| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | January |
| Marks | 13 |
| 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 probability density function question requiring routine integration to find constraints on parameters, then solving simultaneous equations. Part (iii) adds mild problem-solving by comparing median to mean, but all techniques are textbook exercises with no novel insight required. Slightly easier than average due to straightforward algebraic manipulation. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\left[a x + \frac{b x^2}{2}\right]_0^1 = 1\) | M1 | Use total area = 1 |
| B1 | Correct indefinite integral, or convincing area method | |
| \(2a + 2b = 1\) | A1 | 3 |
| (ii) \(\frac{ax^2}{2} + \frac{bx^3}{3}\Bigg | _0^2 = \frac{11}{9}\) | M1 |
| B1 | Correct indefinite integral | |
| \(2a + \frac{8b}{3} = \frac{11}{9}\) | A1 | Correct equation obtained, a.e.f. |
| Solve simultaneously | M1 | Obtain one unknown by correct simultaneous method |
| \(a = \frac{1}{4}, b = \frac{1}{4}\) | A1 | \(a\) correct, 1/6 or a.r.t. 0.167 |
| A1 | 6 | \(b\) correct, 1/3 or a.r.t. 0.333 |
| (iii) e.g. \(P(x < 11/9) = 0.453\), or | M1 | Use \(P(x < 11/9)\), or integrate to find median \(m\) |
| \(\left[a x + \frac{b x^2}{2}\right]_0 = 0.5, m = 1.303 \text{ or } \frac{\sqrt{13}-1}{2}\) | M1 | Substitute into \(\int f(x)dx\), √ on \(a, b\), limits 0 and 11/9 or \(m\) [if finding \(m\), need to solve 3-term quadratic] |
| A1 | Correct numerical answer for probability or \(m\) | |
| Hence median > mean | A1V | 4 |
(i) $\left[a x + \frac{b x^2}{2}\right]_0^1 = 1$ | M1 | Use total area = 1
| B1 | Correct indefinite integral, or convincing area method
$2a + 2b = 1$ | A1 | 3 | Given answer obtained, "1" appearing before last line [if +c, must see it eliminated]
(ii) $\frac{ax^2}{2} + \frac{bx^3}{3}\Bigg|_0^2 = \frac{11}{9}$ | M1 | Use $\int f(x)dx = 11/9$, limits 0, 2
| | | B1 | Correct indefinite integral
$2a + \frac{8b}{3} = \frac{11}{9}$ | A1 | Correct equation obtained, a.e.f.
Solve simultaneously | M1 | Obtain one unknown by correct simultaneous method
$a = \frac{1}{4}, b = \frac{1}{4}$ | A1 | $a$ correct, 1/6 or a.r.t. 0.167
| | | A1 | 6 | $b$ correct, 1/3 or a.r.t. 0.333
(iii) e.g. $P(x < 11/9) = 0.453$, or | M1 | Use $P(x < 11/9)$, or integrate to find median $m$
$\left[a x + \frac{b x^2}{2}\right]_0 = 0.5, m = 1.303 \text{ or } \frac{\sqrt{13}-1}{2}$ | M1 | Substitute into $\int f(x)dx$, √ on $a, b$, limits 0 and 11/9 or $m$ [if finding $m$, need to solve 3-term quadratic]
| | | A1 | Correct numerical answer for probability or $m$
Hence median > mean | A1V | 4 | Correct conclusion, two [Negative answer = M2; median > mean, A2]6 The continuous random variable $X$ has the following probability density function:
$$f ( x ) = \begin{cases} a + b x & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
where $a$ and $b$ are constants.\\
(i) Show that $2 a + 2 b = 1$.\\
(ii) It is given that $\mathrm { E } ( X ) = \frac { 11 } { 9 }$. Use this information to find a second equation connecting $a$ and $b$, and hence find the values of $a$ and $b$.\\
(iii) Determine whether the median of $X$ is greater than, less than, or equal to $\mathrm { E } ( X )$.
\hfill \mbox{\textit{OCR S2 2007 Q6 [13]}}