CAIE S1 2003 June — Question 7 9 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2003
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeHistogram from continuous grouped data
DifficultyEasy -1.2 This is a straightforward S1 question requiring standard procedures: calculating mean and standard deviation from grouped data using midpoints (routine formula application), and drawing a histogram with unequal class widths (requiring frequency density calculation). Both parts are textbook exercises with no problem-solving or conceptual challenges beyond basic statistical literacy.
Spec2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation

7 A random sample of 97 people who own mobile phones was used to collect data on the amount of time they spent per day on their phones. The results are displayed in the table below.
Time spent per
day \(( t\) minutes \()\)
\(0 \leqslant t < 5\)\(5 \leqslant t < 10\)\(10 \leqslant t < 20\)\(20 \leqslant t < 30\)\(30 \leqslant t < 40\)\(40 \leqslant t < 70\)
Number
of people
11203218106
  1. Calculate estimates of the mean and standard deviation of the time spent per day on these mobile phones.
  2. On graph paper, draw a fully labelled histogram to represent the data.

Question 7:
(i)
AnswerMarks Guidance
AnswerMark Guidance
(Using mid-intervals)M1 For using their mid-intervals (not end points or class widths)
\(\text{Mean} = \frac{2.5\times11 + 7.5\times20 + 15\times32 + 25\times18 + 35\times10 + 55\times6}{97} = 18.4\)M1 For using \(\dfrac{\sum fx}{\sum f}\), any \(x\)
\(= 18.4\)A1 For correct answer, cwo; 18.4 no working gets 3/3
\(\text{sd} = \sqrt{\dfrac{2.5^2\times11 + 7.5^2\times20 + 15^2\times32 + 25^2\times18 + 35^2\times10 + 55^2\times6}{97} - \text{mean}^2} = 13.3\)M1 For using \(\dfrac{\sum fx^2}{\sum f} - (\text{their mean})^2\) or equivalent; no \(\sqrt{\,}\) needed, not \((\sum fx)^2/\sum f\)
\(= 13.3\)A1 (5) For correct answer
(ii)
AnswerMarks Guidance
AnswerMark Guidance
Frequency densities: 2.2, 4.0, 3.2, 1.8, 1.0, 0.2M1 For attempting a frequency density of some sort (or scaled frequency); can be upside down but not multiplied
Correct heights on graphA1 For correct heights on the graph
Correct bars on uniform horizontal scale (0 to 5 etc.)B1 For correct bars on uniform horiz. scale, i.e. from 0 to 5 etc.
Frequency density or scaled freq. labelled on vertical axis; time or mins on horizontalB1 (4) Freq. density or scaled freq. labelled on vertical axis, time or mins on horiz.; 'class width' is not enough
## Question 7:

**(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| (Using mid-intervals) | M1 | For using their mid-intervals (not end points or class widths) |
| $\text{Mean} = \frac{2.5\times11 + 7.5\times20 + 15\times32 + 25\times18 + 35\times10 + 55\times6}{97} = 18.4$ | M1 | For using $\dfrac{\sum fx}{\sum f}$, any $x$ |
| $= 18.4$ | A1 | For correct answer, cwo; 18.4 no working gets 3/3 |
| $\text{sd} = \sqrt{\dfrac{2.5^2\times11 + 7.5^2\times20 + 15^2\times32 + 25^2\times18 + 35^2\times10 + 55^2\times6}{97} - \text{mean}^2} = 13.3$ | M1 | For using $\dfrac{\sum fx^2}{\sum f} - (\text{their mean})^2$ or equivalent; no $\sqrt{\,}$ needed, not $(\sum fx)^2/\sum f$ |
| $= 13.3$ | A1 (5) | For correct answer |

**(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Frequency densities: 2.2, 4.0, 3.2, 1.8, 1.0, 0.2 | M1 | For attempting a frequency density of some sort (or scaled frequency); can be upside down but not multiplied |
| Correct heights on graph | A1 | For correct heights on the graph |
| Correct bars on uniform horizontal scale (0 to 5 etc.) | B1 | For correct bars on uniform horiz. scale, i.e. from 0 to 5 etc. |
| Frequency density or scaled freq. labelled on vertical axis; time or mins on horizontal | B1 (4) | Freq. density or scaled freq. labelled on vertical axis, time or mins on horiz.; 'class width' is not enough |
7 A random sample of 97 people who own mobile phones was used to collect data on the amount of time they spent per day on their phones. The results are displayed in the table below.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Time spent per \\
day $( t$ minutes $)$ \\
\end{tabular} & $0 \leqslant t < 5$ & $5 \leqslant t < 10$ & $10 \leqslant t < 20$ & $20 \leqslant t < 30$ & $30 \leqslant t < 40$ & $40 \leqslant t < 70$ \\
\hline
\begin{tabular}{ l }
Number \\
of people \\
\end{tabular} & 11 & 20 & 32 & 18 & 10 & 6 \\
\hline
\end{tabular}
\end{center}

(i) Calculate estimates of the mean and standard deviation of the time spent per day on these mobile phones.\\
(ii) On graph paper, draw a fully labelled histogram to represent the data.

\hfill \mbox{\textit{CAIE S1 2003 Q7 [9]}}