Edexcel S1 2023 January — Question 1 10 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2023
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeComplete frequency table from histogram only
DifficultyModerate -0.3 This is a standard S1 histogram question requiring routine conversions between frequency density and frequency, plus basic statistical calculations (mean, standard deviation, interpolation). While multi-part, each step follows textbook procedures with no novel problem-solving required. The given values (Σft², Q₃) reduce computational burden, making it slightly easier than average.
Spec2.02a Interpret single variable data: tables and diagrams2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation
  1. The histogram shows the times taken, \(t\) minutes, by each of 100 people to swim 500 metres. \includegraphics[max width=\textwidth, alt={}, center]{c316fa29-dedc-4890-bd82-31eb0bb819f9-02_986_1070_342_424}
    1. Use the histogram to complete the frequency table for the times taken by the 100 people to swim 500 metres.
    Time taken ( \(\boldsymbol { t }\) minutes)\(5 - 10\)\(10 - 14\)\(14 - 18\)\(18 - 25\)\(25 - 40\)
    Frequency ( \(\boldsymbol { f }\) )101624
  2. Estimate the number of people who took less than 16 minutes to swim 500 metres.
  3. Find an estimate for the mean time taken to swim 500 metres. Given that \(\sum f t ^ { 2 } = 41033\)
  4. find an estimate for the standard deviation of the times taken to swim 500 metres. Given that \(Q _ { 3 } = 23\)
  5. use linear interpolation to estimate the interquartile range of the times taken to swim 500 metres.
Question 1:
Part (a)
AnswerMarks Guidance
AnswerMark Guidance
Frequencies: 10, 16, 24, 35, 15B1 For 35 and 15; if answers in both table and answer lines, mark answer lines
Part (b)
AnswerMarks Guidance
AnswerMark Guidance
\(10 + 16 + (2 \times 6)\) or \(10 + 16 + \frac{24}{2}\) or \(\frac{x-26}{50-26} = \frac{16-14}{18-14}\)M1 For method shown
\(= 38\)A1 cao
Part (c)
AnswerMarks Guidance
AnswerMark Guidance
\(\sum ft = 7.5\times10 + 12\times16 + 16\times24 + 21.5\times35 + 32.5\times15 = 1891\)M1 A correct method for finding \(\sum ft\); may be implied by 1891; allow one error
\(\text{Mean} = \frac{1891}{100} = 18.91\)A1 Allow 18.9
Part (d)
AnswerMarks Guidance
AnswerMark Guidance
\(\text{SD} = \sqrt{\frac{41033}{100} - \left(\frac{1891}{100}\right)^2}\) or \(\sqrt{\frac{41033 - 100\times 18.91^2}{99}}\)M1 For a correct calculation of SD ft their mean
\(= 7.262...\) or \(7.298...\)A1 awrt 7.26 or awrt 7.3 if using \(n-1\); accept 7.3[0]
Part (e)
AnswerMarks Guidance
AnswerMark Guidance
\(LQ = 10 + \frac{15}{16}(14-10) = 13.75\) or \(14 - \frac{1}{16}(14-10) = 13.75\) or \(\frac{Q_1-10}{14-10} = \frac{25-10}{26-10}\) or \(\frac{Q_1-14}{14-10} = \frac{25-26}{26-10}\)M1 For method shown (or interpolation with 15.25)
\(IQR = 23 - 13.75\)M1 \(UQ - LQ\) ft their LQ provided \(LQ < UQ\)
\(= 9.25\) (or awrt 9.19)A1 For 9.25 or awrt 9.19 if \(n+1\) used 
# Question 1:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Frequencies: 10, 16, 24, 35, 15 | B1 | For 35 and 15; if answers in both table and answer lines, mark answer lines |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $10 + 16 + (2 \times 6)$ or $10 + 16 + \frac{24}{2}$ or $\frac{x-26}{50-26} = \frac{16-14}{18-14}$ | M1 | For method shown |
| $= 38$ | A1 | cao |

## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum ft = 7.5\times10 + 12\times16 + 16\times24 + 21.5\times35 + 32.5\times15 = 1891$ | M1 | A correct method for finding $\sum ft$; may be implied by 1891; allow one error |
| $\text{Mean} = \frac{1891}{100} = 18.91$ | A1 | Allow 18.9 |

## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{SD} = \sqrt{\frac{41033}{100} - \left(\frac{1891}{100}\right)^2}$ or $\sqrt{\frac{41033 - 100\times 18.91^2}{99}}$ | M1 | For a correct calculation of SD ft their mean |
| $= 7.262...$ or $7.298...$ | A1 | awrt 7.26 or awrt 7.3 if using $n-1$; accept 7.3[0] |

## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $LQ = 10 + \frac{15}{16}(14-10) = 13.75$ or $14 - \frac{1}{16}(14-10) = 13.75$ or $\frac{Q_1-10}{14-10} = \frac{25-10}{26-10}$ or $\frac{Q_1-14}{14-10} = \frac{25-26}{26-10}$ | M1 | For method shown (or interpolation with 15.25) |
| $IQR = 23 - 13.75$ | M1 | $UQ - LQ$ ft their LQ provided $LQ < UQ$ |
| $= 9.25$ (or awrt 9.19) | A1 | For 9.25 or awrt 9.19 if $n+1$ used |

---
\begin{enumerate}
  \item The histogram shows the times taken, $t$ minutes, by each of 100 people to swim 500 metres.\\
\includegraphics[max width=\textwidth, alt={}, center]{c316fa29-dedc-4890-bd82-31eb0bb819f9-02_986_1070_342_424}\\
(a) Use the histogram to complete the frequency table for the times taken by the 100 people to swim 500 metres.
\end{enumerate}

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Time taken ( $\boldsymbol { t }$ minutes) & $5 - 10$ & $10 - 14$ & $14 - 18$ & $18 - 25$ & $25 - 40$ \\
\hline
Frequency ( $\boldsymbol { f }$ ) & 10 & 16 & 24 &  &  \\
\hline
\end{tabular}
\end{center}

(b) Estimate the number of people who took less than 16 minutes to swim 500 metres.\\
(c) Find an estimate for the mean time taken to swim 500 metres.

Given that $\sum f t ^ { 2 } = 41033$\\
(d) find an estimate for the standard deviation of the times taken to swim 500 metres.

Given that $Q _ { 3 } = 23$\\
(e) use linear interpolation to estimate the interquartile range of the times taken to swim 500 metres.

\hfill \mbox{\textit{Edexcel S1 2023 Q1 [10]}}
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Q1 10 Q2 10 Q3 11 Q4 13 Q5 17 Q6 14