OCR MEI Paper 2 Specimen — Question 6 4 marks

Exam BoardOCR MEI
ModulePaper 2 (Paper 2)
SessionSpecimen
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeLinear transformation of normal
DifficultyModerate -0.8 This question tests basic understanding of Normal distribution parameters from a graph and straightforward application of linear transformation rules for mean and standard deviation. Part (a) requires reading values from a curve (trivial), while part (b) applies the standard formulas μ_new = 1.8μ + 32 and σ_new = 1.8σ, which are routine A-level statistics knowledge requiring no problem-solving or insight.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04g Normal distribution properties: empirical rule (68-95-99.7), points of inflection5.02c Linear coding: effects on mean and variance

Each day, for many years, the maximum temperature in degrees Celsius at a particular location is recorded. The maximum temperatures for days in October can be modelled by a Normal distribution. The appropriate Normal curve is shown in Fig. 6. \includegraphics{figure_6}
    1. Use the model to write down the mean of the maximum temperatures. [1]
    2. Explain why the curve indicates that the standard deviation is approximately 3 degrees Celsius. [1]
Temperatures can be converted from Celsius to Fahrenheit using the formula \(F = 1.8C + 32\), where \(F\) is the temperature in degrees Fahrenheit and \(C\) is the temperature in degrees Celsius.
  1. For maximum temperature in October in degrees Fahrenheit, estimate
    [2]

Question 6:
AnswerMarks Guidance
6(a) (i)
(ii)S
Mean = 17
Either
Points of inflection are approx. 3 above and below
AnswerMarks Guidance
mean so SD = approx. 3B1
E13.4
2.4AG
OrE1 2.4
Limits are approx. 9 above and below mean so
SD = 9 ÷ 3 = 3
[2]
AG
AnswerMarks Guidance
6(b) Mean in Fahrenheit = 1.8 × 17 + 32 = 62.6
SD in Fahrenheit = 1.8 × 3 = 5.4B1
B1
AnswerMarks
[2]1.1
1.1FT their mean
Question 6:
6 | (a) | (i)
(ii) | S
Mean = 17
Either
Points of inflection are approx. 3 above and below
mean so SD = approx. 3 | B1
E1 | 3.4
2.4 | AG
Or | E1 | 2.4 | AG
Limits are approx. 9 above and below mean so
SD = 9 ÷ 3 = 3
[2]
AG
6 | (b) | Mean in Fahrenheit = 1.8 × 17 + 32 = 62.6
SD in Fahrenheit = 1.8 × 3 = 5.4 | B1
B1
[2] | 1.1
1.1 | FT their mean
Each day, for many years, the maximum temperature in degrees Celsius at a particular location is recorded. The maximum temperatures for days in October can be modelled by a Normal distribution. The appropriate Normal curve is shown in Fig. 6.

\includegraphics{figure_6}

\begin{enumerate}[label=(\alph*)]
\item 
\begin{enumerate}[label=(\roman*)]
\item Use the model to write down the mean of the maximum temperatures. [1]
\item Explain why the curve indicates that the standard deviation is approximately 3 degrees Celsius. [1]
\end{enumerate}
\end{enumerate}

Temperatures can be converted from Celsius to Fahrenheit using the formula $F = 1.8C + 32$, where $F$ is the temperature in degrees Fahrenheit and $C$ is the temperature in degrees Celsius.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item For maximum temperature in October in degrees Fahrenheit, estimate
\begin{itemize}
\item the mean
\item the standard deviation.
\end{itemize} [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Paper 2  Q6 [4]}}