CAIE S2 2016 June — Question 5 10 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2016
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
DifficultyStandard +0.3 This is a straightforward probability density function question requiring standard techniques: integrating to find k, calculating E(T) by integration, finding the median by solving an integral equation, and reading the support from the definition. All parts are routine applications of S2 methods with no novel insight required, making it slightly easier than average.
Spec5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles
The time, \(T\) minutes, taken by people to complete a test has probability density function given by $$\mathrm{f}(t) = \begin{cases} k(10t - t^2) & 5 \leq t \leq 10, \\ 0 & \text{otherwise}, \end{cases}$$ where \(k\) is a constant.
  1. Show that \(k = \frac{3}{250}\). [3]
  2. Find \(\mathrm{E}(T)\). [3]
  3. Find the probability that a randomly chosen value of \(T\) lies between \(\mathrm{E}(T)\) and the median of \(T\). [3]
  4. State the greatest possible length of time taken to complete the test. [1]
Question 5:
AnswerMarks
5 (i)10
2
k ∫ (10t−t )dt = 1
5
10
k   5t2 −t 3 3  5 = 1
k(500 – 1000 – (125 – 125)) = 1
3 3
k × 250= 1
3
(k = 3 AG)
AnswerMarks
250M1
A1
AnswerMarks
A1 [3]Attempt to integrate, ignore limits
Correct integral and limits
No errors seen; No inexact decimals seen
AnswerMarks Guidance
Page 5Mark Scheme Syllabus
Cambridge International A Level – May/June 20169709 73
QuAnswer Marks
(ii)
(iii)
AnswerMarks
(iv)10
3 ∫(10t2 −t3)dt
250
5
10
= 25 3 0   10 3 t3 −t 4 4  5
= 3 (10000 −10000 −(1250 −625)
250 3 4 3 4
= 6.875 or 55/8
"6.875"
P(T < E(T) = 25 3 0   5t2 −t 3 3  5
= 0.5361
“0.5361” – 0.5
P(T between E(T) & median = 0.0361
AnswerMarks
10 (minutes)M1
A1
A1 [3]
M1*
DM1*
A1 [3]
AnswerMarks
B1 [1]Attempt to integrate, ignore limits
Correct integral and limit. Condone missing k
Allow 6.88
ft their E(T)
allow 0.036
Alternative Method
Integrate f(t)limits 5 and m equated to 0.5 M1*
Integrate f(t)limits their 6.736 (provided between
5 and 10) and their 6.875DM1
Allow without "minutes"
6 (i)
(ii)
(iii)
AnswerMarks
(iv)λ= 3.9
–3.9 ×3.94
e
4!
= 0.195
X ~ N(1.6, 1.6 )
75
1.7−1.6(= 0.685)
1.6
75
1– Φ(“0.685”)
= 0.247 (3 sf)
AnswerMarks
X not normally distr. So CLT neededB1
M1
A1 [3]
B1
B1 [2]
M1
M1
A1 [3]
AnswerMarks
B1 [1]M1 allow any λ
SR Combination method
B1 for λ =1.6 AND λ =2.3 used in combination
method (at least 3 combinations)
M1 All correctly combined and added
B1 for N(1.6, ….)stated
B1 for Var = 1.6 stated
75
SR, not stated but all implied in (iii): B1
For standardising (using their values or correct
values .Ignore cc
Correct area consistent with their working
Accept use of 1/2n correction leading to 0.233.
NB Use of Poisson sum Po(120) and N(120,120)
with µ=127.5 leads to 0.247, or 0.233 with cc
Not “it”
AnswerMarks Guidance
Page 6Mark Scheme Syllabus
Cambridge International A Level – May/June 20169709 73
QuAnswer Marks 
Question 5:
--- 5 (i) ---
5 (i) | 10
2
k ∫ (10t−t )dt = 1
5
10
k   5t2 −t 3 3  5 = 1
k(500 – 1000 – (125 – 125)) = 1
3 3
k × 250= 1
3
(k = 3 AG)
250 | M1
A1
A1 [3] | Attempt to integrate, ignore limits
Correct integral and limits
No errors seen; No inexact decimals seen
Page 5 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – May/June 2016 | 9709 | 73
Qu | Answer | Marks | Notes
(ii)
(iii)
(iv) | 10
3 ∫(10t2 −t3)dt
250
5
10
= 25 3 0   10 3 t3 −t 4 4  5
= 3 (10000 −10000 −(1250 −625)
250 3 4 3 4
= 6.875 or 55/8
"6.875"
P(T < E(T) = 25 3 0   5t2 −t 3 3  5
= 0.5361
“0.5361” – 0.5
P(T between E(T) & median = 0.0361
10 (minutes) | M1
A1
A1 [3]
M1*
DM1*
A1 [3]
B1 [1] | Attempt to integrate, ignore limits
Correct integral and limit. Condone missing k
Allow 6.88
ft their E(T)
allow 0.036
Alternative Method
Integrate f(t)limits 5 and m equated to 0.5 M1*
Integrate f(t)limits their 6.736 (provided between
5 and 10) and their 6.875DM1
Allow without "minutes"
6 (i)
(ii)
(iii)
(iv) | λ= 3.9
–3.9 ×3.94
e
4!
= 0.195
X ~ N(1.6, 1.6 )
75
1.7−1.6(= 0.685)
1.6
75
1– Φ(“0.685”)
= 0.247 (3 sf)
X not normally distr. So CLT needed | B1
M1
A1 [3]
B1
B1 [2]
M1
M1
A1 [3]
B1 [1] | M1 allow any λ
SR Combination method
B1 for λ =1.6 AND λ =2.3 used in combination
method (at least 3 combinations)
M1 All correctly combined and added
B1 for N(1.6, ….)stated
B1 for Var = 1.6 stated
75
SR, not stated but all implied in (iii): B1
For standardising (using their values or correct
values .Ignore cc
Correct area consistent with their working
Accept use of 1/2n correction leading to 0.233.
NB Use of Poisson sum Po(120) and N(120,120)
with µ=127.5 leads to 0.247, or 0.233 with cc
Not “it”
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – May/June 2016 | 9709 | 73
Qu | Answer | Marks | Notes
The time, $T$ minutes, taken by people to complete a test has probability density function given by
$$\mathrm{f}(t) = \begin{cases}
k(10t - t^2) & 5 \leq t \leq 10, \\
0 & \text{otherwise},
\end{cases}$$
where $k$ is a constant.

\begin{enumerate}[label=(\roman*)]
\item Show that $k = \frac{3}{250}$. [3]

\item Find $\mathrm{E}(T)$. [3]

\item Find the probability that a randomly chosen value of $T$ lies between $\mathrm{E}(T)$ and the median of $T$. [3]

\item State the greatest possible length of time taken to complete the test. [1]
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2016 Q5 [10]}}
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