OCR Further Statistics AS 2021 November — Question 5 6 marks

Exam BoardOCR
ModuleFurther Statistics AS (Further Statistics AS)
Year2021
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Distribution
TypeVariance of geometric distribution
DifficultyStandard +0.8 This question requires knowledge of the variance formula for geometric distribution (Var(X) = q/p²), solving for p from the variance, then computing P(X≥7) using the complementary probability formula. It combines multiple steps and requires fluency with geometric distribution properties beyond basic recall, making it moderately challenging but still within standard Further Maths scope.
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
5 The discrete random variable \(X\) has a geometric distribution. It is given that \(\operatorname { Var } ( X ) = 20\).
Determine \(\mathrm { P } ( X \geqslant 7 )\).
Question 5:
AnswerMarks Guidance
\(\dfrac{1-p}{p^2} = 20\)M1 Equate \(q/p^2\) to 20
\(20p^2 + p - 1 = 0\)M1 Obtain 3-term quadratic equation
\(p = 0.2\)A1 0.2 or exact equivalent only
\(p = -0.25\) not validB1 Explicitly reject other solution
\(q^6\)M1 Or \(q^7\). Can be implied by \(0.210\ (0.209715\ldots)\)
\(= 0.262\ (= 0.26214\ldots)\)A1 Or \(4096/15625\)
[6]
## Question 5:

$\dfrac{1-p}{p^2} = 20$ | **M1** | Equate $q/p^2$ to 20

$20p^2 + p - 1 = 0$ | **M1** | Obtain 3-term quadratic equation
$p = 0.2$ | **A1** | 0.2 or exact equivalent only
$p = -0.25$ not valid | **B1** | Explicitly reject other solution
$q^6$ | **M1** | Or $q^7$. Can be implied by $0.210\ (0.209715\ldots)$
$= 0.262\ (= 0.26214\ldots)$ | **A1** | Or $4096/15625$
[6]

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5 The discrete random variable $X$ has a geometric distribution. It is given that $\operatorname { Var } ( X ) = 20$.\\
Determine $\mathrm { P } ( X \geqslant 7 )$.

\hfill \mbox{\textit{OCR Further Statistics AS 2021 Q5 [6]}}
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