| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | State or write down basic properties |
| Difficulty | Standard +0.3 This is a structured S2 question covering standard continuous distributions (uniform, triangular, normal). Part (a) uses uniform distribution formulas (routine recall). Part (b) exploits symmetry. Part (c) requires integrating to find c and Var(X) using standard techniques. Parts (d-f) involve straightforward probability calculations and comparison. While multi-part with 19 marks total, each component uses textbook methods without requiring novel insight or complex problem-solving—slightly easier than average A-level. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Mean \(= 5\) by symmetry | B1 M1 A1 | |
| Standard dev. \(= \sqrt{\frac{100}{12}} = 2.89\) | ||
| (b) Mean \(= 5\) | B1; M1 A1 | |
| (c) For area 1 under pdf, \(\frac{1}{2} \times 10 \times 5c = 1\), so \(c = \frac{1}{25}\) | ||
| Variance \(= \int_0^{10} f(x)\, dx - 5^2 = \frac{1}{25}\left[\frac{x^4}{4}\Big | _0^{10} + \frac{10x^3}{3} - \frac{x^4}{4}\Big | _5^{10}\right] - 25 = 4\frac{1}{6}\) |
| so s.d. \(= 2.04\) | A1 | |
| (d) \(P(4 < X < 6) = 2 \times \frac{1}{2} \times 1 \times \left(\frac{4}{25} + \frac{5}{25}\right) = \frac{9}{25}\) or \(0.36\) | M1 A1 A1 | |
| (e) \(P(4 < X < 6) = P\left(-\frac{1}{2.04} < Z < \frac{1}{2.04}\right) = P(-0.49 < Z < 0.49) = 2(0.1879) = 0.376\) | M1 M1 A1 | |
| A1 | ||
| (f) Similar; slightly more concentration near the mean for the Normal model. People are aiming at the middle, so this is probably better | B1 | |
| 19 |
(a) Mean $= 5$ by symmetry | B1 M1 A1 |
Standard dev. $= \sqrt{\frac{100}{12}} = 2.89$ | |
(b) Mean $= 5$ | B1; M1 A1 |
(c) For area 1 under pdf, $\frac{1}{2} \times 10 \times 5c = 1$, so $c = \frac{1}{25}$ | |
Variance $= \int_0^{10} f(x)\, dx - 5^2 = \frac{1}{25}\left[\frac{x^4}{4}\Big|_0^{10} + \frac{10x^3}{3} - \frac{x^4}{4}\Big|_5^{10}\right] - 25 = 4\frac{1}{6}$ | M1 A1 M1 A1 |
so s.d. $= 2.04$ | A1 |
(d) $P(4 < X < 6) = 2 \times \frac{1}{2} \times 1 \times \left(\frac{4}{25} + \frac{5}{25}\right) = \frac{9}{25}$ or $0.36$ | M1 A1 A1 |
(e) $P(4 < X < 6) = P\left(-\frac{1}{2.04} < Z < \frac{1}{2.04}\right) = P(-0.49 < Z < 0.49) = 2(0.1879) = 0.376$ | M1 M1 A1 |
| | A1 |
(f) Similar; slightly more concentration near the mean for the Normal model. People are aiming at the middle, so this is probably better | B1 |
| | 19 |Some children are asked to mark the centre of a scale 10 cm long. The position they choose is indicated by the variable $X$, where $0 \leq X \leq 10$. Initially, $X$ is modelled as a random variable with a continuous uniform distribution.
\begin{enumerate}[label=(\alph*)]
\item Find the mean and the standard deviation of $X$. [3 marks]
\end{enumerate}
It is suggested that a better model would be the distribution with probability density function
$$f(x) = cx, \quad 0 \leq x \leq 5, \quad f(x) = c(10-x), \quad 5 < x \leq 10, \quad f(x) = 0 \text{ otherwise}.$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Write down the mean of $X$. [1 mark]
\item Find $c$, and hence find the standard deviation of $X$ in this model. [7 marks]
\item Find P($4 < X < 6$). [3 marks]
\end{enumerate}
It is then proposed that an even better model for $X$ would be a Normal distribution with the mean and standard deviation found in parts (b) and (c).
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Use these results to find P($4 < X < 6$) in the third model. [4 marks]
\item Compare your answer with (d). Which model do you think is most appropriate? [1 mark]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q7 [19]}}