| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Direct variance calculation from pdf |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring a sketch of a quadratic pdf, standard variance calculation using E(X²) - [E(X)]², and interpretation of the pdf shape. All techniques are routine for this module with no novel problem-solving 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 integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| [Positive parabola sketch] | M1, A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{3}{16}\int_0^4 x^2(x-2)^2\,dx\) | M1 | Attempt \(\int x^2 f(x)\,dx\), limits 0 and 4 |
| \(= \frac{3}{16}\left[\frac{x^5}{5} - x^4 + 4\frac{x^3}{3}\right]_0^4 \quad \left[= 6\tfrac{2}{5}\right]\) | M1, B1 | Method for integration e.g. multiply out [*indept*]; correct indefinite integral \(\frac{3}{16}\left[\frac{x^2(x-2)^3}{3} - \frac{x(x-2)^4}{6} + \frac{(x-2)^5}{30}\right]\) |
| \(\sigma^2 = 6\tfrac{2}{3} - 2^2 = 2\tfrac{2}{3}\) | B1, A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| No because \(x\) represents a value taken by the random variable [*not an event that "occurs"*] | B1 | 1 |
# Question 4:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| [Positive parabola sketch] | M1, A1 | **2** | Positive parabola, all above axis. Correct place, touches $x$-axis, not beyond limits, symmetric ends, not too straight |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{3}{16}\int_0^4 x^2(x-2)^2\,dx$ | M1 | Attempt $\int x^2 f(x)\,dx$, limits 0 and 4 |
| $= \frac{3}{16}\left[\frac{x^5}{5} - x^4 + 4\frac{x^3}{3}\right]_0^4 \quad \left[= 6\tfrac{2}{5}\right]$ | M1, B1 | Method for integration e.g. multiply out [*indept*]; correct indefinite integral $\frac{3}{16}\left[\frac{x^2(x-2)^3}{3} - \frac{x(x-2)^4}{6} + \frac{(x-2)^5}{30}\right]$ |
| $\sigma^2 = 6\tfrac{2}{3} - 2^2 = 2\tfrac{2}{3}$ | B1, A1 | **5** | Subtract $2^2$; final answer 2.4, any equivalent exact form, cwo |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| No because $x$ represents a value taken by the random variable [*not an event that "occurs"*] | B1 | **1** | Show clear understanding that $x$ is a value of $X$. SR: Allow B1 for answer clearly indicating probabilities higher where curve higher, *or* clearly stating all probabilities are effectively zero. "Agree as area under graph increases" or "minimum at 2": B1. "True only between 0 and 4": B0 unless explanation |
---4 A continuous random variable $X$ has probability density function
$$f ( x ) = \begin{cases} \frac { 3 } { 16 } ( x - 2 ) ^ { 2 } & 0 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
(i) Sketch the graph of $y = \mathrm { f } ( x )$.\\
(ii) Calculate the variance of $X$.\\
(iii) A student writes " $X$ is more likely to occur when $x$ takes values further away from 2 ". Explain whether you agree with this statement.
\hfill \mbox{\textit{OCR S2 2011 Q4 [8]}}