| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2003 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Moderate -0.3 This is a standard S2 probability density function question requiring routine integration techniques. Parts (a)-(c) involve straightforward applications of ∫f(x)dx=1, E(X)=∫xf(x)dx, and F(x)=∫f(t)dt. Part (d) uses the CDF from part (c). The algebra is simple since x²+2x+1=(x+1)², making integration elementary. While multi-part with 15 marks total, each step follows textbook procedures without requiring problem-solving insight, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_{-1}^{0} k(x^2 + 2x + 1) \, dx = 1\) | limits needed and \(=1\) | M1 |
| \(\left[k\left(\frac{x^3}{3} + x^2 + x\right)\right]_{-1}^{0} = 1\) | attempt at integration | M1 A1 |
| \(k = 3\) | (*) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \int_{-1}^{0} xf(x) \, dx\) | M1 | |
| \(= \int_{-1}^{0} (3x^3 + 6x^2 + 3x) \, dx\) | limits needed | A1 |
| \(= \left[\frac{3x^4}{4} + 2x^3 + \frac{3x^2}{2}\right]_{-1}^{0}\) | integration and substituting limits | M1 |
| \(= -\frac{1}{4}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_{-1}^{0} (3x^3 + 6x^2 + 3x) \, dx = \left[x^3 + 3x^2 + 3x\right]_{-1}^{0}\) | M1 | |
| \(= x_0 + 3x_0^2 + 3x_0 + 1\) | A1 | |
| \(F(x) = \begin{cases} 0 & x < -1 \\ x^3 + 3x^2 + 3x + 1 & -1 \le x \le 0 \\ 1 & x > 0 \end{cases}\) | B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(-0.3 < X < 0.3) = F(0.3) - F(-0.3)\) | M1 | |
| \(= 1 - 0.343\) | A1 | |
| \(= 0.657\) | A1 |
## (a)
$\int_{-1}^{0} k(x^2 + 2x + 1) \, dx = 1$ | limits needed and $=1$ | M1 |
$\left[k\left(\frac{x^3}{3} + x^2 + x\right)\right]_{-1}^{0} = 1$ | attempt at integration | M1 A1 |
$k = 3$ | (*) | A1 | (4)
## (b)
$E(X) = \int_{-1}^{0} xf(x) \, dx$ | | M1 |
$= \int_{-1}^{0} (3x^3 + 6x^2 + 3x) \, dx$ | limits needed | A1 |
$= \left[\frac{3x^4}{4} + 2x^3 + \frac{3x^2}{2}\right]_{-1}^{0}$ | integration and substituting limits | M1 |
$= -\frac{1}{4}$ | | A1 | (4)
## (c)
$\int_{-1}^{0} (3x^3 + 6x^2 + 3x) \, dx = \left[x^3 + 3x^2 + 3x\right]_{-1}^{0}$ | | M1 |
$= x_0 + 3x_0^2 + 3x_0 + 1$ | | A1 |
$F(x) = \begin{cases} 0 & x < -1 \\ x^3 + 3x^2 + 3x + 1 & -1 \le x \le 0 \\ 1 & x > 0 \end{cases}$ | | B1 B1 | (4)
## (d)
$P(-0.3 < X < 0.3) = F(0.3) - F(-0.3)$ | | M1 |
$= 1 - 0.343$ | | A1 |
$= 0.657$ | | A1 | (3)A continuous random variable $X$ has probability density function f($x$) where
$$\text{f}(x) = \begin{cases}
k(x^2 + 2x + 1) & -1 \leq x \leq 0, \\
0, & \text{otherwise}
\end{cases}$$
where $k$ is a positive integer.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = 3$. [4]
\end{enumerate}
Find
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item E($X$), [4]
\item the cumulative distribution function F($x$), [4]
\item P($-0.3 < X < 0.3$). [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2003 Q7 [15]}}