| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Estimate mean and standard deviation from frequency table |
| Difficulty | Moderate -0.3 This is a standard S1 statistics question covering routine frequency table calculations (mean, standard deviation, median via interpolation) and histogram construction. All techniques are textbook procedures requiring careful arithmetic but no problem-solving insight. The multiple parts and potential for arithmetic errors keep it from being trivially easy, but it remains below average difficulty for A-level. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Hours | \(1 - 10\) | \(11 - 20\) | \(21 - 25\) | \(26 - 30\) | \(31 - 40\) | \(41 - 59\) |
| Frequency | 6 | 15 | 11 | 13 | 8 | 3 |
| Mid-point | 5.5 | 15.5 | 28 | 50 |
| Answer | Marks |
|---|---|
| 23, 35.5 (may be in table) | B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Width of 10 units is 4 cm so width of 5 units is 2 cm | B1 | |
| Height \(= 2.6 \times 4 =\) 10.4 cm | M1 A1 | M1 for their width \(\times\) their height = 20.8 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum fx = 1316.5\), \(\bar{x} = \frac{1316.5}{56}\) = awrt 23.5 | M1 A1 | 1st M1 for reasonable attempt at \(\sum fx\) and /56 |
| \(\sum fx^2 = 37378.25\) (can be implied) | B1 | |
| \(\sigma = \sqrt{\frac{37378.25}{56} - \bar{x}^2}\) = awrt 10.7, allow \(s=10.8\) | M1 A1 | 2nd M1 for method for \(\sigma\) or \(s\), \(\sqrt{}\) required |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q_2 = (20.5) + \frac{(28-21)}{11} \times 5 = 23.68\ldots\) awrt 23.7 or 23.9 | M1 A1 | lcb can be 20, 20.5 or 21; width 4 or 5; fraction correct for M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q_3 - Q_2 = 5.6\), \(Q_2 - Q_1 = 7.9\) (or \(\bar{x} < Q_2\)) | M1 | M1 for attempting skewness test using quartiles or mean/median |
| \([7.9 > 5.6\) so] negative skew | A1 |
# Question 5:
## Part (a)
| 23, 35.5 (may be in table) | B1 B1 | |
## Part (b)
| Width of 10 units is 4 cm so width of 5 units is **2 cm** | B1 | |
|---|---|---|
| Height $= 2.6 \times 4 =$ **10.4 cm** | M1 A1 | M1 for their width $\times$ their height = 20.8 |
## Part (c)
| $\sum fx = 1316.5$, $\bar{x} = \frac{1316.5}{56}$ = awrt **23.5** | M1 A1 | 1st M1 for reasonable attempt at $\sum fx$ and /56 |
|---|---|---|
| $\sum fx^2 = 37378.25$ (can be implied) | B1 | |
| $\sigma = \sqrt{\frac{37378.25}{56} - \bar{x}^2}$ = awrt **10.7**, allow $s=10.8$ | M1 A1 | 2nd M1 for method for $\sigma$ or $s$, $\sqrt{}$ required |
## Part (d)
| $Q_2 = (20.5) + \frac{(28-21)}{11} \times 5 = 23.68\ldots$ awrt **23.7 or 23.9** | M1 A1 | lcb can be 20, 20.5 or 21; width 4 or 5; fraction correct for M1 |
## Part (e)
| $Q_3 - Q_2 = 5.6$, $Q_2 - Q_1 = 7.9$ (or $\bar{x} < Q_2$) | M1 | M1 for attempting skewness test using quartiles or mean/median |
|---|---|---|
| $[7.9 > 5.6$ so] **negative skew** | A1 | |
---\begin{enumerate}
\item A teacher selects a random sample of 56 students and records, to the nearest hour, the time spent watching television in a particular week.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
Hours & $1 - 10$ & $11 - 20$ & $21 - 25$ & $26 - 30$ & $31 - 40$ & $41 - 59$ \\
\hline
Frequency & 6 & 15 & 11 & 13 & 8 & 3 \\
\hline
Mid-point & 5.5 & 15.5 & & 28 & & 50 \\
\hline
\end{tabular}
\end{center}
(a) Find the mid-points of the 21-25 hour and 31-40 hour groups.
A histogram was drawn to represent these data. The 11-20 group was represented by a bar of width 4 cm and height 6 cm .\\
(b) Find the width and height of the 26-30 group.\\
(c) Estimate the mean and standard deviation of the time spent watching television by these students.\\
(d) Use linear interpolation to estimate the median length of time spent watching television by these students.
The teacher estimated the lower quartile and the upper quartile of the time spent watching television to be 15.8 and 29.3 respectively.\\
(e) State, giving a reason, the skewness of these data.\\
\hfill \mbox{\textit{Edexcel S1 Q5 [14]}}