Question 5 - A-Level Maths

CAIE S2 2006 June — Question 5 10 marks

Exam Board	CAIE
Module	S2 (Statistics 2)
Year	2006
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	Single-piece PDF with k
Difficulty	Moderate -0.3 This is a standard continuous probability distribution question requiring routine application of well-established formulas (integration for normalization, mean, variance, and quartiles). While it involves multiple parts and some algebraic manipulation, each step follows directly from textbook methods with no novel insight required. The integration is straightforward (polynomial powers), making it slightly easier than average for A-level statistics questions.
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

5 The random variable $X$ has probability density function given by $$f ( x ) = \begin{cases} 4 x ^ { k } & 0 \leqslant x \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$ where $k$ is a positive constant.

Show that $k = 3$.
Show that the mean of $X$ is 0.8 and find the variance of $X$.
Find the upper quartile of $X$.
Find the interquartile range of $X$.

Show mark scheme Show mark scheme source

Part (i):

$\int_0 4x^3 dx = 1$

$\left[\frac{4x^{k+1}}{k+1}\right]_0 = 1$

Answer	Marks	Guidance
$k = 3$ AG	M1, A1 (2 marks)	Equating to $1$ and attempting to integrate. Correct answer legitimately obtained

Part (ii):

Answer	Marks	Guidance
$E(X) = \int_0^1 4x^4 dx = \left[\frac{4x^5}{5}\right]_0 = 0.8$ AG	M1, A1	Attempting to evaluate $\int_0^1 4x^4 dx$ with or without limits. Correct answer legitimately obtained
Var$(X) = \int_0^1 4x^6 dx - 0.8^2 = \left[\frac{4x^6}{6}\right]_0 - 0.8^2 = 0.0267$	M1, A1 (4 marks)	Attempt at integral of $x^2f(x) - 0.8^2$. Correct answer (accept $.027$)

Part (iii):

$\int_0^{q_3} 4x^3 dx = 0.75$

Answer	Marks	Guidance
$q_3 = \sqrt[4]{0.75} (= 0.931) = 0.931$	M1, A1ft (2 marks)	Finding UQ by solving integral $= 0.75$. Correct UQ

Part (iv):

$q_1 = \sqrt[4]{0.25} (= 0.707)$

Answer	Marks	Guidance
IQ range $= 0.223$ $(0.22349)$	B1, B1ft (2 marks)	Correct LQ. Accept $0.223$ or $0.224$

**Part (i):**
$\int_0 4x^3 dx = 1$

$\left[\frac{4x^{k+1}}{k+1}\right]_0 = 1$

$k = 3$ AG | M1, A1 (2 marks) | Equating to $1$ and attempting to integrate. Correct answer legitimately obtained

**Part (ii):**
$E(X) = \int_0^1 4x^4 dx = \left[\frac{4x^5}{5}\right]_0 = 0.8$ AG | M1, A1 | Attempting to evaluate $\int_0^1 4x^4 dx$ with or without limits. Correct answer legitimately obtained

Var$(X) = \int_0^1 4x^6 dx - 0.8^2 = \left[\frac{4x^6}{6}\right]_0 - 0.8^2 = 0.0267$ | M1, A1 (4 marks) | Attempt at integral of $x^2f(x) - 0.8^2$. Correct answer (accept $.027$)

**Part (iii):**
$\int_0^{q_3} 4x^3 dx = 0.75$

$q_3 = \sqrt[4]{0.75} (= 0.931) = 0.931$ | M1, A1ft (2 marks) | Finding UQ by solving integral $= 0.75$. Correct UQ

**Part (iv):**
$q_1 = \sqrt[4]{0.25} (= 0.707)$

IQ range $= 0.223$ $(0.22349)$ | B1, B1ft (2 marks) | Correct LQ. Accept $0.223$ or $0.224$

Show LaTeX source

5 The random variable $X$ has probability density function given by

$$f ( x ) = \begin{cases} 4 x ^ { k } & 0 \leqslant x \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$

where $k$ is a positive constant.\\
(i) Show that $k = 3$.\\
(ii) Show that the mean of $X$ is 0.8 and find the variance of $X$.\\
(iii) Find the upper quartile of $X$.\\
(iv) Find the interquartile range of $X$.

\hfill \mbox{\textit{CAIE S2 2006 Q5 [10]}}

This paper (7 questions)

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Q1 3 Q2 4 Q3 6 Q4 6 Q5 10 Q6 10 Q7 11

CAIE S2 2006 June — Question 5 10 marks

Part (i):

\(\int_0 4x^3 dx = 1\)

\(\left[\frac{4x^{k+1}}{k+1}\right]_0 = 1\)

Part (ii):

Part (iii):

\(\int_0^{q_3} 4x^3 dx = 0.75\)

Part (iv):

\(q_1 = \sqrt[4]{0.25} (= 0.707)\)

This paper (7 questions)

Answer	Marks	Guidance
\(E(X) = \int_0^1 4x^4 dx = \left[\frac{4x^5}{5}\right]_0 = 0.8\) AG	M1, A1	Attempting to evaluate \(\int_0^1 4x^4 dx\) with or without limits. Correct answer legitimately obtained
Var\((X) = \int_0^1 4x^6 dx - 0.8^2 = \left[\frac{4x^6}{6}\right]_0 - 0.8^2 = 0.0267\)	M1, A1 (4 marks)	Attempt at integral of \(x^2f(x) - 0.8^2\). Correct answer (accept \(.027\))