| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Conditional probability with uniform |
| Difficulty | Moderate -0.8 This is a straightforward application of standard uniform distribution formulas. Parts (a)-(c) require direct recall of mean, variance, and probability formulas for uniform distributions with no problem-solving. Part (d) involves conditional probability but is computationally simple (finding P(X<6|X>4) = 2/5). All calculations are routine with no conceptual challenges beyond basic S2 content. |
| Spec | 2.03d Calculate conditional probability: from first principles5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \frac{9+3}{2} = 6\) | B1 | |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Var}(X) = \frac{(9-3)^2}{12} = 3\) | M1A1 | M1 for \(\frac{(9-3)^2}{12}\) or \(\frac{(9+3)^2}{12}\) |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X>7) = (9-7) \times \frac{1}{6} = \frac{1}{3}\) | M1A1 | M1 for \(\frac{(9-7)}{6}\) or \(1-\frac{(7-3)}{6}\) or \(\int_7^9 \frac{1}{6}\,dx\) or \(1-\int_3^7 \frac{1}{6}\,dx\); A1 also accepts \(0.\dot{3}\), \(0.3\overline{3}\), awrt \(0.333\) |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X<6 \mid X>4) = \frac{P(4 | M1 | M1 for \(\frac{P(4 |
| \(= \dfrac{\tfrac{2}{6}}{\tfrac{5}{6}} = \dfrac{2}{5}\) | A1 | A1 for \(\frac{P(4 |
| (3) [8] |
# Question 1:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \frac{9+3}{2} = 6$ | B1 | |
| | **(1)** | |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Var}(X) = \frac{(9-3)^2}{12} = 3$ | M1A1 | M1 for $\frac{(9-3)^2}{12}$ or $\frac{(9+3)^2}{12}$ |
| | **(2)** | |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X>7) = (9-7) \times \frac{1}{6} = \frac{1}{3}$ | M1A1 | M1 for $\frac{(9-7)}{6}$ or $1-\frac{(7-3)}{6}$ or $\int_7^9 \frac{1}{6}\,dx$ or $1-\int_3^7 \frac{1}{6}\,dx$; A1 also accepts $0.\dot{3}$, $0.3\overline{3}$, awrt $0.333$ |
| | **(2)** | |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X<6 \mid X>4) = \frac{P(4<X<6)}{P(X>4)}$ | M1 | M1 for $\frac{P(4<X<6)}{P(X>4)}$ or $\frac{P(X<6)}{P(X>4)}$ or $\frac{2/6}{5/6}$ or $\frac{3/6}{5/6}$ or $1-\frac{P(X>6)}{P(X>4)}$ or $\frac{6-4}{9-4}$ or $\frac{3}{5}$ |
| $= \dfrac{\tfrac{2}{6}}{\tfrac{5}{6}} = \dfrac{2}{5}$ | A1 | A1 for $\frac{P(4<X<6)}{P(X>4)}$ or $\frac{2/6}{5/6}$ or $1-\frac{P(X>6)}{P(X>4)}$ or $\frac{6-4}{9-4}$; an answer of $\frac{2}{5}$ gains all 3 marks; $\leq$ and $\geq$ accepted in formulae |
| | **(3) [8]** | |\begin{enumerate}
\item The time in minutes that Elaine takes to checkout at her local supermarket follows a continuous uniform distribution defined over the interval [3,9].
\end{enumerate}
Find\\
(a) Elaine's expected checkout time,\\
(b) the variance of the time taken to checkout at the supermarket,\\
(c) the probability that Elaine will take more than 7 minutes to checkout.
Given that Elaine has already spent 4 minutes at the checkout,\\
(d) find the probability that she will take a total of less than 6 minutes to checkout.\\
\hfill \mbox{\textit{Edexcel S2 2012 Q1 [8]}}