| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2003 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Interquartile range and percentiles |
| Difficulty | Moderate -0.8 This is a straightforward application of continuous uniform distribution properties. Part (a) requires writing down the standard PDF formula and sketching a rectangle. Parts (b)-(d) involve direct substitution into uniform distribution formulas (CDF calculations and quartiles). Part (e) requires solving a simple linear equation involving probabilities. All techniques are routine for S2 level with no novel problem-solving required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = \begin{cases} 0.05 & 180 \le x \le 200 \\ 0 & \text{otherwise} \end{cases}\) | B1 B1 | |
| Graph with labels and 3 parts | B1 B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \le 183) = 3 \times 0.05\) | M1 | |
| \(= 0.15\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X = 183) = 0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{IQR} = 10\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.05(200 - x) = 0.05(x - 180) \times 2\) | M1; A1 | |
| \(200 - x = 2x - 360\) | ||
| \(x = 186\frac{2}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{3}\) of all cups of lemonade dispensed contains \(186\frac{2}{3}\) ml or less (or \(\frac{2}{3}\) of all cups of lemonade dispensed contains \(186\frac{2}{3}\) ml or more) | B1 B1 ft |
## (a)
$f(x) = \begin{cases} 0.05 & 180 \le x \le 200 \\ 0 & \text{otherwise} \end{cases}$ | | B1 B1 |
Graph with labels and 3 parts | | B1 B1 B1 | (4)
## (b)(i)
$P(X \le 183) = 3 \times 0.05$ | | M1 |
$= 0.15$ | | A1 |
## (b)(ii)
$P(X = 183) = 0$ | | B1 | (3)
## (c)
$\text{IQR} = 10$ | | B1 | (1)
## (d)
$0.05(200 - x) = 0.05(x - 180) \times 2$ | | M1; A1 |
$200 - x = 2x - 360$ | |
$x = 186\frac{2}{3}$ | | A1 | (3)
## (e)
$\frac{1}{3}$ of all cups of lemonade dispensed contains $186\frac{2}{3}$ ml or less (or $\frac{2}{3}$ of all cups of lemonade dispensed contains $186\frac{2}{3}$ ml or more) | | B1 B1 ft | (2)
---A drinks machine dispenses lemonade into cups. It is electronically controlled to cut off the flow of lemonade randomly between 180 ml and 200 ml. The random variable $X$ is the volume of lemonade dispensed into a cup.
\begin{enumerate}[label=(\alph*)]
\item Specify the probability density function of $X$ and sketch its graph. [4]
\item Find the probability that the machine dispenses
\begin{enumerate}[label=(\roman*)]
\item less than 183 ml,
\item exactly 183 ml.
\end{enumerate} [3]
\item Calculate the inter-quartile range of $X$. [1]
\item Determine the value of $x$ such that P($X \geq x$) = 2P($X \leq x$). [3]
\item Interpret in words your value of $x$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2003 Q5 [13]}}