CAIE S1 Specimen — Question 3 6 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
SessionSpecimen
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicData representation
TypeCalculate frequency density from frequency
DifficultyEasy -1.8 This is a straightforward application of the formula frequency density = frequency ÷ class width. Students only need to identify class widths (accounting for 'correct to nearest minute') and apply basic arithmetic. This is routine recall with minimal problem-solving, making it easier than typical A-level questions.
Spec2.02b Histogram: area represents frequency

3 Robert has a part-time job delivering newspapers. On a number of days he noted the time, correct to the nearest minute, that it took him to do his job. Robert used his results to draw up the following table; two of the values in the table are denoted by \(a\) and \(b\).
Time \(( t\) minutes \()\)\(60 - 62\)\(63 - 64\)\(65 - 67\)\(68 - 71\)
Frequency (number of days)396\(b\)
Frequency density1\(a\)21.5
  1. Find the values of \(a\) and \(b\).
  2. Draw a histogram to represent Robert's times.
    \includegraphics[max width=\textwidth, alt={}]{34ae4f06-d485-4138-82d8-902b70f08995-04_206_100_1516_441}"\(\_\_\_\_\)□ □\includegraphics[max width=\textwidth, alt={}]{34ae4f06-d485-4138-82d8-902b70f08995-04_204_28_1518_1197}\(\_\_\_\_\)

Question 3(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(a = 9/\text{cw}\)M1 Using \(\text{fd} = f/\text{cw}\)
\(= 9/2 = 4.5\)A1 Correct \(a\)
\(1.5 = b/4\) so \(b = 6\)A1 Correct \(b\)
Total: 3
Question 3(ii):
AnswerMarks Guidance
AnswerMarks Guidance
Correct heights ft their \(b\)B1\(\sqrt{}\) Correct heights ft their \(b\)
Correct widths, i.e. 3, 2, 3, 4 starting either 60 or 59.5B1 Correct widths
Labels: fd, time or minutes and squiggle and bars from 59.5 to 71.5B1 Labels fd, time or minutes and squiggle and bars from 59.5 to 71.5
Total: 3
## Question 3(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 9/\text{cw}$ | M1 | Using $\text{fd} = f/\text{cw}$ |
| $= 9/2 = 4.5$ | A1 | Correct $a$ |
| $1.5 = b/4$ so $b = 6$ | A1 | Correct $b$ |
| **Total: 3** | | |

---

## Question 3(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct heights ft their $b$ | B1$\sqrt{}$ | Correct heights ft their $b$ |
| Correct widths, i.e. 3, 2, 3, 4 starting either 60 or 59.5 | B1 | Correct widths |
| Labels: fd, time or minutes and squiggle and bars from 59.5 to 71.5 | B1 | Labels fd, time or minutes and squiggle and bars from 59.5 to 71.5 |
| **Total: 3** | | |

---
3 Robert has a part-time job delivering newspapers. On a number of days he noted the time, correct to the nearest minute, that it took him to do his job. Robert used his results to draw up the following table; two of the values in the table are denoted by $a$ and $b$.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Time $( t$ minutes $)$ & $60 - 62$ & $63 - 64$ & $65 - 67$ & $68 - 71$ \\
\hline
Frequency (number of days) & 3 & 9 & 6 & $b$ \\
\hline
Frequency density & 1 & $a$ & 2 & 1.5 \\
\hline
\end{tabular}
\end{center}

(i) Find the values of $a$ and $b$.\\

(ii) Draw a histogram to represent Robert's times.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline
 & \includegraphics[max width=\textwidth, alt={}]{34ae4f06-d485-4138-82d8-902b70f08995-04_206_100_1516_441}
 &  & " &  & □ & $\_\_\_\_$ &  &  &  &  &  &  &  & □ &  &  &  &  & □ □ &  & \includegraphics[max width=\textwidth, alt={}]{34ae4f06-d485-4138-82d8-902b70f08995-04_204_28_1518_1197}
 &  &  & $\_\_\_\_$ &  &  &  &  &  &  &  & □ &  &  &  &  &  &  &  &  &  &  &  &  \\
\hline
 &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  \\
\hline
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\hline
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\hline
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\hline
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\hline
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\hline
\end{tabular}
\end{center}

\hfill \mbox{\textit{CAIE S1  Q3 [6]}}