| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2018 |
| Session | March |
| Marks | 9 |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Verify algebraic PDF formula |
| Difficulty | Standard +0.3 This is a straightforward Further Statistics question requiring standard verification that ∫f(x)dx = 1 (simple polynomial integration) and solving P(X<a) = 15/16 by integrating and solving a cubic equation. While it involves multiple steps, each technique is routine for Further Maths students, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^2 \frac{1}{4}x(4 - x^2)dx = 1\) | M1 | Attempt to show this |
| A1 | Algebra needed for "show that" | "By calculator": M1A0 |
| \(f(x) \geq 0\) for \(0 \leq x \leq 2\) | B1 | Explicitly stated – not just diagram drawn |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^a \frac{1}{4}x(4 - x^2)dx = \frac{15}{16}\) | M1 | Integral, correct limits somewhere, and equate to \(\frac{15}{16}\) |
| A1 | Correct indefinite integral | |
| \(\left[\frac{x^2}{2} - \frac{x^4}{16}\right]_0^a = \frac{15}{16}\) | M1 | Reduce to quadratic |
| \(a^2 = 3\) or \(5\) | M1 | Solve for \(a^2\) |
| \(a \neq -\sqrt{3}\) or \(\pm\sqrt{5}\) (outside range) | A1 | Explicitly rejected |
| \(a = \sqrt{3}\) and no other solutions | A1 | Needn't obtain previous A1 |
## Part (i)
$\int_0^2 \frac{1}{4}x(4 - x^2)dx = 1$ | M1 | Attempt to show this
| A1 | Algebra needed for "show that" | "By calculator": M1A0
$f(x) \geq 0$ for $0 \leq x \leq 2$ | B1 | Explicitly stated – not just diagram drawn
## Part (ii)
$\int_0^a \frac{1}{4}x(4 - x^2)dx = \frac{15}{16}$ | M1 | Integral, correct limits somewhere, and equate to $\frac{15}{16}$
| A1 | Correct indefinite integral
$\left[\frac{x^2}{2} - \frac{x^4}{16}\right]_0^a = \frac{15}{16}$ | M1 | Reduce to quadratic
$a^2 = 3$ or $5$ | M1 | Solve for $a^2$ | BC
$a \neq -\sqrt{3}$ or $\pm\sqrt{5}$ (outside range) | A1 | Explicitly rejected | No reason needed
$a = \sqrt{3}$ and no other solutions | A1 | Needn't obtain previous A17 The function $\mathrm { f } ( x )$ is defined by
$$f ( x ) = \begin{cases} \frac { 1 } { 4 } x \left( 4 - x ^ { 2 } \right) & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$
(i) Show that $\mathrm { f } ( x )$ satisfies the conditions for a probability density function.\\
(ii) Find the value of $a$ such that $\mathrm { P } ( X < a ) = \frac { 15 } { 16 }$.
\hfill \mbox{\textit{OCR Further Statistics 2018 Q7 [9]}}