CAIE S2 2003 November — Question 7 11 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2003
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Distribution
TypeFind constant in PDF
DifficultyModerate -0.8 This is a straightforward exponential distribution question requiring only standard techniques: integrating the PDF to find k (routine normalization), using the CDF for the quartile (direct substitution), and recalling/computing the mean formula. All three parts are textbook exercises with no problem-solving or novel insight required, making it easier than average.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles
7 The lifetime, \(x\) years, of the power light on a freezer, which is left on continuously, can be modelled by the continuous random variable with density function given by $$\mathrm { f } ( x ) = \begin{cases} k \mathrm { e } ^ { - 3 x } & x > 0 \\ 0 & \text { otherwise } \end{cases}$$ where \(k\) is a constant.
  1. Show that \(k = 3\).
  2. Find the lower quartile.
  3. Find the mean lifetime.
(i) \(\int_0^{\infty} ke^{-3x} dx = 1\)
AnswerMarks Guidance
\(0 - \frac{-k}{3} = 1 \Rightarrow k = 3\)M1, A1 For attempting to integrate from 0 to \(\infty\) and putting the integral = 1; For obtaining given answer correctly
[2]
(ii) \(\int_0^{q1} 3e^{-3x} dx = 0.25\)
\(\left[-e^{-3x}\right]_0^{q1} = 0.25\)
\(-e^{-3q1} + 1 = 0.25\)
\(0.75 = e^{-3q1}\)
AnswerMarks Guidance
\(q1 = 0.0959\)M1, M1, M1, A1 For equating \(\int 3e^{-3x} dx\) to 0.25 (no limits needed); For attempting to integrate and substituting (sensible) limits and rearranging; For correct answer
[3]
(iii) Mean \(= \int_0^{\infty} 3xe^{-3x} dx\)
AnswerMarks Guidance
\(= \left[xe^{-3x}\right]_0^{\infty} - \int_0^{\infty} -e^{-3x} dx = 0 + \left[-\frac{e^{-3x}}{3}\right]_0^{\infty} = 0.333\) or \(\frac{1}{3}\)B1, M1, A1, M1, A1, A1 For correct statement for mean; For attempting to integrate \(3xe^{-3x}\) (no limits needed); For \(-xe^{-3x}\) or \(-xe^{-3x}/3\); For attempt \(\int -e^{-3x} dx\) (their integral); For \(0 + \left[-\frac{e^{-3x}}{3}\right]_0^{\infty}\); For correct answer
[6] 
**(i)** $\int_0^{\infty} ke^{-3x} dx = 1$

$0 - \frac{-k}{3} = 1 \Rightarrow k = 3$ | M1, A1 | For attempting to integrate from 0 to $\infty$ and putting the integral = 1; For obtaining given answer correctly

| [2]

**(ii)** $\int_0^{q1} 3e^{-3x} dx = 0.25$

$\left[-e^{-3x}\right]_0^{q1} = 0.25$

$-e^{-3q1} + 1 = 0.25$

$0.75 = e^{-3q1}$

$q1 = 0.0959$ | M1, M1, M1, A1 | For equating $\int 3e^{-3x} dx$ to 0.25 (no limits needed); For attempting to integrate and substituting (sensible) limits and rearranging; For correct answer

| [3]

**(iii)** Mean $= \int_0^{\infty} 3xe^{-3x} dx$

$= \left[xe^{-3x}\right]_0^{\infty} - \int_0^{\infty} -e^{-3x} dx = 0 + \left[-\frac{e^{-3x}}{3}\right]_0^{\infty} = 0.333$ or $\frac{1}{3}$ | B1, M1, A1, M1, A1, A1 | For correct statement for mean; For attempting to integrate $3xe^{-3x}$ (no limits needed); For $-xe^{-3x}$ or $-xe^{-3x}/3$; For attempt $\int -e^{-3x} dx$ (their integral); For $0 + \left[-\frac{e^{-3x}}{3}\right]_0^{\infty}$; For correct answer

| [6]
7 The lifetime, $x$ years, of the power light on a freezer, which is left on continuously, can be modelled by the continuous random variable with density function given by

$$\mathrm { f } ( x ) = \begin{cases} k \mathrm { e } ^ { - 3 x } & x > 0 \\ 0 & \text { otherwise } \end{cases}$$

where $k$ is a constant.\\
(i) Show that $k = 3$.\\
(ii) Find the lower quartile.\\
(iii) Find the mean lifetime.

\hfill \mbox{\textit{CAIE S2 2003 Q7 [11]}}
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