OCR MEI Further Statistics A AS 2024 June — Question 1 7 marks

Exam BoardOCR MEI
ModuleFurther Statistics A AS (Further Statistics A AS)
Year2024
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeOne unknown from sum constraint only
DifficultyEasy -1.2 This is a straightforward textbook exercise testing basic probability distribution properties: summing probabilities to 1, calculating E(X) and Var(X) using standard formulas, and applying linear transformation rules. All steps are routine recall with no problem-solving or insight required, making it easier than average.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance
1 The probability distribution for a discrete random variable \(X\) is given in the table below.
\(x\)0123
\(\mathrm { P } ( \mathrm { X } = \mathrm { x } )\)\(2 c\)\(3 c\)\(0.5 - c\)\(c\)
  1. Find the value of \(c\).
  2. Find the value of each of the following.
    The random variable \(Y\) is defined by \(Y = 2 X - 3\).
  3. Find the value of each of the following.
Question 1:
AnswerMarks Guidance
1(a) 2c + 3c + 0.5 – c + c = 1
=> 0.5 + 5c = 1
AnswerMarks
=> c = 0.5/5 = 0.1M1
A1
AnswerMarks Guidance
[2]1.1
1.1Using p = 1.
1(b) E(X) = 1.4 cao
Var(X) = (020.2 +) 120.3 + 220.4 + 320.1
– “1.4”2
or (022c +) 123c + 22(0.5 – c) + 32c – “1.4”2
AnswerMarks
Var(X) = 2.8 – 1.42 = 0.84B1
M1
A1FT
AnswerMarks
[3]1.1
1.1
AnswerMarks Guidance
1.1FT their c provided Var(X) > 0 Could see
02c + 13c + 2(0.5 – c) + 3c
(= 1 + 4c)
or 00.2 + 10.3 + 20.4 + 30.1
Correct form for finding variance,
either numerical or in terms of c.
NB Var(X) = 2 + 8c – (1 + 4c)2
If B0 because E(X) = 1 + 4c then A1
can be awarded for
Var(X) = (1+ 4c)(1 – 4c) or 1 – 16c2
AnswerMarks Guidance
1(c) E(Y) = –0.2
Var(Y) = 3.36B1FT
B1FT
AnswerMarks
[2]1.1
1.1FT 2“1.4” – 3
FT 22“0.84” provided Var(X) > 0SC1 for both E(Y) = 8c – 1
and Var(Y) = 4 – 64c2 oe
Question 1:
1 | (a) | 2c + 3c + 0.5 – c + c = 1
=> 0.5 + 5c = 1
=> c = 0.5/5 = 0.1 | M1
A1
[2] | 1.1
1.1 | Using p = 1.
1 | (b) | E(X) = 1.4 cao
Var(X) = (020.2 +) 120.3 + 220.4 + 320.1
– “1.4”2
or (022c +) 123c + 22(0.5 – c) + 32c – “1.4”2
Var(X) = 2.8 – 1.42 = 0.84 | B1
M1
A1FT
[3] | 1.1
1.1
1.1 | FT their c provided Var(X) > 0 | Could see
02c + 13c + 2(0.5 – c) + 3c
(= 1 + 4c)
or 00.2 + 10.3 + 20.4 + 30.1
Correct form for finding variance,
either numerical or in terms of c.
NB Var(X) = 2 + 8c – (1 + 4c)2
If B0 because E(X) = 1 + 4c then A1
can be awarded for
Var(X) = (1+ 4c)(1 – 4c) or 1 – 16c2
1 | (c) | E(Y) = –0.2
Var(Y) = 3.36 | B1FT
B1FT
[2] | 1.1
1.1 | FT 2“1.4” – 3
FT 22“0.84” provided Var(X) > 0 | SC1 for both E(Y) = 8c – 1
and Var(Y) = 4 – 64c2 oe
1 The probability distribution for a discrete random variable $X$ is given in the table below.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( \mathrm { X } = \mathrm { x } )$ & $2 c$ & $3 c$ & $0.5 - c$ & $c$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the value of $c$.
\item Find the value of each of the following.

\begin{itemize}
  \item $\mathrm { E } ( X )$
  \item $\operatorname { Var } ( X )$
\end{itemize}

The random variable $Y$ is defined by $Y = 2 X - 3$.
\item Find the value of each of the following.

\begin{itemize}
  \item E(Y)
  \item $\operatorname { Var } ( Y )$
\end{itemize}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2024 Q1 [7]}}
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