| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Find parameters from given statistics |
| Difficulty | Moderate -0.3 This is a straightforward S2 question testing standard formulas for continuous uniform distribution (mean = (α+β)/2, variance = (β-α)²/12) and basic conditional probability. Part (a) involves solving two simultaneous equations, part (b) uses the law of total probability with given probabilities, and part (c) applies Bayes' theorem. All techniques are routine for S2 with no novel insight required, making it slightly easier than average. |
| Spec | 2.03d Calculate conditional probability: from first principles5.03a Continuous random variables: pdf and cdf |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{\beta+\alpha}{2} = 4\), \(\frac{(\beta-\alpha)^2}{12} = 12\) | B1 | Both correct oe |
| \(\beta + \alpha = 8\) and \((\beta-\alpha) = 12\) or \(\alpha^2 - 8\alpha - 20 = 0\) or \(\beta^2 - 8\beta - 20 = 0\) | B1 | Pair of correct linear equations or correct single equation in \(\alpha\) or \(\beta\) |
| \(2\beta = 20\) | M1d | Dep on 1st B; correct method eliminating \(\alpha\) or \(\beta\) |
| \(\beta = 10\) | A1 | Must state \(\beta = 10\) |
| \(\alpha = -2\) | A1 | Must state \(\alpha = -2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P(\text{David late}) = 0.05 + 0.95 \times \left(\frac{\text{"10"}-5}{\text{"12"}}\right)\) | M1, B1ft | B1ft: \(\frac{10-5}{12}\) or \(\frac{5}{12}\) or awrt 0.417 |
| \(= \frac{107}{240}\) or \(0.4458\ldots\) | A1 | awrt 0.446 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P(\text{missed train} \mid \text{late}) = \frac{0.05}{0.446}\) | M1 | \(\frac{0.05}{\text{their (b)}}\) |
| \(= \frac{12}{107}\) or \(0.1121\ldots\) | A1 | awrt 0.112 |
# Question 4:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{\beta+\alpha}{2} = 4$, $\frac{(\beta-\alpha)^2}{12} = 12$ | B1 | Both correct oe |
| $\beta + \alpha = 8$ and $(\beta-\alpha) = 12$ or $\alpha^2 - 8\alpha - 20 = 0$ or $\beta^2 - 8\beta - 20 = 0$ | B1 | Pair of correct linear equations or correct single equation in $\alpha$ or $\beta$ |
| $2\beta = 20$ | M1d | Dep on 1st B; correct method eliminating $\alpha$ or $\beta$ |
| $\beta = 10$ | A1 | Must state $\beta = 10$ |
| $\alpha = -2$ | A1 | Must state $\alpha = -2$ |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(\text{David late}) = 0.05 + 0.95 \times \left(\frac{\text{"10"}-5}{\text{"12"}}\right)$ | M1, B1ft | B1ft: $\frac{10-5}{12}$ or $\frac{5}{12}$ or awrt 0.417 |
| $= \frac{107}{240}$ or $0.4458\ldots$ | A1 | awrt 0.446 |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(\text{missed train} \mid \text{late}) = \frac{0.05}{0.446}$ | M1 | $\frac{0.05}{\text{their (b)}}$ |
| $= \frac{12}{107}$ or $0.1121\ldots$ | A1 | awrt 0.112 |
---\begin{enumerate}
\item David aims to catch the train to work each morning. The scheduled departure time of the train is 0830
\end{enumerate}
The number of minutes after 0830 that the train departs may be modelled by the random variable $X$. Given that $X$ has a continuous uniform distribution over $[ \alpha , \beta ]$ and that $\mathrm { E } ( X ) = 4$ and $\operatorname { Var } ( X ) = 12$\\
(a) find the value of $\alpha$ and the value of $\beta$.
Each morning, the probability that David oversleeps is 0.05
If David oversleeps he will be late for work.
If he does not oversleep he will be in time to catch the train, but will be late for work if the train departs after 0835\\
(b) Find the probability that David will be late for work.
Given that David is late for work,\\
(c) find the probability that he overslept.\\
\hfill \mbox{\textit{Edexcel S2 2018 Q4 [10]}}