OCR S3 2007 June — Question 1 4 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2007
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeSingle-piece PDF with k
DifficultyModerate -0.8 This is a straightforward application of the fundamental property that a PDF must integrate to 1. Students need to split the integral at x=1, evaluate two standard integrals (constant and 1/x²), and solve a simple equation for a. It's routine calculus with no conceptual challenges or problem-solving required.
Spec5.03a Continuous random variables: pdf and cdf

1 The continuous random variable \(X\) has probability density function given by $$\mathrm { f } ( x ) = \begin{cases} a & 0 \leqslant x \leqslant 1 , \\ \frac { a } { x ^ { 2 } } & x > 1 , \\ 0 & \text { otherwise. } \end{cases}$$ Find the value of the constant \(a\).

AnswerMarks Guidance
\(\int_0^a adx + \int_1^a \frac{a}{x^2}dx = 1\)M1 For sum of integrals = 1
\([ax]_0^a + \left[-\frac{a}{x^3}\right]_1^a = 1\)A1 For second integral
\(a + a = 1\) and \(a = \frac{1}{2}\)A1 For second \(a\)
A1Or from \(F(x) = 1\), M1A1 then \(a = \frac{1}{2}\), A1
4 marks total
$\int_0^a adx + \int_1^a \frac{a}{x^2}dx = 1$ | M1 | For sum of integrals = 1

$[ax]_0^a + \left[-\frac{a}{x^3}\right]_1^a = 1$ | A1 | For second integral

$a + a = 1$ and $a = \frac{1}{2}$ | A1 | For second $a$

| A1 | Or from $F(x) = 1$, M1A1 then $a = \frac{1}{2}$, A1

| | 4 marks total |
1 The continuous random variable $X$ has probability density function given by

$$\mathrm { f } ( x ) = \begin{cases} a & 0 \leqslant x \leqslant 1 , \\ \frac { a } { x ^ { 2 } } & x > 1 , \\ 0 & \text { otherwise. } \end{cases}$$

Find the value of the constant $a$.

\hfill \mbox{\textit{OCR S3 2007 Q1 [4]}}