OCR MEI Further Statistics Major 2022 June — Question 3 6 marks

Exam BoardOCR MEI
ModuleFurther Statistics Major (Further Statistics Major)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeTwo unknowns from sum and expectation
DifficultyStandard +0.3 This question requires setting up two simultaneous equations (probabilities sum to 1, and expectation equals 1.8) and solving for a and b, followed by routine application of linear transformation formulas for expectation and variance. While it involves algebraic manipulation including a quadratic term (b²), the approach is standard and methodical with no novel insight required. It's slightly easier than average because the steps are predictable and the techniques are well-practiced in Further Maths Stats courses.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions

3 The table shows the probability distribution of the random variable \(X\), where \(a\) and \(b\) are constants.
\(r\)01234
\(\mathrm { P } ( X = r )\)\(a\)\(b\)0.240.32\(b ^ { 2 }\)
  1. Given that \(\mathrm { E } ( X ) = 1.8\), determine the values of \(a\) and \(b\). The random variable \(Y\) is given by \(Y = 10 - 3 X\).
  2. Using the values of \(a\) and \(b\) which you found in part (a), find each of the following.

Question 3:
AnswerMarks Guidance
3(a) b + 0.48 + 0.96 +4b2 = 1.8
b = 0.2
AnswerMarks
a + b + b2 + 0.24 + 0.32 = 1 ⇒ a = 0.2M1
A1
B1
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1BC
FT
2
AnswerMarks Guidance
3(b) E(Y) = 10 – 3 × 1.8 = 4.6
Var(X) = 1.44
AnswerMarks
Var(Y) = 9 × 1.44 = 12.96B1
B1
B1
AnswerMarks
[3]1.1
3.1a
AnswerMarks
1.10.44−𝑏𝑏−𝑏𝑏
BC Can get B1 implied by correct answer to Var(Y)
FT their Var (X)
Question 3:
3 | (a) | b + 0.48 + 0.96 +4b2 = 1.8
b = 0.2
a + b + b2 + 0.24 + 0.32 = 1 ⇒ a = 0.2 | M1
A1
B1
[3] | 3.1a
1.1
1.1 | BC
FT
2
3 | (b) | E(Y) = 10 – 3 × 1.8 = 4.6
Var(X) = 1.44
Var(Y) = 9 × 1.44 = 12.96 | B1
B1
B1
[3] | 1.1
3.1a
1.1 | 0.44−𝑏𝑏−𝑏𝑏
BC Can get B1 implied by correct answer to Var(Y)
FT their Var (X)
3 The table shows the probability distribution of the random variable $X$, where $a$ and $b$ are constants.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = r )$ & $a$ & $b$ & 0.24 & 0.32 & $b ^ { 2 }$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Given that $\mathrm { E } ( X ) = 1.8$, determine the values of $a$ and $b$.

The random variable $Y$ is given by $Y = 10 - 3 X$.
\item Using the values of $a$ and $b$ which you found in part (a), find each of the following.

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

\hfill \mbox{\textit{OCR MEI Further Statistics Major 2022 Q3 [6]}}