| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| 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 routine integration techniques: finding k by integrating to 1, computing E(X), deriving the CDF, and solving for a quartile. All steps follow textbook methods 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^2 kx\,dx + \int_2^4 \frac{k}{2}(4-x)^2\,dx = 1\) | M1 | |
| \(2k + \frac{8k}{6}\ (= 1)\) | B1 | Ignore limits for this mark. |
| \(k = \frac{3}{10}\) AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^2 kx^2\,dx + \int_2^4 \frac{kx}{2}(4-x)^2\,dx\) | M1 | |
| \(\frac{1}{10}x^3,\ \frac{3}{20}(8x^2 - \frac{8x^3}{3} + \frac{x^4}{4})\) | B1,B1 | \(\frac{-x(4-x)^3}{3} - \frac{(4-x)^4}{12}\) or \(\frac{x^2(4-x)^2}{2} + \frac{4x^3}{3} - \frac{x^4}{4}\bigg |
| 1.8 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3x^2}{20} \quad (0 \leq x < 2)\) | B1 | |
| \(\text{"0.6"} + \frac{k}{2}\int_2^X (4-x)^2\,dx\) oe | M1 | Or \(-\frac{(4-x)^3}{20} + c\) and attempt to find c. |
| \(1 - \frac{(4-x)^3}{20}\) oe, \((2 \leq x \leq 4)\) | A1 | \(0.05x^3 - 0.6x^2 + 2.4x - 2.2\) |
| \(0\ \ x<0,\quad 1\ \ x>4\) | B1 | 0, 1 and all ranges. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - \dfrac{(4-x)^3}{20} = 0.75\) | M1* | |
| \((4-x)^3 = 5\) | \*M1 | Valid attempt to solve. Must produce soln. |
| \(2.29\) | A1 | |
| [3] |
## Question 7:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^2 kx\,dx + \int_2^4 \frac{k}{2}(4-x)^2\,dx = 1$ | M1 | |
| $2k + \frac{8k}{6}\ (= 1)$ | B1 | Ignore limits for this mark. |
| $k = \frac{3}{10}$ AG | A1 | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^2 kx^2\,dx + \int_2^4 \frac{kx}{2}(4-x)^2\,dx$ | M1 | |
| $\frac{1}{10}x^3,\ \frac{3}{20}(8x^2 - \frac{8x^3}{3} + \frac{x^4}{4})$ | B1,B1 | $\frac{-x(4-x)^3}{3} - \frac{(4-x)^4}{12}$ or $\frac{x^2(4-x)^2}{2} + \frac{4x^3}{3} - \frac{x^4}{4}\bigg|_{4}^{4}$ from int by parts. Ignore limits for these marks. |
| 1.8 | A1 | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3x^2}{20} \quad (0 \leq x < 2)$ | B1 | |
| $\text{"0.6"} + \frac{k}{2}\int_2^X (4-x)^2\,dx$ oe | M1 | Or $-\frac{(4-x)^3}{20} + c$ and attempt to find c. |
| $1 - \frac{(4-x)^3}{20}$ oe, $(2 \leq x \leq 4)$ | A1 | $0.05x^3 - 0.6x^2 + 2.4x - 2.2$ |
| $0\ \ x<0,\quad 1\ \ x>4$ | B1 | 0, 1 and all ranges. |
## Question (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - \dfrac{(4-x)^3}{20} = 0.75$ | M1* | |
| $(4-x)^3 = 5$ | \*M1 | Valid attempt to solve. Must produce soln. | $Q^3 - 12Q^2 + 48Q - 59 = 0$ |
| $2.29$ | A1 | |
| | **[3]** | |7 A continuous random variable $X$ has probability density function
$$f ( x ) = \left\{ \begin{array} { c c }
k x & 0 \leqslant x < 2 \\
\frac { k ( 4 - x ) ^ { 2 } } { 2 } & 2 \leqslant x \leqslant 4 \\
0 & \text { otherwise }
\end{array} \right.$$
where $k$ is a constant.\\
(i) Show that $k = \frac { 3 } { 10 }$.\\
(ii) Find $\mathrm { E } ( X )$.\\
(iii) Find the cumulative distribution function of $X$.\\
(iv) Find the upper quartile of $X$, correct to 3 significant figures.
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR S3 2015 Q7 [14]}}