OCR MEI Further Statistics Major 2024 June — Question 1 5 marks

Exam BoardOCR MEI
ModuleFurther Statistics Major (Further Statistics Major)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeCalculate E(X) from given distribution
DifficultyEasy -1.8 This is a straightforward textbook exercise requiring only direct application of standard formulas for E(X), Var(X), and linear transformations. No problem-solving insight needed—just arithmetic computation of ΣrP(r) and Σr²P(r), then applying E(aX+b) and SD(aX+b) rules. Significantly easier than average A-level questions.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance

1 The number of insurance policy sales made per month by a salesperson is modelled by the random variable \(X\), with probability distribution shown in the table.
\(r\)0123456
\(\mathrm { P } ( \mathrm { X } = \mathrm { r } )\)0.050.10.250.30.150.10.05
  1. Find each of the following.
    The salesperson is paid a basic salary of \(\pounds 1000\) per month plus \(\pounds 500\) for each policy that is sold.
  2. Find the mean and standard deviation of the salesperson's monthly salary.

Question 1:
AnswerMarks Guidance
1(a) E(X) = 2.9
Var(X) = 2.09B1
B1
AnswerMarks
[2]1.1
1.1
AnswerMarks Guidance
1(b) Mean = £2450
Variance of salary = 5002×2.09 (= 522500)
AnswerMarks
Standard deviation = £723 [£722.84]B1FT
M1
A1FT
AnswerMarks
[3]3.3
1.1
AnswerMarks
1.1For 5002×𝑡ℎ𝑒𝑖𝑟 variance Do not allow M1 for use of 2.9
unless stated to be Var(X). Do not allow M1 for simply
5002×Var(𝑋) without substituting in a value for Var(X)
Allow 722.8. Do not allow to greater accuracy than the
nearest penny. Do not allow answer given only in surds.
Question 1:
1 | (a) | E(X) = 2.9
Var(X) = 2.09 | B1
B1
[2] | 1.1
1.1
1 | (b) | Mean = £2450
Variance of salary = 5002×2.09 (= 522500)
Standard deviation = £723 [£722.84] | B1FT
M1
A1FT
[3] | 3.3
1.1
1.1 | For 5002×𝑡ℎ𝑒𝑖𝑟 variance Do not allow M1 for use of 2.9
unless stated to be Var(X). Do not allow M1 for simply
5002×Var(𝑋) without substituting in a value for Var(X)
Allow 722.8. Do not allow to greater accuracy than the
nearest penny. Do not allow answer given only in surds.
1 The number of insurance policy sales made per month by a salesperson is modelled by the random variable $X$, with probability distribution shown in the table.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$\mathrm { P } ( \mathrm { X } = \mathrm { r } )$ & 0.05 & 0.1 & 0.25 & 0.3 & 0.15 & 0.1 & 0.05 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find each of the following.

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

The salesperson is paid a basic salary of $\pounds 1000$ per month plus $\pounds 500$ for each policy that is sold.
\item Find the mean and standard deviation of the salesperson's monthly salary.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Major 2024 Q1 [5]}}