OCR Further Statistics 2023 June — Question 3 6 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2023
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Random Variables
TypeLinear combinations of independent variables
DifficultyStandard +0.3 Part (a) requires standard application of expectation and variance formulas for linear combinations of independent discrete uniform variables, involving straightforward algebra to solve simultaneous equations. Part (b) tests conceptual understanding of CLT conditions but requires only a brief explanation. This is slightly easier than average as it's mostly routine application of well-practiced formulas with minimal problem-solving insight needed.
Spec5.02e Discrete uniform distribution5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05a Sample mean distribution: central limit theorem

3 The discrete random variable \(W\) has the distribution \(\mathrm { U } ( 11 )\). The independent discrete random variable \(V\) has the distribution \(\mathrm { U } ( 5 )\).
  1. It is given that, for constants \(m\) and \(n\), with \(m > 0\), \(\mathrm { E } ( \mathrm { mW } + \mathrm { nV } ) = 0\) and \(\operatorname { Var } ( \mathrm { mW } + \mathrm { nV } ) = 1\). Determine the exact values of \(m\) and \(n\). The random variable \(T\) is the mean of three independent observations of \(W\).
  2. Explain whether the Central Limit Theorem can be used to say that the distribution of \(T\) is approximately normal.

Question 3:
AnswerMarks Guidance
3(a) E(W) = 6, Var(W) = 10, E(V) = 3, Var(V) = 2
6m + 3n = 0
10m2 + 2n2 = 1
10m2 + 2.(4m2) = 1 ⇒ m = 2 (or 1 )
6 1 8
n = 2 (or 2 )
− −
AnswerMarks
3 1 8B1
B1ft
B1ft
M1
A1
AnswerMarks
[5]1.1
2.1
2.1
1.1
AnswerMarks
2.2aAll four, can be implied by subsequent working
Ft on their means
Ft on their variances
Solve simultaneously to get one letter, allow n > 0 if found first
Both correct (n = –2m), any exact form (not ISW on both signs as
question requires m > 0)
AnswerMarks
(b)No, as the number of independent random variables
being averaged is not big enough for the Central
AnswerMarks Guidance
Limit Theorem to apply.B1
[1]2.4 State “no”, with correct reason, e.g. “3 is not big enough” or “sample is
not big enough” but not “n is not big enough” unless explicit that n = 3.
Allow “no as 3 < 25”, but if a number is quoted it must be 25 or 30.
No wrong extras.
Question 3:
3 | (a) | E(W) = 6, Var(W) = 10, E(V) = 3, Var(V) = 2
6m + 3n = 0
10m2 + 2n2 = 1
10m2 + 2.(4m2) = 1 ⇒ m = 2 (or 1 )
6 1 8
n = 2 (or 2 )
− −
3 1 8 | B1
B1ft
B1ft
M1
A1
[5] | 1.1
2.1
2.1
1.1
2.2a | All four, can be implied by subsequent working
Ft on their means
Ft on their variances
Solve simultaneously to get one letter, allow n > 0 if found first
Both correct (n = –2m), any exact form (not ISW on both signs as
question requires m > 0)
(b) | No, as the number of independent random variables
being averaged is not big enough for the Central
Limit Theorem to apply. | B1
[1] | 2.4 | State “no”, with correct reason, e.g. “3 is not big enough” or “sample is
not big enough” but not “n is not big enough” unless explicit that n = 3.
Allow “no as 3 < 25”, but if a number is quoted it must be 25 or 30.
No wrong extras.
3 The discrete random variable $W$ has the distribution $\mathrm { U } ( 11 )$. The independent discrete random variable $V$ has the distribution $\mathrm { U } ( 5 )$.
\begin{enumerate}[label=(\alph*)]
\item It is given that, for constants $m$ and $n$, with $m > 0$, $\mathrm { E } ( \mathrm { mW } + \mathrm { nV } ) = 0$ and $\operatorname { Var } ( \mathrm { mW } + \mathrm { nV } ) = 1$.

Determine the exact values of $m$ and $n$.

The random variable $T$ is the mean of three independent observations of $W$.
\item Explain whether the Central Limit Theorem can be used to say that the distribution of $T$ is approximately normal.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Statistics 2023 Q3 [6]}}