Edexcel S1 2002 November — Question 7 18 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2002
SessionNovember
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeDraw box plot from raw data
DifficultyModerate -0.8 This is a straightforward S1 statistics question testing standard procedures: reading a stem-and-leaf diagram, finding quartiles, identifying outliers, drawing a box plot, and calculating mean/standard deviation from given summations. All parts require routine application of formulas with no problem-solving insight. The most demanding aspect is the multi-step box plot construction, but this remains a textbook exercise well below average A-level difficulty.
Spec2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers2.02i Select/critique data presentation

The following stem and leaf diagram shows the aptitude scores \(x\) obtained by all the applicants for a particular job.
Aptitude score\(3|1\) means 31
31 2 9(3)
42 4 6 8 9(5)
51 3 3 5 6 7 9(7)
60 1 3 3 3 5 6 8 8 9(10)
71 2 2 2 4 5 5 5 6 8 8 8 8 9(14)
80 1 2 3 5 8 8 9(8)
90 1 2(3)
  1. Write down the modal aptitude score. [1]
  2. Find the three quartiles for these data. [3]
Outliers can be defined to be outside the limits \(Q_1 - 1.0(Q_3 - Q_1)\) and \(Q_3 + 1.0(Q_3 - Q_1)\).
  1. On a graph paper, draw a box plot to represent these data. [7]
For these data, \(\Sigma x = 3363\) and \(\Sigma x^2 = 238305\).
  1. Calculate, to 2 decimal places, the mean and the standard deviation for these data. [3]
  2. Use two different methods to show that these data are negatively skewed. [4]

(a)
AnswerMarks
Mode \(= 78\)B1
(1 mark)
(b)
AnswerMarks
\(Q_1 = 56\); \(Q_2 = 70\); \(Q_3 = 78\)B1; B1; B1
(3 marks)
(c)
AnswerMarks
\((Q_3 - Q_1) = 22\)M1 A1
\(Q_1 - 1.0(Q_3 - Q_1) = 34 \Rightarrow 31 \text{ & } 31 \text{ are outliers}\)A1
\(Q_3 + 1.0(Q_3 - Q_1) = 100 \Rightarrow \text{no outliers}\)
(3 marks)
(d)
AnswerMarks
(accurate sketch on graph paper required)M1
boxplot
scales and labelsB1
\(Q_1, Q_2, Q_3\)A1
31, 32, 34 (39), 92A1
(4 marks)
(e)
AnswerMarks
\(\mu = \frac{3363}{50} = 67.26\)B1
\(\sigma^2 = \frac{238305}{50} - (67.26)^2 = 242.1924\)M1
\(\sigma = \sqrt{242.1924} = 15.56253\ldots\)A1
(3 marks)
(f)
AnswerMarks
\((Q_3 - Q_2) < (Q_2 - Q_1)\), i.e. \(8 < 14 \Rightarrow\) negative skewM1, A1
Mean \(<\) Median \(<\) Mode, i.e. \(67.26 < 70 < 78 \Rightarrow\) negative skewM1, A1
(4 marks)
Total: 18 marks
## (a)
Mode $= 78$ | B1 |
| (1 mark) |

## (b)
$Q_1 = 56$; $Q_2 = 70$; $Q_3 = 78$ | B1; B1; B1 |
| (3 marks) |

## (c)
$(Q_3 - Q_1) = 22$ | M1 A1 |
$Q_1 - 1.0(Q_3 - Q_1) = 34 \Rightarrow 31 \text{ & } 31 \text{ are outliers}$ | A1 |
$Q_3 + 1.0(Q_3 - Q_1) = 100 \Rightarrow \text{no outliers}$ | |
| (3 marks) |

## (d)
(accurate sketch on graph paper required) | M1 |
boxplot | |
scales and labels | B1 |
$Q_1, Q_2, Q_3$ | A1 |
31, 32, 34 (39), 92 | A1 |
| (4 marks) |

## (e)
$\mu = \frac{3363}{50} = 67.26$ | B1 |
$\sigma^2 = \frac{238305}{50} - (67.26)^2 = 242.1924$ | M1 |
$\sigma = \sqrt{242.1924} = 15.56253\ldots$ | A1 |
| (3 marks) |

## (f)
$(Q_3 - Q_2) < (Q_2 - Q_1)$, i.e. $8 < 14 \Rightarrow$ negative skew | M1, A1 |
Mean $<$ Median $<$ Mode, i.e. $67.26 < 70 < 78 \Rightarrow$ negative skew | M1, A1 |
| (4 marks) |

**Total: 18 marks**
The following stem and leaf diagram shows the aptitude scores $x$ obtained by all the applicants for a particular job.

\begin{center}
\begin{tabular}{c|l|r}
Aptitude score & & $3|1$ means 31 \\
\hline
3 & 1 2 9 & (3) \\
4 & 2 4 6 8 9 & (5) \\
5 & 1 3 3 5 6 7 9 & (7) \\
6 & 0 1 3 3 3 5 6 8 8 9 & (10) \\
7 & 1 2 2 2 4 5 5 5 6 8 8 8 8 9 & (14) \\
8 & 0 1 2 3 5 8 8 9 & (8) \\
9 & 0 1 2 & (3) \\
\end{tabular}
\end{center}

\begin{enumerate}[label=(\alph*)]
\item Write down the modal aptitude score. [1]

\item Find the three quartiles for these data. [3]
\end{enumerate}

Outliers can be defined to be outside the limits $Q_1 - 1.0(Q_3 - Q_1)$ and $Q_3 + 1.0(Q_3 - Q_1)$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item On a graph paper, draw a box plot to represent these data. [7]
\end{enumerate}

For these data, $\Sigma x = 3363$ and $\Sigma x^2 = 238305$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Calculate, to 2 decimal places, the mean and the standard deviation for these data. [3]

\item Use two different methods to show that these data are negatively skewed. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2002 Q7 [18]}}