| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2004 |
| Session | June |
| Marks | 17 |
| 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 standard S2 question testing routine application of continuous probability distributions. Parts (a)-(c) require straightforward integration and algebraic manipulation using well-practiced formulas (E(X), CDF construction, median from F(x)=0.5). Part (d) is simple comparison. While multi-part with 17 marks total, each component is textbook-standard with no novel insight required, making it slightly easier than average. |
| 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 |
|---|---|---|
| \(= 1.268 = 1.27\) to 3 sf or \(\frac{571}{450}\) or \(\frac{121}{450}\) | M1M1, A1A1, awrt1.27 A1 | (5) |
| Answer | Marks |
|---|---|
| \(F(x_0) = \int_0^{x_0} \frac{1}{3}dx = \frac{1}{3}x_0\) for \(0 \leq x < 1\) | M1A1 |
| Answer | Marks |
|---|---|
| \(F(x_0) = \frac{1}{3} + \int_0^{x_0} \frac{8x^3}{45}dx\) for \(1 \leq x \leq 2\) | M1 |
| Answer | Marks |
|---|---|
| \(= \frac{1}{3} + \left[\frac{8x^4}{180}\right]_0^{x_0}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{1}{45}\left(2x_0^4 + 13\right)\) | A1 | |
| \[F(x) = \begin{cases} 0 & x < 0 \\ \frac{1}{3}x & 0 \leq x < 1 \\ \frac{1}{45}(2x^4 + 13) & 1 \leq x \leq 2 \\ 1 & x > 2 \end{cases}\] | B1B1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| \(m = 1.48\) to 3 sf | M1A1ift, awrt1.48 A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| mean < median. Negative Skew | B1, dep B1 | (2) |
## Part (a)
$E(X) = \int_0^1 \frac{1}{3}x dx + \int_1^2 \frac{8x^4}{45} dx$
$= \left[\frac{1}{6}x^2\right]_0^1 + \left[\frac{8x^5}{225}\right]_1^2$
$= 1.268 = 1.27$ to 3 sf or $\frac{571}{450}$ or $\frac{121}{450}$ | M1M1, A1A1, awrt1.27 A1 | (5)
$\int xf(x)dx$, 2 terms added; Expressions, limits
## Part (b)
$F(x_0) = \int_0^{x_0} \frac{1}{3}dx = \frac{1}{3}x_0$ for $0 \leq x < 1$ | M1A1 |
Variable upper limit; $\int f(x)dx$, $\frac{1}{3}x_0$
$F(x_0) = \frac{1}{3} + \int_0^{x_0} \frac{8x^3}{45}dx$ for $1 \leq x \leq 2$ | M1 |
Their fraction + v.u.l on; $\int f(x)dx$ &2 terms
$= \frac{1}{3} + \left[\frac{8x^4}{180}\right]_0^{x_0}$ | A1 |
$\frac{8x^4}{180}$
$= \frac{1}{45}\left(2x_0^4 + 13\right)$ | A1 |
$$F(x) = \begin{cases} 0 & x < 0 \\ \frac{1}{3}x & 0 \leq x < 1 \\ \frac{1}{45}(2x^4 + 13) & 1 \leq x \leq 2 \\ 1 & x > 2 \end{cases}$$ | B1B1 | (7)
Middle pair, ends
## Part (c)
$F(m) = 0.5$
$\frac{1}{45}(2x^4 + 13) = \frac{1}{2}$
$m^4 = 4.75$
$m = 1.48$ to 3 sf | M1A1ift, awrt1.48 A1 | (3)
Their function=0.5
## Part (d)
mean < median. Negative Skew | B1, dep B1 | (2)
(Total 17 marks)A random variable $X$ has probability density function given by
$$f(x) = \begin{cases}
\frac{1}{3}, & 0 \leq x \leq 1, \\
\frac{8x^3}{45}, & 1 \leq x \leq 2, \\
0, & \text{otherwise}.
\end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Calculate the mean of $X$. [5]
\item Specify fully the cumulative distribution function F$(x)$. [7]
\item Find the median of $X$. [3]
\item Comment on the skewness of the distribution of $X$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2004 Q7 [17]}}