Edexcel AS Paper 2 Specimen — Question 1 4 marks

Exam BoardEdexcel
ModuleAS Paper 2 (AS Paper 2)
SessionSpecimen
Marks4
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TopicMeasures of Location and Spread
TypeDescribing sampling methods
DifficultyEasy -1.8 This is a very straightforward question testing basic definitions and calculations. Part (a) requires simple recall of 'systematic sampling', part (b) needs awareness that the dataset has missing values, and part (c) is a routine standard deviation calculation using the formula. No problem-solving or conceptual depth required.
Spec2.01c Sampling techniques: simple random, opportunity, etc2.02g Calculate mean and standard deviation
  1. Sara is investigating the variation in daily maximum gust, \(t \mathrm { kn }\), for Camborne in June and July 1987.
She used the large data set to select a sample of size 20 from the June and July data for 1987. Sara selected the first value using a random number from 1 to 4 and then selected every third value after that.
  1. State the sampling technique Sara used.
  2. From your knowledge of the large data set explain why this process may not generate a sample of size 20 . The data Sara collected are summarised as follows $$n = 20 \quad \sum t = 374 \quad \sum t ^ { 2 } = 7600$$
  3. Calculate the standard deviation.
Question 1
1(a)
AnswerMarks Guidance
Systematic (sample) caoB1 1.2
1(b)
AnswerMarks Guidance
In LDS some days have gaps because the data was not recordedB1 2.4
Guidance:
B1: A correct explanation
1(c)
AnswerMarks Guidance
\(\frac{374}{t} \geq 18.7\)M1 1.1a
\(\frac{20 \times 7600}{t^2} \left[ = 30.31 \right]\)
\(t \geq 20\)
\(= 5.5054\ldots\) awrt 5.51
\(\frac{7600}{20t^2}\)
(Accept use of \(s = 5.6484\ldots\))
AnswerMarks Guidance
\(t \geq 19\)A1 1.1b
(4 marks)
Guidance:
M1: For a correct expression for \(\sigma_t\) and/or \(s_t\)
ft an incorrect evaluation of \(t\)
A1: For \(\sigma_t =\) awrt 5.51 or \(s_t =\) awrt 5.65
# Question 1

## 1(a)
Systematic (sample) cao | B1 | 1.2

## 1(b)
In LDS some days have gaps because the data was not recorded | B1 | 2.4

**Guidance:**
B1: A correct explanation

## 1(c)
$\frac{374}{t} \geq 18.7$ | M1 | 1.1a

$\frac{20 \times 7600}{t^2} \left[ = 30.31 \right]$

$t \geq 20$

$= 5.5054\ldots$ awrt 5.51

$\frac{7600}{20t^2}$

(Accept use of $s = 5.6484\ldots$)

$t \geq 19$ | A1 | 1.1b

(4 marks)

**Guidance:**

M1: For a correct expression for $\sigma_t$ and/or $s_t$

ft an incorrect evaluation of $t$

A1: For $\sigma_t =$ awrt 5.51 or $s_t =$ awrt 5.65
\begin{enumerate}
  \item Sara is investigating the variation in daily maximum gust, $t \mathrm { kn }$, for Camborne in June and July 1987.
\end{enumerate}

She used the large data set to select a sample of size 20 from the June and July data for 1987. Sara selected the first value using a random number from 1 to 4 and then selected every third value after that.\\
\begin{enumerate}[label=(\alph*)]
\item State the sampling technique Sara used.
\item From your knowledge of the large data set explain why this process may not generate a sample of size 20 .

The data Sara collected are summarised as follows

$$n = 20 \quad \sum t = 374 \quad \sum t ^ { 2 } = 7600$$
\item Calculate the standard deviation.
\end{enumerate}

\hfill \mbox{\textit{Edexcel AS Paper 2  Q1 [4]}}
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