| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Standard +0.3 This is a straightforward S1 question requiring standard probability distribution calculations: finding k using ΣP(X=x)=1, computing E(X) from definition, and calculating sample statistics. All steps are routine applications of formulas with simple arithmetic. The 'comment on suitability' requires only basic comparison of means, making this slightly easier than average. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Number of attempts | 1 | 2 | 3 | 4 |
| Number of children | 43 | 26 | 13 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum fx = 146\); mean \(= \frac{146}{85} = 1.72\) (3sf) | M1 A1 | |
| \(\sum fx^2 = 312\) | M1 | |
| std. dev. \(= \sqrt{\frac{312}{85} - (1.7176)^2} = 0.849\) (3sf) | M1 A1 | |
| \(\sum P(x) = 19k + 16k + 11k + 4k = 50k = 1\) \(\therefore k = \frac{1}{50}\) | M2 A1 | |
| \(\sum xP(x) = \frac{19}{50} + \frac{32}{50} + \frac{33}{50} + \frac{16}{50} = 2\) | M1 A1 | |
| e.g. mean of model not very close \(\therefore\) not very suitable | B1 | (11) |
| $\sum fx = 146$; mean $= \frac{146}{85} = 1.72$ (3sf) | M1 A1 | |
| $\sum fx^2 = 312$ | M1 | |
| std. dev. $= \sqrt{\frac{312}{85} - (1.7176)^2} = 0.849$ (3sf) | M1 A1 | |
| $\sum P(x) = 19k + 16k + 11k + 4k = 50k = 1$ $\therefore k = \frac{1}{50}$ | M2 A1 | |
| $\sum xP(x) = \frac{19}{50} + \frac{32}{50} + \frac{33}{50} + \frac{16}{50} = 2$ | M1 A1 | |
| e.g. mean of model not very close $\therefore$ not very suitable | B1 | (11) |5. A group of children were each asked to try and complete a task to test hand-eye coordination. Each child repeated the task until he or she had been successful or had made four attempts.
The number of attempts made by the children in the group are summarised in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
Number of attempts & 1 & 2 & 3 & 4 \\
\hline
Number of children & 43 & 26 & 13 & 3 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Calculate the mean and standard deviation of the number of attempts made by each child.
It is suggested that the number of attempts made by each child could be modelled by a discrete random variable $X$ with the probability function
$$P ( X = x ) = \left\{ \begin{array} { c c }
k \left( 20 - x ^ { 2 } \right) , & x = 1,2,3,4 \\
0 , & \text { otherwise }
\end{array} \right.$$
\item Show that $k = \frac { 1 } { 50 }$.
\item Find $\mathrm { E } ( X )$.
\item Comment on the suitability of this model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q5 [11]}}