Question 6 - A-Level Maths

CAIE S2 2014 June — Question 6 10 marks

Exam Board	CAIE
Module	S2 (Statistics 2)
Year	2014
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	Standard +0.3 This is a standard continuous probability distribution question requiring routine integration techniques: finding k by integrating to 1, calculating a probability by integration, and finding E(T). All three parts follow textbook procedures with straightforward polynomial integration, making it slightly easier than average for A-level.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03c Calculate mean/variance: by integration

6 The time, $T$ hours, spent by people on a visit to a museum has probability density function $$\mathrm { f } ( t ) = \begin{cases} k t \left( 16 - t ^ { 2 } \right) & 0 \leqslant t \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$ where $k$ is a constant.

Show that $k = \frac { 1 } { 64 }$.
Calculate the probability that two randomly chosen people each spend less than 1 hour on a visit to the museum.
Find the mean time spent on a visit to the museum.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(i) $k\int_0^4 (16t - t^3)dt = 1$	M1	Int f(t) = 1 ignore limits
$k\left[8t^2 - \frac{t^4}{4}\right]_0^4 = 1$ o.e.	A1	correct integration with correct limits
$k(128 - 64) = 1$ o.e. $k \times 64 = 1$, $k = \frac{1}{64}$ AG	A1 [3]	must be convinced (AG)
(ii) $\frac{1}{64}\int_0^1 (16t - t^3)dt$	M1	Int f(t) between 0 and 1 (accept 0 and a value $< 1$, 1 and 4)
$= \frac{1}{64}\left[8t^2 - \frac{t^4}{4}\right]_0^1 = \frac{1}{64}\left[8 - \frac{1}{4}\right] = \frac{31}{256}$ or $0.121094$	A1	correct integration and correct limits (ignore "k")
$\left(\frac{31}{256}\right)^2 = 0.0147$ (3 s.f.) o.e.	A1 [3]	ft their "$\frac{31}{256}$"
(iii) $\frac{1}{64}\int_0^4 (16t^2 - t^4)dt$	M1	Int tf(t) ignore limits
$= \frac{1}{64}\left[\frac{16t^3}{3} - \frac{t^5}{5}\right]_0^4 = \frac{1}{64}\left[\frac{1024}{3} - \frac{1024}{5}\right] = \frac{32}{15}$ or $2.13$ (3 s.f.) o.e.	A1 [3]	correct integration and correct limits (ignore "k")

**(i)** $k\int_0^4 (16t - t^3)dt = 1$ | M1 | Int f(t) = 1 ignore limits

$k\left[8t^2 - \frac{t^4}{4}\right]_0^4 = 1$ o.e. | A1 | correct integration with correct limits

$k(128 - 64) = 1$ o.e. $k \times 64 = 1$, $k = \frac{1}{64}$ AG | A1 [3] | must be convinced (AG)

**(ii)** $\frac{1}{64}\int_0^1 (16t - t^3)dt$ | M1 | Int f(t) between 0 and 1 (accept 0 and a value $< 1$, 1 and 4)

$= \frac{1}{64}\left[8t^2 - \frac{t^4}{4}\right]_0^1 = \frac{1}{64}\left[8 - \frac{1}{4}\right] = \frac{31}{256}$ or $0.121094$ | A1 | correct integration and correct limits (ignore "k")

$\left(\frac{31}{256}\right)^2 = 0.0147$ (3 s.f.) o.e. | A1 [3] | ft their "$\frac{31}{256}$"

**(iii)** $\frac{1}{64}\int_0^4 (16t^2 - t^4)dt$ | M1 | Int tf(t) ignore limits

$= \frac{1}{64}\left[\frac{16t^3}{3} - \frac{t^5}{5}\right]_0^4 = \frac{1}{64}\left[\frac{1024}{3} - \frac{1024}{5}\right] = \frac{32}{15}$ or $2.13$ (3 s.f.) o.e. | A1 [3] | correct integration and correct limits (ignore "k")

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Show LaTeX source

6 The time, $T$ hours, spent by people on a visit to a museum has probability density function

$$\mathrm { f } ( t ) = \begin{cases} k t \left( 16 - t ^ { 2 } \right) & 0 \leqslant t \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$

where $k$ is a constant.\\
(i) Show that $k = \frac { 1 } { 64 }$.\\
(ii) Calculate the probability that two randomly chosen people each spend less than 1 hour on a visit to the museum.\\
(iii) Find the mean time spent on a visit to the museum.

\hfill \mbox{\textit{CAIE S2 2014 Q6 [10]}}

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Q1 3 Q2 5 Q3 5 Q4 8 Q5 9 Q6 10 Q7 10

Answer	Marks	Guidance
(i) \(k\int_0^4 (16t - t^3)dt = 1\)	M1	Int f(t) = 1 ignore limits
\(k\left[8t^2 - \frac{t^4}{4}\right]_0^4 = 1\) o.e.	A1	correct integration with correct limits
\(k(128 - 64) = 1\) o.e. \(k \times 64 = 1\), \(k = \frac{1}{64}\) AG	A1 [3]	must be convinced (AG)
(ii) \(\frac{1}{64}\int_0^1 (16t - t^3)dt\)	M1	Int f(t) between 0 and 1 (accept 0 and a value \(< 1\), 1 and 4)
\(= \frac{1}{64}\left[8t^2 - \frac{t^4}{4}\right]_0^1 = \frac{1}{64}\left[8 - \frac{1}{4}\right] = \frac{31}{256}\) or \(0.121094\)	A1	correct integration and correct limits (ignore "k")
\(\left(\frac{31}{256}\right)^2 = 0.0147\) (3 s.f.) o.e.	A1 [3]	ft their "\(\frac{31}{256}\)"
(iii) \(\frac{1}{64}\int_0^4 (16t^2 - t^4)dt\)	M1	Int tf(t) ignore limits
\(= \frac{1}{64}\left[\frac{16t^3}{3} - \frac{t^5}{5}\right]_0^4 = \frac{1}{64}\left[\frac{1024}{3} - \frac{1024}{5}\right] = \frac{32}{15}\) or \(2.13\) (3 s.f.) o.e.	A1 [3]	correct integration and correct limits (ignore "k")