CAIE S2 2011 June — Question 5 9 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2011
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of a Poisson distribution
TypeTest with normal approximation
DifficultyStandard +0.3 This is a straightforward application of Poisson distribution properties (scaling parameters and addition) and a standard one-tailed hypothesis test. Part (i) requires scaling two Poisson parameters to 2 minutes and adding them, then calculating P(X<3). Part (ii) is a routine hypothesis test with clear hypotheses, test statistic calculation, and comparison to critical value. While it requires careful parameter scaling and understanding of Poisson properties, these are standard S2 techniques with no novel problem-solving required, making it slightly easier than average.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson5.05b Unbiased estimates: of population mean and variance
The number of adult customers arriving in a shop during a 5-minute period is modelled by a random variable with distribution \(\text{Po}(6)\). The number of child customers arriving in the same shop during a 10-minute period is modelled by an independent random variable with distribution \(\text{Po}(4.5)\).
  1. Find the probability that during a randomly chosen 2-minute period, the total number of adult and child customers who arrive in the shop is less than 3. [3]
  2. During a sale, the manager claims that more adult customers are arriving than usual. In a randomly selected 30-minute period during the sale, 49 adult customers arrive. Test the manager's claim at the 2.5\% significance level. [6]
Question 5:
(ii) ---
5 (i)
AnswerMarks
(ii)Po(3.3)
e –3.3 (1 + 3.3 + 3.32 )
2
= 0.359
X~Po(36)
X~N(36, 36)
48.5−36
36
= 2.08(3)
comp with 1.96
AnswerMarks
Evidence to support claimB1
M1
A1 [3]
B1
B1
M1
A1
M1
A1√
AnswerMarks
[6]seen or implied
Poisson P(0) + P(1) + P(2). Allow + P(3)
Allow wrong λ.
Accept equiv method.
Allow with no or wrong cc or no √
2.08(3) or 0.0186/0.0187 if area comparison
Valid comparison
Correct conclusion (ft their z)
[Total: 9]
6 (i)
(ii)
(iii)
AnswerMarks
(iv)1 1
H0: P(6) = H1: P(6) >
6 6
 5 10 +10× 5 9 ×1 + 10  × 5 8 ×1 2 + 10  × 5 7 × 1 3
6 6 6 2 6 6 3 6 6
1 – ( 5  10 +10× 5  9 ×1 + 10  × 5  8 × 1  2
6 6 6 2 6 6
10  5  7 1  3
+  ×  ×  )
3 6 6
= 0.0697 (3 sfs)
Die biased towards a six but result < 4 so no
evidence of bias
P(0, 1, 2 or 3 sixes)
1 10 1 9 1 10  1 8 1 2 10  1 7 1 3
(  +10×  × +  ×  ×  +  ×  ×  )
2 2 2 2 2 2 3 2 2
AnswerMarks
= 0.172 or 11/64B1 [1]
M1
M1
A1 [3]
B1 [1]
B1
M1
AnswerMarks
A1 [3]Condone undefined p
(1 –) P(0,1,2,3) o.e. using B(10,1/6)
allow end errors
Attempt at fully correct expression for
1 – P(0,1,2,3) o.e.
Accept 0.0698
or equiv. Must be in context
Stated or attempted. Can be implied
Attempt at P(0,1,2,3) with p = 1/2, allow
end errors.
[Total: 8]
AnswerMarks Guidance
Page 6Mark Scheme: Teachers’ version Syllabus
GCE AS/A LEVEL – May/June 20119709 72
7 (i)
(ii)
(iii)
AnswerMarks
(iv)∫ 1 k(1−x)dx = 1
−1
(k[ x− x2 ] 1 = 1)
2 −1
2k = 1
(k = 1 AG)
2
( ∫ 1 1(1−x)dx = 1 [ x− x2 ] 1 )
2 2 2 0.5
0.5
1
= or 0.0625
16
∫ 1 1(x−x2)dx
2
−1
1 x2 x3 1
= [ – ]
2 2 3 −1
1
= – or –0.333
3
∫ a 1(1−x)dx = 0.25
2
−1
(1 [ x− x2 ] a = 0.25)
2 2 −1
(1 (a – a2 – (–1 – 1 ) = 0.25)
2 2 2
a2 – 2a – 2 = 0
AnswerMarks
a = 1 – √3 or –0.732M1
A1 [2]
B1 [1]
M1
A1
A1 [3]
M1
A1
AnswerMarks
A1 [3]Attempt integ f(x) = 1 with correct limits
∫xf(x)dx
ignore limits
Correct integrand and limits
Correct limits (or integral from a to
1 = 0.75)
any correct QE with “= 0”(or in
completed square form (a – 1) 2 = 3)
Not a = 1 ± √3; Not –0.732 or 2.732
[Total: 9]
Question 5:
--- 5 (i)
(ii) ---
5 (i)
(ii) | Po(3.3)
e –3.3 (1 + 3.3 + 3.32 )
2
= 0.359
X~Po(36)
X~N(36, 36)
48.5−36
36
= 2.08(3)
comp with 1.96
Evidence to support claim | B1
M1
A1 [3]
B1
B1
M1
A1
M1
A1√
[6] | seen or implied
Poisson P(0) + P(1) + P(2). Allow + P(3)
Allow wrong λ.
Accept equiv method.
Allow with no or wrong cc or no √
2.08(3) or 0.0186/0.0187 if area comparison
Valid comparison
Correct conclusion (ft their z)
[Total: 9]
6 (i)
(ii)
(iii)
(iv) | 1 1
H0: P(6) = H1: P(6) >
6 6
 5 10 +10× 5 9 ×1 + 10  × 5 8 ×1 2 + 10  × 5 7 × 1 3
6 6 6 2 6 6 3 6 6
1 – ( 5  10 +10× 5  9 ×1 + 10  × 5  8 × 1  2
6 6 6 2 6 6
10  5  7 1  3
+  ×  ×  )
3 6 6
= 0.0697 (3 sfs)
Die biased towards a six but result < 4 so no
evidence of bias
P(0, 1, 2 or 3 sixes)
1 10 1 9 1 10  1 8 1 2 10  1 7 1 3
(  +10×  × +  ×  ×  +  ×  ×  )
2 2 2 2 2 2 3 2 2
= 0.172 or 11/64 | B1 [1]
M1
M1
A1 [3]
B1 [1]
B1
M1
A1 [3] | Condone undefined p
(1 –) P(0,1,2,3) o.e. using B(10,1/6)
allow end errors
Attempt at fully correct expression for
1 – P(0,1,2,3) o.e.
Accept 0.0698
or equiv. Must be in context
Stated or attempted. Can be implied
Attempt at P(0,1,2,3) with p = 1/2, allow
end errors.
[Total: 8]
Page 6 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE AS/A LEVEL – May/June 2011 | 9709 | 72
7 (i)
(ii)
(iii)
(iv) | ∫ 1 k(1−x)dx = 1
−1
(k[ x− x2 ] 1 = 1)
2 −1
2k = 1
(k = 1 AG)
2
( ∫ 1 1(1−x)dx = 1 [ x− x2 ] 1 )
2 2 2 0.5
0.5
1
= or 0.0625
16
∫ 1 1(x−x2)dx
2
−1
1 x2 x3 1
= [ – ]
2 2 3 −1
1
= – or –0.333
3
∫ a 1(1−x)dx = 0.25
2
−1
(1 [ x− x2 ] a = 0.25)
2 2 −1
(1 (a – a2 – (–1 – 1 ) = 0.25)
2 2 2
a2 – 2a – 2 = 0
a = 1 – √3 or –0.732 | M1
A1 [2]
B1 [1]
M1
A1
A1 [3]
M1
A1
A1 [3] | Attempt integ f(x) = 1 with correct limits
∫xf(x)dx
ignore limits
Correct integrand and limits
Correct limits (or integral from a to
1 = 0.75)
any correct QE with “= 0”(or in
completed square form (a – 1) 2 = 3)
Not a = 1 ± √3; Not –0.732 or 2.732
[Total: 9]
The number of adult customers arriving in a shop during a 5-minute period is modelled by a random variable with distribution $\text{Po}(6)$. The number of child customers arriving in the same shop during a 10-minute period is modelled by an independent random variable with distribution $\text{Po}(4.5)$.

\begin{enumerate}[label=(\roman*)]
\item Find the probability that during a randomly chosen 2-minute period, the total number of adult and child customers who arrive in the shop is less than 3. [3]
\item During a sale, the manager claims that more adult customers are arriving than usual. In a randomly selected 30-minute period during the sale, 49 adult customers arrive. Test the manager's claim at the 2.5\% significance level. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2011 Q5 [9]}}
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