| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate statistics from discrete frequency table |
| Difficulty | Moderate -0.3 This is a straightforward S1 statistics question requiring standard calculations of mean and standard deviation from a frequency table (using calculator or formulas), knowledge that a discrete uniform distribution on 1-9 has mean 5, and a simple comparison to conclude the model is unsuitable. The calculations are routine and the conceptual demand is minimal—slightly easier than average A-level maths questions. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.02e Discrete uniform distribution |
| Value chosen | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| Number of children | 3 | 4 | 5 | 10 | 12 | 13 | 7 | 4 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum fx = 303\) | M1 | |
| mean \(= \frac{303}{60} = 5.05\) | M1 A1 | |
| \(\sum fx^2 = 1753\) | M1 | |
| std. dev. \(= \sqrt{\frac{1753}{60} - (5.05)^2} = 1.93\) | M1 A1 | |
| (by symmetry) 5 | M1 A1 | |
| actual std. dev. much lower than in model; tendency to pick numbers nearer the middle | B1 B1 | (10) |
| $\sum fx = 303$ | M1 | |
| mean $= \frac{303}{60} = 5.05$ | M1 A1 | |
| $\sum fx^2 = 1753$ | M1 | |
| std. dev. $= \sqrt{\frac{1753}{60} - (5.05)^2} = 1.93$ | M1 A1 | |
| (by symmetry) 5 | M1 A1 | |
| actual std. dev. much lower than in model; tendency to pick numbers nearer the middle | B1 B1 | (10) |A group of 60 children were each asked to choose an integer value between 1 and 9 inclusive. Their choices are summarised in the table below.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
Value chosen & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
Number of children & 3 & 4 & 5 & 10 & 12 & 13 & 7 & 4 & 2 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\alph*)]
\item Calculate the mean and standard deviation of the values chosen. [6]
\end{enumerate}
It is suggested that the value chosen could be modelled by a discrete uniform distribution.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Write down the mean that this model would predict. [2]
\end{enumerate}
Given also that the standard deviation according to this model would be 2.58,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item explain why this model is not suitable and suggest why this is the case. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q3 [10]}}