Question 7 - A-Level Maths

CAIE S2 2008 November — Question 7 12 marks

Exam Board	CAIE
Module	S2 (Statistics 2)
Year	2008
Session	November
Marks	12
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 straightforward continuous probability distribution question requiring standard techniques: finding k by integration, calculating mean and median, and finding a probability. All steps are routine applications of formulas with no conceptual challenges beyond basic S2 content. The integration of √t is elementary, making this slightly easier than average.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03c Calculate mean/variance: by integration

7 The time in hours taken for clothes to dry can be modelled by the continuous random variable with probability density function given by $$f ( t ) = \begin{cases} k \sqrt { } t & 1 \leqslant t \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$ where $k$ is a constant.

Show that $k = \frac { 3 } { 14 }$.
Find the mean time taken for clothes to dry.
Find the median time taken for clothes to dry.
Find the probability that the time taken for clothes to dry is between the mean time and the median time.

Show mark scheme Show mark scheme source

(i) $\int_1^4 kt^{0.5} dt = 1$

$\left[\frac{2kt^{1.5}}{3}\right]_1^4 = 1$

$\frac{16k}{3} - \frac{2k}{3} = 1$

Answer	Marks	Guidance
$k = 3/14$ AG	M1	Equating to 1 and attempting to integrate (ignore limits)
A1	Correct integration ignore limits
A1 [3]	Correct answer legitimately obtained

(ii) mean time $= \int kt^{1.5} dt$

Answer	Marks	Guidance
$= \left[\frac{2}{5}kt^{2.5}\right]_1^4 = \left[\frac{64k}{5} - \frac{2k}{5}\right] = 2.66$ hours	M1	Attempting to evaluate $\int kt^{1.5} dt$ (ignore limits)
A1	Correct integration and correct limits
M1	Substituting correct limits in their integration (need not be correct)
AJ... [4]	Correct answer

(iii) $\int_1^m kt^{0.5} dt = 0.5$

$\frac{m^{1.5}}{7} - \frac{1}{7} = 0.5$

Answer	Marks	Guidance
$m = 2.73$ hours	M1	Attempt to evaluate $\int kt^{0.5} dt$ (accept k missing)
M1	Attempt to solve an equation in m, = 0.5
A1 [3]	Correct answer (aef)
(iv) $\int_{2.657}^{2.726} kt^{0.5} dt = \left[\frac{2.726^{1.5}}{7} - \frac{2.657^{1.5}}{7}\right] = 0.0243$	M1	Attempt to integrate using their mean and median as limits
A1 [2]	Correct answer accept between 0.0241 and 0.0257

(i) $\int_1^4 kt^{0.5} dt = 1$

$\left[\frac{2kt^{1.5}}{3}\right]_1^4 = 1$

$\frac{16k}{3} - \frac{2k}{3} = 1$

$k = 3/14$ AG | M1 | Equating to 1 and attempting to integrate (ignore limits)
| A1 | Correct integration ignore limits
| A1 [3] | Correct answer legitimately obtained

(ii) mean time $= \int kt^{1.5} dt$

$= \left[\frac{2}{5}kt^{2.5}\right]_1^4 = \left[\frac{64k}{5} - \frac{2k}{5}\right] = 2.66$ hours | M1 | Attempting to evaluate $\int kt^{1.5} dt$ (ignore limits)
| A1 | Correct integration and correct limits
| M1 | Substituting correct limits in their integration (need not be correct)
| AJ... [4] | Correct answer

(iii) $\int_1^m kt^{0.5} dt = 0.5$

$\frac{m^{1.5}}{7} - \frac{1}{7} = 0.5$

$m = 2.73$ hours | M1 | Attempt to evaluate $\int kt^{0.5} dt$ (accept k missing)
| M1 | Attempt to solve an equation in m, = 0.5
| A1 [3] | Correct answer (aef)

(iv) $\int_{2.657}^{2.726} kt^{0.5} dt = \left[\frac{2.726^{1.5}}{7} - \frac{2.657^{1.5}}{7}\right] = 0.0243$ | M1 | Attempt to integrate using their mean and median as limits
| A1 [2] | Correct answer accept between 0.0241 and 0.0257

Show LaTeX source

7 The time in hours taken for clothes to dry can be modelled by the continuous random variable with probability density function given by

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

where $k$ is a constant.\\
(i) Show that $k = \frac { 3 } { 14 }$.\\
(ii) Find the mean time taken for clothes to dry.\\
(iii) Find the median time taken for clothes to dry.\\
(iv) Find the probability that the time taken for clothes to dry is between the mean time and the median time.

\hfill \mbox{\textit{CAIE S2 2008 Q7 [12]}}

This paper (7 questions)

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Q1 4 Q2 5 Q3 5 Q4 7 Q5 8 Q6 9 Q7 12

CAIE S2 2008 November — Question 7 12 marks

(i) \(\int_1^4 kt^{0.5} dt = 1\)

\(\left[\frac{2kt^{1.5}}{3}\right]_1^4 = 1\)

\(\frac{16k}{3} - \frac{2k}{3} = 1\)

(ii) mean time \(= \int kt^{1.5} dt\)

(iii) \(\int_1^m kt^{0.5} dt = 0.5\)

\(\frac{m^{1.5}}{7} - \frac{1}{7} = 0.5\)

This paper (7 questions)

Answer	Marks	Guidance
\(k = 3/14\) AG	M1	Equating to 1 and attempting to integrate (ignore limits)
A1	Correct integration ignore limits
A1 [3]	Correct answer legitimately obtained

Answer	Marks	Guidance
\(= \left[\frac{2}{5}kt^{2.5}\right]_1^4 = \left[\frac{64k}{5} - \frac{2k}{5}\right] = 2.66\) hours	M1	Attempting to evaluate \(\int kt^{1.5} dt\) (ignore limits)
A1	Correct integration and correct limits
M1	Substituting correct limits in their integration (need not be correct)
AJ... [4]	Correct answer

Answer	Marks	Guidance
\(m = 2.73\) hours	M1	Attempt to evaluate \(\int kt^{0.5} dt\) (accept k missing)
M1	Attempt to solve an equation in m, = 0.5
A1 [3]	Correct answer (aef)
(iv) \(\int_{2.657}^{2.726} kt^{0.5} dt = \left[\frac{2.726^{1.5}}{7} - \frac{2.657^{1.5}}{7}\right] = 0.0243\)	M1	Attempt to integrate using their mean and median as limits
A1 [2]	Correct answer accept between 0.0241 and 0.0257