| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Piecewise PDF with k |
| Difficulty | Standard +0.3 This is a standard S3 piecewise PDF question requiring integration to find k using the total probability condition, then calculating E(X). The integrals involve routine techniques (sin x and linear functions) with straightforward limits. While it requires careful algebraic manipulation and multiple integration steps, it follows a completely standard template with no novel problem-solving required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^{\frac{1}{2}\pi} k\sin x\,dx + \int_{\frac{1}{2}\pi}^{\pi} k\!\left(2 - \frac{2x}{\pi}\right)dx = 1\) | M1 | |
| \(\left[-k\cos x\right]_0^{\frac{1}{2}\pi} + k\!\left[2x - \frac{x^2}{\pi}\right]_{\frac{1}{2}\pi}^{\pi} = 1\) | B1 | Both integrals correct, ignore limits. |
| M1 | Substitute limits and attempt to simplify | |
| \(= \frac{4}{4+\pi}\) AG | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k\int_0^{\frac{1}{2}\pi} x\sin x\,dx + k\int_{\frac{1}{2}\pi}^{\pi} x\!\left(2 - \frac{2x}{\pi}\right)dx\) | M1 | |
| \(k\left[-x\cos x + \sin x\right]_0^{\frac{1}{2}\pi} + k\!\left[x^2 - \frac{2x^3}{3\pi}\right]_{\frac{1}{2}\pi}^{\pi}\) | M1 | Correct method for both integrals. Allow 1 error for M1. |
| A1 | Both integrals correct, ignore limits. | |
| \(= 1.48\) (3sf) | A1 | |
| [4] |
## Question 6:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^{\frac{1}{2}\pi} k\sin x\,dx + \int_{\frac{1}{2}\pi}^{\pi} k\!\left(2 - \frac{2x}{\pi}\right)dx = 1$ | M1 | |
| $\left[-k\cos x\right]_0^{\frac{1}{2}\pi} + k\!\left[2x - \frac{x^2}{\pi}\right]_{\frac{1}{2}\pi}^{\pi} = 1$ | B1 | Both integrals correct, ignore limits. |
| | M1 | Substitute limits and attempt to simplify |
| $= \frac{4}{4+\pi}$ AG | A1 | |
| **[4]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k\int_0^{\frac{1}{2}\pi} x\sin x\,dx + k\int_{\frac{1}{2}\pi}^{\pi} x\!\left(2 - \frac{2x}{\pi}\right)dx$ | M1 | |
| $k\left[-x\cos x + \sin x\right]_0^{\frac{1}{2}\pi} + k\!\left[x^2 - \frac{2x^3}{3\pi}\right]_{\frac{1}{2}\pi}^{\pi}$ | M1 | Correct method for both integrals. Allow 1 error for M1. |
| | A1 | Both integrals correct, ignore limits. |
| $= 1.48$ (3sf) | A1 | |
| **[4]** | | |
---6 The continuous random variable $X$ has probability density function given by
$$\mathrm { f } ( x ) = \left\{ \begin{array} { c l }
k \sin x & 0 \leqslant x \leqslant \frac { 1 } { 2 } \pi , \\
k \left( 2 - \frac { 2 x } { \pi } \right) & \frac { 1 } { 2 } \pi \leqslant x \leqslant \pi , \\
0 & \text { otherwise, }
\end{array} \right.$$
where $k$ is a constant.\\
(i) Show that $k = \frac { 4 } { 4 + \pi }$.\\
(ii) Find $\mathrm { E } ( X )$, correct to 3 significant figures, showing all necessary working.
\hfill \mbox{\textit{OCR S3 2014 Q6 [8]}}