Question 6 - A-Level Maths

CAIE S2 2009 November — Question 6 8 marks

Exam Board	CAIE
Module	S2 (Statistics 2)
Year	2009
Session	November
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	PDF with multiple constants
Difficulty	Standard +0.3 This is a standard S2 PDF question requiring routine integration to find k, then E(X), and median comparison. All steps are textbook procedures with straightforward polynomial integration—slightly easier than average due to the mechanical nature of the calculations and clear structure.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03c Calculate mean/variance: by integration 5.03f Relate pdf-cdf: medians and percentiles

6 The continuous random variable $X$ has probability density function given by $$f ( x ) = \begin{cases} \frac { 1 } { 3 } x ( k - x ) & 1 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$

Show that the value of $k$ is $\frac { 32 } { 9 }$.
Find $\mathrm { E } ( X )$.
Is the median less than or greater than 1.5? Justify your answer numerically.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(i) $\int_1^x (k-x) dx = 1$ $\left[\frac{kx^3}{6} - \frac{x^3}{9}\right]_1^2 = 1$ $\left[\frac{4k}{6} - \frac{8}{9}\right] - \left[\frac{k}{6} - \frac{1}{9}\right] = 1$ $k = 32/9$ AG	M1, A1, A1 [3]	Equating to 1 and attempting to integrate, ignore limits here. Correct integration and correct limits. 32/9 correctly obtained.
(ii) $E(X) = \int_1^2 (k-x) dx = \left[\frac{kx^3}{9} - \frac{x^3}{12}\right]_1^2 = 1.52$ (491/324)	M1, A1 [2]	attempting to evaluate $\int_1^2 (k-x) dx$. correct answer.
(iii) $\int_1^{1.5} (k-x) dx = \left[\frac{kx}{6} - \frac{x^2}{9}\right]_1 = 0.477$ (or 0.523) this is $< 0.5$ so 1.5 is $<$ median	M1, A1 ft [3]	attempt to integrate using 1 and 1.5 or 1.5 and 2. correct reasoning and conclusion, ft their integration
Alternative method	M1, A1, A1	$\int f(x) dx$ limits 1 to $m$ (or $m$ to 2) and equate to 0.5 attempted. Correct cubic in $m$ in simplified form (no $k$'s). $m = 1.52(2)$, so median $> 1.5$, or, 2 trials showing that median $> 1.5$.

(i) $\int_1^x (k-x) dx = 1$ $\left[\frac{kx^3}{6} - \frac{x^3}{9}\right]_1^2 = 1$ $\left[\frac{4k}{6} - \frac{8}{9}\right] - \left[\frac{k}{6} - \frac{1}{9}\right] = 1$ $k = 32/9$ AG | M1, A1, A1 [3] | Equating to 1 and attempting to integrate, ignore limits here. Correct integration and correct limits. 32/9 correctly obtained.

(ii) $E(X) = \int_1^2 (k-x) dx = \left[\frac{kx^3}{9} - \frac{x^3}{12}\right]_1^2 = 1.52$ (491/324) | M1, A1 [2] | attempting to evaluate $\int_1^2 (k-x) dx$. correct answer.

(iii) $\int_1^{1.5} (k-x) dx = \left[\frac{kx}{6} - \frac{x^2}{9}\right]_1 = 0.477$ (or 0.523) this is $< 0.5$ so 1.5 is $<$ median | M1, A1 ft [3] | attempt to integrate using 1 and 1.5 or 1.5 and 2. correct reasoning and conclusion, ft their integration

Alternative method | M1, A1, A1 | $\int f(x) dx$ limits 1 to $m$ (or $m$ to 2) and equate to 0.5 attempted. Correct cubic in $m$ in simplified form (no $k$'s). $m = 1.52(2)$, so median $> 1.5$, or, 2 trials showing that median $> 1.5$.

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6 The continuous random variable $X$ has probability density function given by

$$f ( x ) = \begin{cases} \frac { 1 } { 3 } x ( k - x ) & 1 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$

(i) Show that the value of $k$ is $\frac { 32 } { 9 }$.\\
(ii) Find $\mathrm { E } ( X )$.\\
(iii) Is the median less than or greater than 1.5? Justify your answer numerically.

\hfill \mbox{\textit{CAIE S2 2009 Q6 [8]}}

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Q1 4 Q2 6 Q3 7 Q4 7 Q5 8 Q6 8 Q7 10

Answer	Marks	Guidance
(i) \(\int_1^x (k-x) dx = 1\) \(\left[\frac{kx^3}{6} - \frac{x^3}{9}\right]_1^2 = 1\) \(\left[\frac{4k}{6} - \frac{8}{9}\right] - \left[\frac{k}{6} - \frac{1}{9}\right] = 1\) \(k = 32/9\) AG	M1, A1, A1 [3]	Equating to 1 and attempting to integrate, ignore limits here. Correct integration and correct limits. 32/9 correctly obtained.
(ii) \(E(X) = \int_1^2 (k-x) dx = \left[\frac{kx^3}{9} - \frac{x^3}{12}\right]_1^2 = 1.52\) (491/324)	M1, A1 [2]	attempting to evaluate \(\int_1^2 (k-x) dx\). correct answer.
(iii) \(\int_1^{1.5} (k-x) dx = \left[\frac{kx}{6} - \frac{x^2}{9}\right]_1 = 0.477\) (or 0.523) this is \(< 0.5\) so 1.5 is \(<\) median	M1, A1 ft [3]	attempt to integrate using 1 and 1.5 or 1.5 and 2. correct reasoning and conclusion, ft their integration
Alternative method	M1, A1, A1	\(\int f(x) dx\) limits 1 to \(m\) (or \(m\) to 2) and equate to 0.5 attempted. Correct cubic in \(m\) in simplified form (no \(k\)'s). \(m = 1.52(2)\), so median \(> 1.5\), or, 2 trials showing that median \(> 1.5\).