OCR Further Statistics AS 2018 June — Question 6 5 marks

Exam BoardOCR
ModuleFurther Statistics AS (Further Statistics AS)
Year2018
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Distribution
TypeFinding binomial parameters from properties
DifficultyStandard +0.8 This requires students to work backwards from mean and standard deviation to find n and p using the formulas E(T)=np and SD(T)=√(np(1-p)). While the formulas are standard, solving the simultaneous equations requires algebraic manipulation and careful arithmetic with non-integer values, making it moderately challenging but still within typical A-level scope.
Spec5.02d Binomial: mean np and variance np(1-p)
6 In this question you must show detailed reasoning. The random variable \(T\) has a binomial distribution. It is known that \(\mathrm { E } ( T ) = 5.625\) and the standard deviation of \(T\) is 1.875 . Find the values of the parameters of the distribution.
Question 6:
\(np = 5.625\)
\(np(1-p) = 1.875^2 \quad [= 3.515625]\)
\(\Rightarrow (1-p) = \dfrac{5}{8}\) or \(0.625\)
\(\Rightarrow p = \dfrac{3}{8}\) or \(0.375\)
AnswerMarks Guidance
\(n = \mathbf{15}\)B1 B1 M1 A1 A1 [5] B1: Allow \(np(1-p) = 1.875\). Allow \(q\) for \(1-p\). M1: Eliminate one letter. A1: Or exact equivalent. A1: Exact answer only 
## Question 6:
$np = 5.625$

$np(1-p) = 1.875^2 \quad [= 3.515625]$

$\Rightarrow (1-p) = \dfrac{5}{8}$ or $0.625$

$\Rightarrow p = \dfrac{3}{8}$ or $0.375$

$n = \mathbf{15}$ | **B1** B1 M1 A1 A1 [5] | B1: Allow $np(1-p) = 1.875$. Allow $q$ for $1-p$. M1: Eliminate one letter. A1: Or exact equivalent. A1: Exact answer only

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6 In this question you must show detailed reasoning.
The random variable $T$ has a binomial distribution. It is known that $\mathrm { E } ( T ) = 5.625$ and the standard deviation of $T$ is 1.875 . Find the values of the parameters of the distribution.

\hfill \mbox{\textit{OCR Further Statistics AS 2018 Q6 [5]}}
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