| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Probability distribution from formula |
| Difficulty | Moderate -0.3 This is a straightforward probability distribution question requiring understanding of 'inversely proportional' and standard calculations of mean and variance. Part (a) involves setting up k/x = probability and using Σp=1 to find k, which is routine. Part (b) applies standard formulas E(X) and Var(X) with simple arithmetic. The context is slightly unusual but the mathematical techniques are basic S1 content with no problem-solving insight required, making it slightly easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P(X = x) = \frac{k}{x}\) where \(\frac{k}{1} + \frac{k}{2} + \frac{k}{3} + \frac{k}{4} = 1\) | M1 M1 | |
| \(25k = 12\) so \(k = \frac{12}{25}\) | A1 | |
| \(P(1) = \frac{12}{25}\), \(P(2) = \frac{6}{25}\), \(P(3) = \frac{4}{25}\), \(P(4) = \frac{3}{25}\) | A1 A1 A1 | |
| (b) \(E(X) = \frac{4 \times 12}{25} = 1.92\) | M1 A1 | |
| \(E(X^2) = \frac{12 + 24 + 36 + 48}{25} = 4.8\) | M1 A1 A1 | |
| \(\text{Var}(X) = 4.8 - 1.92^2 = 1.11\) | M1 A1 A1 | Total: 10 marks |
(a) $P(X = x) = \frac{k}{x}$ where $\frac{k}{1} + \frac{k}{2} + \frac{k}{3} + \frac{k}{4} = 1$ | M1 M1 |
$25k = 12$ so $k = \frac{12}{25}$ | A1 |
$P(1) = \frac{12}{25}$, $P(2) = \frac{6}{25}$, $P(3) = \frac{4}{25}$, $P(4) = \frac{3}{25}$ | A1 A1 A1 |
(b) $E(X) = \frac{4 \times 12}{25} = 1.92$ | M1 A1 |
$E(X^2) = \frac{12 + 24 + 36 + 48}{25} = 4.8$ | M1 A1 A1 |
$\text{Var}(X) = 4.8 - 1.92^2 = 1.11$ | M1 A1 A1 | **Total: 10 marks**A regular tetrahedron has its faces numbered 1, 2, 3 and 4. It is weighted so that when it is thrown, the probability of each face being in contact with the table is inversely proportional to the number on that face. This number is represented by the random variable $X$.
\begin{enumerate}[label=(\alph*)]
\item Show that $P(X = 1) = \frac{12}{25}$ and find the probabilities of the other values of $X$. [5 marks]
\item Calculate the mean and the variance of $X$. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q3 [10]}}