| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find multiple parameters from system |
| Difficulty | Standard +0.3 This is a standard S2 continuous probability distribution question requiring routine integration to find k, then using E(Y) to find a. Part (a)(i) requires basic reasoning about where the pdf is positive, (a)(ii) and (b) involve straightforward integration with algebraic manipulation, and parts (c)-(d) are simple applications. While multi-part with 15 marks total, each step follows textbook methods without requiring novel insight or complex problem-solving. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (ai) \(f(y) \geq 0\) or \(f(3) \geq 0\) | M1 | (6) |
| \(ky(a - y) \geq 0\) or \(3k(a - 3) \geq 0\) or \((a - y) \geq 0\) or \((a - 3) \geq 0\) | A1 cso | |
| \(a \geq 3\) | ||
| (aii) \(\int_0^a k(ay - y^2)dy = 1\) | M1 | (6) |
| \[k\left[\frac{ay^2}{2} - \frac{y^3}{3}\right]_0^a = 1\] | A1 | |
| \[k\left(\frac{9a}{2} - 9\right) = 1\] | M1 | |
| \[k\left[\frac{9a - 18}{2}\right] = 1\] | ||
| \[k = \frac{2}{9(a - 2)}\] | M1 | |
| \[k = \frac{2}{9(a - 2)} \quad * \] | A1 cso | |
| (b) \(\int_0^a k(ay^2 - y^3)dy = 1.75\) | M1 | (6) |
| \[k\left[\frac{ay^3}{3} - \frac{y^4}{4}\right]_0^a = 1.75\] | A1 | |
| \[k\left(9a - \frac{81}{4}\right) = 1.75\] | ||
| \[2\left(9a - \frac{81}{4}\right) = 15.75(a - 2)\] | M1dep | |
| \[2.25a = -31.5 + \frac{81}{2}\] | ||
| \[a = 4\] | A1cso | |
| \[k = \frac{1}{9}\] | B1 | |
| (c) Sketch showing: correct shape. No straight lines. No need for patios. Completely correct graph. Needs to go through origin and the curve ends at 3. | B1; B1 | (2) |
| (d) mode = 2 | B1 | (1) |
**(ai)** $f(y) \geq 0$ or $f(3) \geq 0$ | M1 | (6) | M1 for putting $f(y) > 0$ or $f(3) > 0$ or $ky(a - y) \geq 0$ or $3k(a - 3) \geq 0$ or $(a - y) \geq 0$ or $(a - 3) \geq 0$ or state in words the probability can not be negative o.e.
$ky(a - y) \geq 0$ or $3k(a - 3) \geq 0$ or $(a - y) \geq 0$ or $(a - 3) \geq 0$ | A1 cso | | **A1** need one of $ky(a - y) \geq 0$ or $3k(a - 3) \geq 0$ or $(a - y) \geq 0$ or $(a - 3) \geq 0$ **and** $a \geq 3$
$a \geq 3$ | | |
**(aii)** $\int_0^a k(ay - y^2)dy = 1$ | M1 | (6) | M1 attempting to integrate (at least one $y^n \to y^{n+1}$) (ignore limits); **A1** Correct integration. Limits not needed. And equals 1 not needed.; M1 dependent on the previous M being awarded. Putting equal to 1 and have the correct limits. Limits do not need to be substituted.
$$k\left[\frac{ay^2}{2} - \frac{y^3}{3}\right]_0^a = 1$$ | A1 | | **A1 cso**
$$k\left(\frac{9a}{2} - 9\right) = 1$$ | M1 | |
$$k\left[\frac{9a - 18}{2}\right] = 1$$ | | |
$$k = \frac{2}{9(a - 2)}$$ | M1 | |
$$k = \frac{2}{9(a - 2)} \quad * $$ | A1 cso | |
**(b)** $\int_0^a k(ay^2 - y^3)dy = 1.75$ | M1 | (6) | M1 for attempting to find $\int yf(y)dy$ (at least one $y^n \to y^{n+1}$) (ignore limits); **A1** correct Integration; M1 $\int yf(y) = 1.75$ and limits 0,3 M1dep $\int yf(y) = 1.75$ and limits [0,3]
$$k\left[\frac{ay^3}{3} - \frac{y^4}{4}\right]_0^a = 1.75$$ | A1 | |
$$k\left(9a - \frac{81}{4}\right) = 1.75$$ | | |
$$2\left(9a - \frac{81}{4}\right) = 15.75(a - 2)$$ | M1dep | |
$$2.25a = -31.5 + \frac{81}{2}$$ | | |
$$a = 4$$ | A1cso | |
$$k = \frac{1}{9}$$ | B1 | |
**(c)** Sketch showing: correct shape. No straight lines. No need for patios. Completely correct graph. Needs to go through origin and the curve ends at 3. | B1; B1 | (2) | **Special case:** If draw full parabola from 0 to 4 get B1 B0 Allow full marks if the portion between $x = 3$ and $x = 4$ is dotted and the rest of the curve solid. [Graph showing parabola from origin, peaking around x=2, returning toward x-axis at x=3]
**(d)** mode = 2 | B1 | (1) | B1 cao 2
---The random variable $Y$ has probability density function f(y) given by
$$\text{f}(y) = \begin{cases}
ky(a - y) & 0 \leqslant y \leqslant 3 \\
0 & \text{otherwise}
\end{cases}$$
where $k$ and $a$ are positive constants.
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item Explain why $a \geqslant 3$
\item Show that $k = \frac{2}{9(a-2)}$
\end{enumerate}
[6]
\end{enumerate}
Given that E(Y) = 1.75
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item show that $a = 4$ and write down the value of $k$. [6]
\end{enumerate}
For these values of $a$ and $k$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item sketch the probability density function, [2]
\item write down the mode of $Y$. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2010 Q7 [15]}}