Edexcel S1 2016 June — Question 4 12 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2016
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeCoding to simplify calculation
DifficultyModerate -0.8 This is a routine S1 statistics question testing standard procedures: finding midpoints, using given summations to calculate mean and standard deviation with coding, linear interpolation for median, and reverse-coding. All techniques are straightforward textbook applications with no problem-solving insight required. The multi-part structure (6 parts) adds length but not conceptual difficulty.
Spec2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02j Clean data: missing data, errors

4. A researcher recorded the time, \(t\) minutes, spent using a mobile phone during a particular afternoon, for each child in a club. The researcher coded the data using \(v = \frac { t - 5 } { 10 }\) and the results are summarised in the table below.
Coded Time (v)Frequency ( \(\boldsymbol { f }\) )Coded Time Midpoint (m)
\(0 \leqslant v < 5\)202.5
\(5 \leqslant v < 10\)24\(a\)
\(10 \leqslant v < 15\)1612.5
\(15 \leqslant v < 20\)1417.5
\(20 \leqslant v < 30\)6\(b\)
$$\text { (You may use } \sum f m = 825 \text { and } \sum f m ^ { 2 } = 12012.5 \text { ) }$$
  1. Write down the value of \(a\) and the value of \(b\).
  2. Calculate an estimate of the mean of \(v\).
  3. Calculate an estimate of the standard deviation of \(v\).
  4. Use linear interpolation to estimate the median of \(v\).
  5. Hence describe the skewness of the distribution. Give a reason for your answer.
  6. Calculate estimates of the mean and the standard deviation of the time spent using a mobile phone during the afternoon by the children in this club.

AnswerMarks Guidance
PartAnswer/Working Marks
(a)7.5 and 25 B1
(b)Mean \(= 10.3125\) B1
(c)\(\sigma = \sqrt{\frac{120125}{80} - '10.3125'^2}\) M1
\(= 6.6188...\) \((s = 6.6605...)\)A1 A1 for awrt 6.62 (Allow \(s =\) awrt 6.66)
(d)\(\text{Median} = \{5\} + \frac{20}{24} \times 5\) or \( \{10\}
\(= 9.16666\)A1 A1 for awrt 9.17 or (if using \(n+1\)) for awrt 9.27
(e)Mean > median \(\therefore\) positive skew M1A1
(f)\(t = 10v + 5\) M1
Mean \(= 10 \times 10.3125 + 5 = 108.125\)A1
\(\sigma = 10 \times 6.6188\)M1 2nd M1 for \(10 \times\) 'their sd'
\(= 66.188...\) (66.605 from \(s\))A1 A1 awrt 66.2
Total: 12 marks
Notes: For part (d), if using \(n+1\), A1 for awrt 9.27
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | 7.5 and 25 | B1 | B1 both values correct (may be seen in table) |
| (b) | Mean $= 10.3125$ | B1 | B1 for awrt 10.3 (Do not allow improper fractions) |
| (c) | $\sigma = \sqrt{\frac{120125}{80} - '10.3125'^2}$ | M1 | M1 for correct expression including square root (allow ft from their mean) |
| | $= 6.6188...$ $(s = 6.6605...)$ | A1 | A1 for awrt 6.62 (Allow $s =$ awrt 6.66) |
| (d) | $\text{Median} = \{5\} + \frac{20}{24} \times 5$ or $|\{10\}| - \frac{4}{24} \times 5$ | M1 | M1 for correct fraction: $\frac{20}{24} \times 5$ or if using $n+1$ for $\frac{20.5}{24} \times 5$ |
| | $= 9.16666$ | A1 | A1 for awrt 9.17 or (if using $n+1$) for awrt 9.27 |
| (e) | Mean > median $\therefore$ positive skew | M1A1 | M1 for correct comparison of 'their b' and 'their d' (must have answer to both b and d). Comparison may be part of bigger expression e.g. $3(\text{mean} - \text{median})/\text{s.d.}$ Allow use of $Q_3 - Q_2 > Q_2 - Q_1$ only if $Q_1 = 5$ and $Q_3 = 15$ are both seen. A1 for positive skew (which must follow from their values) |
| (f) | $t = 10v + 5$ | M1 | 1st M1 for $10 \times$ 'their mean'$+5$ |
| | Mean $= 10 \times 10.3125 + 5 = 108.125$ | A1 | |
| | $\sigma = 10 \times 6.6188$ | M1 | 2nd M1 for $10 \times$ 'their sd' |
| | $= 66.188...$ (66.605 from $s$) | A1 | A1 awrt 66.2 |

**Total: 12 marks**

Notes: For part (d), if using $n+1$, A1 for awrt 9.27

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4. A researcher recorded the time, $t$ minutes, spent using a mobile phone during a particular afternoon, for each child in a club.

The researcher coded the data using $v = \frac { t - 5 } { 10 }$ and the results are summarised in the table below.

\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Coded Time (v) & Frequency ( $\boldsymbol { f }$ ) & Coded Time Midpoint (m) \\
\hline
$0 \leqslant v < 5$ & 20 & 2.5 \\
\hline
$5 \leqslant v < 10$ & 24 & $a$ \\
\hline
$10 \leqslant v < 15$ & 16 & 12.5 \\
\hline
$15 \leqslant v < 20$ & 14 & 17.5 \\
\hline
$20 \leqslant v < 30$ & 6 & $b$ \\
\hline
\end{tabular}
\end{center}

$$\text { (You may use } \sum f m = 825 \text { and } \sum f m ^ { 2 } = 12012.5 \text { ) }$$
\begin{enumerate}[label=(\alph*)]
\item Write down the value of $a$ and the value of $b$.
\item Calculate an estimate of the mean of $v$.
\item Calculate an estimate of the standard deviation of $v$.
\item Use linear interpolation to estimate the median of $v$.
\item Hence describe the skewness of the distribution. Give a reason for your answer.
\item Calculate estimates of the mean and the standard deviation of the time spent using a mobile phone during the afternoon by the children in this club.

\begin{center}

\end{center}
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2016 Q4 [12]}}