CAIE FP2 2019 November — Question 7 7 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2019
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Distribution
TypeFind threshold for given probability
DifficultyStandard +0.3 This is a straightforward application of the exponential distribution with standard bookwork (mean = 1/a) and routine probability calculations. Part (iii) requires setting up an inequality with complements and logarithms, but follows a predictable pattern for 'at least one' problems. Slightly easier than average due to the direct application of formulas with minimal problem-solving insight required.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf

The time, \(T\) days, before an electrical component develops a fault has distribution function F given by $$\mathrm{F}(t) = \begin{cases} 1 - e^{-at} & t \geqslant 0, \\ 0 & \text{otherwise}, \end{cases}$$ where \(a\) is a positive constant. The mean value of \(T\) is 200.
  1. Write down the value of \(a\). [1]
  2. Find the probability that an electrical component of this type develops a fault in less than 150 days. [2]
A piece of equipment contains \(n\) of these components, which develop faults independently of each other. The probability that, after 150 days, at least one of the \(n\) components has not developed a fault is greater than 0.99.
  1. Find the smallest possible value of \(n\). [4]

Question 7:

AnswerMarks Guidance
7(i)a = 1/200 or 0⋅005 B1
1

AnswerMarks Guidance
7(ii)p = P(T < 150) = F(150) = 1 – e-150a M1
p = 1 – e - 0⋅75 = 0⋅528A1
2

AnswerMarks Guidance
7(iii)1 – p n > 0⋅99 M1
0⋅01 > (1 – e - 0⋅75) n
AnswerMarks Guidance
or 0⋅01 > 0⋅528 nA1
n > log 0⋅01 / log 0⋅528M1 Rearrange and take logs to give bound
n > 7⋅20 [or 7⋅21] so n = 8
AnswerMarks Guidance
minA1 Find n
min
4
AnswerMarks Guidance
QuestionAnswer Marks
Question 7:
--- 7(i) ---
7(i) | a = 1/200 or 0⋅005 | B1 | State a or find a by equating mean value to 1/a
1
--- 7(ii) ---
7(ii) | p = P(T < 150) = F(150) = 1 – e-150a | M1 | Find P(T < 150)
p = 1 – e - 0⋅75 = 0⋅528 | A1
2
--- 7(iii) ---
7(iii) | 1 – p n > 0⋅99 | M1 | Formulate condition for n
0⋅01 > (1 – e - 0⋅75) n
or 0⋅01 > 0⋅528 n | A1
n > log 0⋅01 / log 0⋅528 | M1 | Rearrange and take logs to give bound
n > 7⋅20 [or 7⋅21] so n = 8
min | A1 | Find n
min
4
Question | Answer | Marks | Guidance
The time, $T$ days, before an electrical component develops a fault has distribution function F given by
$$\mathrm{F}(t) = \begin{cases} 1 - e^{-at} & t \geqslant 0, \\ 0 & \text{otherwise}, \end{cases}$$
where $a$ is a positive constant. The mean value of $T$ is 200.

\begin{enumerate}[label=(\roman*)]
\item Write down the value of $a$.
[1]

\item Find the probability that an electrical component of this type develops a fault in less than 150 days.
[2]
\end{enumerate}

A piece of equipment contains $n$ of these components, which develop faults independently of each other. The probability that, after 150 days, at least one of the $n$ components has not developed a fault is greater than 0.99.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the smallest possible value of $n$.
[4]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP2 2019 Q7 [7]}}