| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Symmetry property of PDF |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard PDF properties: integration to find k (routine calculus), recognizing symmetry for the mean, variance calculation using standard formulas, and a probability calculation. Part (b) rewards spotting the symmetry of the distribution, but even without this insight, students can calculate E(X) directly. All techniques are standard textbook exercises with no novel problem-solving required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int_0^k \dfrac{3}{32}x(k-x)\,dx = 1\) | M1 | M1 for attempting to multiply out bracket and integrate \(f(x)\); both \(x^n \to x^{n+1}\) |
| \(\dfrac{3}{32}\left[\dfrac{kx^2}{2} - \dfrac{x^3}{3}\right]_0^k = 1\) | A1 | Correct integration; need \(\dfrac{3}{32}\!\left(\dfrac{kx^2}{2} - \dfrac{x^3}{3}\right)\) oe |
| \(\dfrac{3k^3}{64} - \dfrac{3k^3}{96} = 1\) | M1 dep | Dependent on previous M; correct use of limits set equal to 1 |
| \(3k^3 - 2k^3 = 64\), \(k^3 = 64\), \(k = 4\) | A1 cso | cso; or for verifying \(\dfrac{3}{32}\!\left(\dfrac{4^3}{2}-\dfrac{4^3}{3}\right)=1\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = 2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X^2) = \int_0^4 \dfrac{3}{32}x^3(4-x)\,dx\) | M1 | Attempt to multiply out bracket and integrate \(\int x^2 f(x)\); both \(x^n \to x^{n+1}\) |
| \(= \left[\dfrac{3x^4}{32} - \dfrac{3x^5}{160}\right]_0^4 = \left[\dfrac{3\times 4^4}{32} - \dfrac{3\times 4^5}{160}\right] = 4.8\) | A1 | |
| \(\text{Var}(X) = 4.8 - 4\) | M1 | 2nd M1 for their \(E(X^2) - (\text{their mean})^2\) |
| \(= 0.8\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int_{1.5}^{2.5} \dfrac{3}{32}x(4-x)\,dx = \left[\dfrac{3x^2}{16} - \dfrac{x^3}{32}\right]_{1.5}^{2.5}\) or \(\int_0^{1.5}\dfrac{3}{32}x(4-x)\,dx = \left[\dfrac{3x^2}{16}-\dfrac{x^3}{32}\right]_0^{1.5}\) | M1 | M1 multiply out brackets, attempt to integrate, with either limits (their \(b \pm 0.5\)) or (their \(b\) and 0) |
| \(= \dfrac{47}{128} = 0.3671875\) or \(= \dfrac{81}{256} = 0.31640625\) | ||
| \(1 - \dfrac{47}{128} = \dfrac{81}{128}\) awrt 0.633 or \(2 \times \dfrac{81}{256} = \dfrac{81}{128}\) awrt 0.633 | M1depA1 | 2nd M1dep on 1st M1: \(1-(\) using limits their \(b \pm 0.5)\) or \(2\times\) (using limits their \(b\) and 0) |
# Question 5:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_0^k \dfrac{3}{32}x(k-x)\,dx = 1$ | M1 | M1 for attempting to multiply out bracket and integrate $f(x)$; both $x^n \to x^{n+1}$ |
| $\dfrac{3}{32}\left[\dfrac{kx^2}{2} - \dfrac{x^3}{3}\right]_0^k = 1$ | A1 | Correct integration; need $\dfrac{3}{32}\!\left(\dfrac{kx^2}{2} - \dfrac{x^3}{3}\right)$ oe |
| $\dfrac{3k^3}{64} - \dfrac{3k^3}{96} = 1$ | M1 dep | Dependent on previous M; correct use of limits set equal to 1 |
| $3k^3 - 2k^3 = 64$, $k^3 = 64$, $k = 4$ | A1 cso | cso; or for verifying $\dfrac{3}{32}\!\left(\dfrac{4^3}{2}-\dfrac{4^3}{3}\right)=1$ oe |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = 2$ | B1 | |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X^2) = \int_0^4 \dfrac{3}{32}x^3(4-x)\,dx$ | M1 | Attempt to multiply out bracket and integrate $\int x^2 f(x)$; both $x^n \to x^{n+1}$ |
| $= \left[\dfrac{3x^4}{32} - \dfrac{3x^5}{160}\right]_0^4 = \left[\dfrac{3\times 4^4}{32} - \dfrac{3\times 4^5}{160}\right] = 4.8$ | A1 | |
| $\text{Var}(X) = 4.8 - 4$ | M1 | 2nd M1 for their $E(X^2) - (\text{their mean})^2$ |
| $= 0.8$ | A1 | |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_{1.5}^{2.5} \dfrac{3}{32}x(4-x)\,dx = \left[\dfrac{3x^2}{16} - \dfrac{x^3}{32}\right]_{1.5}^{2.5}$ or $\int_0^{1.5}\dfrac{3}{32}x(4-x)\,dx = \left[\dfrac{3x^2}{16}-\dfrac{x^3}{32}\right]_0^{1.5}$ | M1 | M1 multiply out brackets, attempt to integrate, with either limits (their $b \pm 0.5$) or (their $b$ and 0) |
| $= \dfrac{47}{128} = 0.3671875$ or $= \dfrac{81}{256} = 0.31640625$ | | |
| $1 - \dfrac{47}{128} = \dfrac{81}{128}$ awrt 0.633 or $2 \times \dfrac{81}{256} = \dfrac{81}{128}$ awrt 0.633 | M1depA1 | 2nd M1dep on 1st M1: $1-($ using limits their $b \pm 0.5)$ or $2\times$ (using limits their $b$ and 0) |
---\begin{enumerate}
\item The queueing time, $X$ minutes, of a customer at a till of a supermarket has probability density function
\end{enumerate}
$$f ( x ) = \left\{ \begin{array} { c c }
\frac { 3 } { 32 } x ( k - x ) & 0 \leqslant x \leqslant k \\
0 & \text { otherwise }
\end{array} \right.$$
(a) Show that the value of $k$ is 4\\
(b) Write down the value of $\mathrm { E } ( X )$.\\
(c) Calculate $\operatorname { Var } ( X )$.\\
(d) Find the probability that a randomly chosen customer's queueing time will differ from the mean by at least half a minute.\\
\hfill \mbox{\textit{Edexcel S2 2012 Q5 [12]}}