AQA Further AS Paper 2 Statistics Specimen — Question 3 4 marks

Exam BoardAQA
ModuleFurther AS Paper 2 Statistics (Further AS Paper 2 Statistics)
SessionSpecimen
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeMultiple unknowns from expectation and variance
DifficultyStandard +0.3 This is a standard Further Maths statistics problem requiring three simultaneous equations from probability axioms, expectation, and variance definitions. While it involves algebraic manipulation and substitution across multiple equations, the approach is methodical and follows directly from standard formulas without requiring novel insight or complex problem-solving strategies.
Spec5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance

3 The discrete random variable \(R\) has the following probability distribution.
\(\boldsymbol { r }\)- 20\(a\)4
\(\mathbf { P } ( \boldsymbol { R } = \boldsymbol { r } )\)0.3\(b\)\(c\)0.1
It is known that \(\mathrm { E } ( R ) = 0.2\) and \(\operatorname { Var } ( R ) = 3.56\) Find the values of \(a , b\) and \(c\).
[0pt] [4 marks]

Question 3:
AnswerMarks Guidance
\(0.4 + b + c = 1 \Rightarrow b + c = 0.6\)M1 Uses sum of probabilities \(= 1\)
\(E(R) = 0.2 \Rightarrow (-2 \times 0.3) + (0 \times b) + (a \times c) + (4 \times 0.1) = 0.2 \Rightarrow ac = 0.4\)M1 Uses formula for \(E(R)\)
\(E(X^2) - (E(X))^2 = (4 \times 0.3) + (0 \times b) + (a^2 \times c) + (16 \times 0.1) - (0.2)^2 = 3.56 \Rightarrow a^2c = 0.8\)M1 Uses formula for variance \(E(X^2) - (E(X))^2\)
From (2) and (3): \(a = 2\), hence \(c = 0.2\) and \(b = 0.4\)A1 Obtains \(a\), \(b\) and \(c\); CAO
## Question 3:

$0.4 + b + c = 1 \Rightarrow b + c = 0.6$ | M1 | Uses sum of probabilities $= 1$

$E(R) = 0.2 \Rightarrow (-2 \times 0.3) + (0 \times b) + (a \times c) + (4 \times 0.1) = 0.2 \Rightarrow ac = 0.4$ | M1 | Uses formula for $E(R)$

$E(X^2) - (E(X))^2 = (4 \times 0.3) + (0 \times b) + (a^2 \times c) + (16 \times 0.1) - (0.2)^2 = 3.56 \Rightarrow a^2c = 0.8$ | M1 | Uses formula for variance $E(X^2) - (E(X))^2$

From (2) and (3): $a = 2$, hence $c = 0.2$ and $b = 0.4$ | A1 | Obtains $a$, $b$ and $c$; CAO

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3 The discrete random variable $R$ has the following probability distribution.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
$\boldsymbol { r }$ & - 2 & 0 & $a$ & 4 \\
\hline
$\mathbf { P } ( \boldsymbol { R } = \boldsymbol { r } )$ & 0.3 & $b$ & $c$ & 0.1 \\
\hline
\end{tabular}
\end{center}

It is known that $\mathrm { E } ( R ) = 0.2$ and $\operatorname { Var } ( R ) = 3.56$\\
Find the values of $a , b$ and $c$.\\[0pt]
[4 marks]\\

\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics  Q3 [4]}}