Edexcel S1 2023 June — Question 3 9 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeCalculate variance/SD from coded sums
DifficultyModerate -0.8 This is a straightforward S1 statistics question testing standard formulas for coding transformations and basic box plot interpretation. Parts (a) and (b) require direct application of mean/variance formulas with linear coding (textbook exercises), part (c) is simple quartile reading, and part (d) involves routine calculation of outlier boundaries using the given formula. No problem-solving or conceptual insight required—purely procedural recall.
Spec2.02a Interpret single variable data: tables and diagrams2.02g Calculate mean and standard deviation2.02h Recognize outliers5.02c Linear coding: effects on mean and variance
Jim records the length, \(l\) mm, of 81 salmon. The data are coded using \(x = l - 600\) and the following summary statistics are obtained. $$n = 81 \quad \sum x = 3711 \quad \sum x^2 = 475181$$
  1. Find the mean length of these salmon. [3]
  2. Find the variance of the lengths of these salmon. [2]
The weight, \(w\) grams, of each of the 81 salmon is recorded to the nearest gram. The recorded results for the 81 salmon are summarised in the box plot below. \includegraphics{figure_2}
  1. Find the maximum number of salmon that have weights in the interval $$4600 < w \leqslant 7700$$ [1]
Raj says that the box plot is incorrect as Jim has not included outliers. For these data an outlier is defined as a value that is more than \(1.5 \times\) IQR above the upper quartile \quad or \quad \(1.5 \times\) IQR below the lower quartile
  1. Show that there are no outliers. [3]
Question 3:
AnswerMarks
3(a)3711
x = =45.814... 
AnswerMarks
81l=3711+81600 =52311 
 M1
 l = "45.814..."+600
AnswerMarks
 "52311"
l =
 
AnswerMarks
81M1
 l = 645.81... awrt 646
AnswerMarks
 A1
(3)
AnswerMarks
(b)2
475181 3711
  2 = − =3767 
 x   
AnswerMarks
81  81 2
34088381 "52311"
 Var(L)= − 
AnswerMarks
81  81 M1
=3767.43...2 =3767.43...
AnswerMarks Guidance
l= 3767.43… awrt 3770 A1
(2)
AnswerMarks Guidance
(c)40 B1cao
(1)
AnswerMarks Guidance
(d)IQR = 5400 – 3800 [= 1600] M1
5400+1.5"1600" [= 7800] or3800−1.5"1600"[= 1400]M1
7800 > 7700 and 1400 < 1600 therefore there are no outliersA1
(3)
AnswerMarks
NotesTotal 9
(a)M1 for a correct method to find x or l Allow 45.8 or better. Ignore labels
3711
M1 for a correct method to find l ft their x if it is clearly labelled or it comes from or ft
81
their lif it is clearly labelled or comes from 3711+81600
17437 52311
A1 awrt 646 or or oe
27 81
AnswerMarks
(b)M1 correct method to find Var (X) implied by awrt 3770 or a correct method to find Var (L) ft their
l or Allow calculation of sd =awrt 61.4Ignore labels
x
A1 awrt 3770 labelled clearly as Var(L) or Var (L) = Var(X) or  = stated or variance is not
l x
2
34088381 "52311"
changed by coding is stated or they have gained the answer from − 
81  81 
AnswerMarks
(c)B1 cao
(d)M1 correct method to find IQR. May be implied by a correct limit.
NB 1.5(5400−3800)=2400
M1 for a correct method to find the upper or the lower outlier boundary.
A1 both 7800 and 1400 correct and 7700 and 1600 (as the minimum not IQR) seen and explicitly
stating no outliers
AnswerMarks Guidance
QuScheme Marks 
Question 3:
--- 3(a) ---
3(a) | 3711
x = =45.814... 
81 | l=3711+81600 =52311 
  | M1
 l = "45.814..."+600
  | "52311"
l =
 
81 | M1
 l = 645.81... awrt 646
  | A1
(3)
(b) | 2
475181 3711
  2 = − =3767 
 x   
81  81  | 2
34088381 "52311"
 Var(L)= − 
81  81  | M1
=3767.43...2 =3767.43...
l | = 3767.43… awrt 3770 | A1
(2)
(c) | 40 | B1cao
(1)
(d) | IQR = 5400 – 3800 [= 1600] | M1
5400+1.5"1600" [= 7800] or3800−1.5"1600"[= 1400] | M1
7800 > 7700 and 1400 < 1600 therefore there are no outliers | A1
(3)
Notes | Total 9
(a) | M1 for a correct method to find x or l Allow 45.8 or better. Ignore labels
3711
M1 for a correct method to find l ft their x if it is clearly labelled or it comes from or ft
81
their lif it is clearly labelled or comes from 3711+81600
17437 52311
A1 awrt 646 or or oe
27 81
(b) | M1 correct method to find Var (X) implied by awrt 3770 or a correct method to find Var (L) ft their
l or Allow calculation of sd =awrt 61.4Ignore labels
x
A1 awrt 3770 labelled clearly as Var(L) or Var (L) = Var(X) or  = stated or variance is not
l x
2
34088381 "52311"
changed by coding is stated or they have gained the answer from − 
81  81 
(c) | B1 cao
(d) | M1 correct method to find IQR. May be implied by a correct limit.
NB 1.5(5400−3800)=2400
M1 for a correct method to find the upper or the lower outlier boundary.
A1 both 7800 and 1400 correct and 7700 and 1600 (as the minimum not IQR) seen and explicitly
stating no outliers
Qu | Scheme | Marks
Jim records the length, $l$ mm, of 81 salmon. The data are coded using $x = l - 600$ and the following summary statistics are obtained.

$$n = 81 \quad \sum x = 3711 \quad \sum x^2 = 475181$$

\begin{enumerate}[label=(\alph*)]
\item Find the mean length of these salmon. [3]
\item Find the variance of the lengths of these salmon. [2]
\end{enumerate}

The weight, $w$ grams, of each of the 81 salmon is recorded to the nearest gram. The recorded results for the 81 salmon are summarised in the box plot below.

\includegraphics{figure_2}

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the maximum number of salmon that have weights in the interval
$$4600 < w \leqslant 7700$$ [1]
\end{enumerate}

Raj says that the box plot is incorrect as Jim has not included outliers.

For these data an outlier is defined as a value that is more than

$1.5 \times$ IQR above the upper quartile \quad or \quad $1.5 \times$ IQR below the lower quartile

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Show that there are no outliers. [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2023 Q3 [9]}}
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