the geometric distribution. Part (i) requires stating standard assumptions (constant probability, independence), while parts (ii)-(iv) involve direct application of standard formulas with no problem-solving or conceptual challenge. This is easier than average A-level content.
Spec5.02f 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

1 A learner driver keeps taking the driving test until she passes. The number of attempts taken, up to and including the pass, is denoted by \(X\).
  1. State two assumptions needed for \(X\) to be well modelled by a geometric distribution. Assume now that \(X \sim \operatorname { Geo } ( 0.4 )\).
  2. Find \(\mathrm { P } ( X < 6 )\).
  3. Find \(\mathrm { E } ( X )\).
  4. Find \(\operatorname { Var } ( X )\).

Part (i)
AnswerMarks Guidance
Outcome of each attempt is independent of the outcome of any other attemptB1 Must be contextualised for any marks - NOT "probability of passing at all is independent"
Probability of passing on any attempt is the sameB1
Part (ii)
AnswerMarks Guidance
\(1 - 0.6^7\) or \(1 - 0.6^6\)M1 Allow \(1 - 0.6^7\) or \(1 - 0.6^6\)
\(= 0.92224\)A1 Answer, awrt 0.922; Or \(p\sum_{r=0}^{6} q^r\) allow \(\pm 1\) term for M1
Part (iii)
AnswerMarks Guidance
\(E(X) = 1/p = 2\frac{1}{2}\)B1 Any exact equivalent
Part (iv)
AnswerMarks Guidance
\(\text{Var}(X) = q/p^2 = 0.6/0.4^2\)M1 Accept 0.6÷0.16
\(3\frac{3}{4}\)A1 Any exact equivalent
### Part (i)
Outcome of each attempt is independent of the outcome of any other attempt | **B1** | Must be contextualised for any marks - NOT "probability of passing at all is independent"

Probability of passing on any attempt is the same | **B1** |

### Part (ii)
$1 - 0.6^7$ or $1 - 0.6^6$ | **M1** | Allow $1 - 0.6^7$ or $1 - 0.6^6$

$= 0.92224$ | **A1** | Answer, awrt 0.922; Or $p\sum_{r=0}^{6} q^r$ allow $\pm 1$ term for M1

### Part (iii)
$E(X) = 1/p = 2\frac{1}{2}$ | **B1** | Any exact equivalent

### Part (iv)
$\text{Var}(X) = q/p^2 = 0.6/0.4^2$ | **M1** | Accept 0.6÷0.16

$3\frac{3}{4}$ | **A1** | Any exact equivalent

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1 A learner driver keeps taking the driving test until she passes. The number of attempts taken, up to and including the pass, is denoted by $X$.\\
(i) State two assumptions needed for $X$ to be well modelled by a geometric distribution.

Assume now that $X \sim \operatorname { Geo } ( 0.4 )$.\\
(ii) Find $\mathrm { P } ( X < 6 )$.\\
(iii) Find $\mathrm { E } ( X )$.\\
(iv) Find $\operatorname { Var } ( X )$.

\hfill \mbox{\textit{OCR FS1 AS 2018 Q1 [7]}}