| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Calculate and compare mean, median, mode |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard pdf/cdf techniques: calculating E(X) and Var(X) using integration, finding the cdf by integration, solving F(x)=0.5 for the median, and identifying the mode. All steps are routine applications of formulas with no conceptual challenges. Part (f) requires minimal interpretation. Slightly easier than average due to the mechanical nature of the calculations. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(E(X) = \int_4^{10} \frac{x^2}{312} \, dx = \left[\frac{x^3}{1248}\right]_4^{10} = 7.81\) | M1 A1 A1 | |
| (b) \(\text{Var}(X) = \int_4^{10} \frac{x^4}{312} \, dx - 7.81^2 = \left[\frac{x^5}{1560}\right]_4^{10} - 7.808^2 = 2.49\) | M1 A1 M1 A1 | |
| (c) \(F(x) = 0\) \((x < 4)\), \(F(x) = \int_4^x \frac{u^2}{312} \, du = \frac{x^3 - 64}{936}\) \((4 \leq x \leq 10)\), \(F(x) = 1\) \((x > 10)\) | B1 M1 A1 A1 | |
| (d) For median \(m\), \(\frac{m^3 - 64}{936} = \frac{1}{2}\) \(m^3 = 532\) \(m = 8.10\) | M1 A1 A1 | |
| (e) By inspection, mode = 10 | B1 | |
| (f) \(2(8.1 - 7.81) = 0.58\) \(10 - 8.1 = 1.9\) Not very similar; mode is an extreme point and is not centrally located | B1 B1 B1 | 18 marks total |
(a) $E(X) = \int_4^{10} \frac{x^2}{312} \, dx = \left[\frac{x^3}{1248}\right]_4^{10} = 7.81$ | M1 A1 A1 |
(b) $\text{Var}(X) = \int_4^{10} \frac{x^4}{312} \, dx - 7.81^2 = \left[\frac{x^5}{1560}\right]_4^{10} - 7.808^2 = 2.49$ | M1 A1 M1 A1 |
(c) $F(x) = 0$ $(x < 4)$, $F(x) = \int_4^x \frac{u^2}{312} \, du = \frac{x^3 - 64}{936}$ $(4 \leq x \leq 10)$, $F(x) = 1$ $(x > 10)$ | B1 M1 A1 A1 |
(d) For median $m$, $\frac{m^3 - 64}{936} = \frac{1}{2}$ $m^3 = 532$ $m = 8.10$ | M1 A1 A1 |
(e) By inspection, mode = 10 | B1 |
(f) $2(8.1 - 7.81) = 0.58$ $10 - 8.1 = 1.9$ Not very similar; mode is an extreme point and is not centrally located | B1 B1 B1 | 18 marks totalA continuous random variable $X$ has a probability density function given by
$$f(x) = \frac{x^2}{312} \quad 4 \leq x \leq 10,$$
$$f(x) = 0 \quad \text{otherwise}.$$
\begin{enumerate}[label=(\alph*)]
\item Find E$(X)$. [3 marks]
\item Find the variance of $X$. [4 marks]
\item Find the cumulative distribution function F$(x)$, for all values of $x$. [5 marks]
\item Hence find the median value of $X$. [3 marks]
\item Write down the modal value of $X$. [1 mark]
\end{enumerate}
It is sometimes suggested that, for most distributions,
$$2 \times (\text{median} - \text{mean}) \approx \text{mode} - \text{median}.$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{5}
\item Show that this result is not satisfied in this case, and suggest a reason why. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q7 [18]}}