Edexcel S1 — Question 6 15 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeUse linear interpolation for median or quartiles
DifficultyModerate -0.3 This is a standard S1 cumulative frequency question requiring routine techniques: naming a graph type, linear interpolation for median/quartiles/percentiles, drawing a box plot, and basic comparison. While multi-part with 15 marks total, each component uses straightforward textbook methods with no novel problem-solving or conceptual challenges beyond standard curriculum expectations.
Spec2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread

1000 houses were sold in a small town in a one-year period. The selling prices were as given in the following table:
Selling PriceNumber of HousesSelling PriceNumber of Houses
Up to £50 00060Up to £150 000642
Up to £75 000227Up to £200 000805
Up to £100 000305Up to £500 000849
Up to £125 000414Up to £750 0001000
  1. Name (do not draw) a suitable type of graph for illustrating this data. [1 mark]
  2. Use interpolation to find estimates of the median and the quartiles. [6 marks]
  3. Estimate the 37th percentile. [2 marks]
Given further that the lowest price was £42 000 and the range of the prices was £690 000,
  1. draw a box plot to represent the data. Show your scale clearly. [4 marks]
In another town the median price was £149 000, and the interquartile range was £90 000.
  1. Briefly compare the prices in the two towns using this information. [2 marks]

AnswerMarks Guidance
(a) Cumulative frequency graph drawnB1
(b) \(Q_1 = 75\,000 + \frac{86}{228} \times 25\,000 = £82\,372\)M1 A1
\(Q_2 = 125\,000 + \frac{23}{78} \times 25\,000 = £134\,430\)M1 A1
\(Q_3 = 150\,000 + \frac{108}{163} \times 50\,000 = £183\,129\)M1 A1
(c) \(100\,000 + \frac{65}{109} \times 25\,000 = £114\,908\)M1 A1
(d) Box plot drawn; upper limit \(£732\,000\)B4
(e) First town IQR = \(£100\,757\). Second town has higher median and smaller IQR, so is more expensive overall and more consistentB1 B1 15 marks total
(a) Cumulative frequency graph drawn | B1 |

(b) $Q_1 = 75\,000 + \frac{86}{228} \times 25\,000 = £82\,372$ | M1 A1 |

$Q_2 = 125\,000 + \frac{23}{78} \times 25\,000 = £134\,430$ | M1 A1 |

$Q_3 = 150\,000 + \frac{108}{163} \times 50\,000 = £183\,129$ | M1 A1 |

(c) $100\,000 + \frac{65}{109} \times 25\,000 = £114\,908$ | M1 A1 |

(d) Box plot drawn; upper limit $£732\,000$ | B4 |

(e) First town IQR = $£100\,757$. Second town has higher median and smaller IQR, so is more expensive overall and more consistent | B1 B1 | 15 marks total
1000 houses were sold in a small town in a one-year period. The selling prices were as given in the following table:

\begin{tabular}{|c|c||c|c|}
\hline
Selling Price & Number of Houses & Selling Price & Number of Houses \\
\hline
Up to £50 000 & 60 & Up to £150 000 & 642 \\
Up to £75 000 & 227 & Up to £200 000 & 805 \\
Up to £100 000 & 305 & Up to £500 000 & 849 \\
Up to £125 000 & 414 & Up to £750 000 & 1000 \\
\hline
\end{tabular}

\begin{enumerate}[label=(\alph*)]
\item Name (do not draw) a suitable type of graph for illustrating this data. [1 mark]
\item Use interpolation to find estimates of the median and the quartiles. [6 marks]
\item Estimate the 37th percentile. [2 marks]
\end{enumerate}

Given further that the lowest price was £42 000 and the range of the prices was £690 000,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item draw a box plot to represent the data. Show your scale clearly. [4 marks]
\end{enumerate}

In another town the median price was £149 000, and the interquartile range was £90 000.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Briefly compare the prices in the two towns using this information. [2 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1  Q6 [15]}}