Pre-U Pre-U 9795/2 2014 June — Question 2 8 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2014
SessionJune
Marks8
TopicCentral limit theorem
TypeFinding n from sample mean distribution
DifficultyChallenging +1.2 This is a standard sampling distribution problem requiring knowledge that X̄ ~ N(μ, σ²/n) and using normal tables to find n from a probability inequality, then a straightforward probability calculation. While it involves multiple steps and careful algebraic manipulation with the standard error, it's a textbook application of sampling theory without requiring novel insight—slightly above average difficulty due to the algebraic setup and two-part structure, but well within expected Further Maths content.
Spec5.05a Sample mean distribution: central limit theorem5.05d Confidence intervals: using normal distribution
The mean of a random sample of \(n\) observations drawn from a normal distribution with mean \(\mu\) and variance \(\sigma^2\) is denoted by \(\bar{X}\). It is given that P(\(\mu - 0.5\sigma < \bar{X} < \mu + 0.5\sigma\)) > 0.95.
  1. Find the smallest possible value of \(n\). [5]
  2. With this value of \(n\), find P(\(\bar{X} > \mu - 0.1\sigma\)). [3]
Question 2:
(ii) ---
2 (i)
AnswerMarks
(ii)σ2
X ~ N(µ, ) stated or implied
n
µ+0.5σ −µ
>1.96
σ / n
⇒ 0.5 n > 1.96 ⇒ n > 15.355 ⇒ Least n is 16
µ−0.1σ −µ
z= =−0.4
σ /4
AnswerMarks
⇒ P( X > µ – 0.1σ) = 0.6551.96
FT on their integer n, and
AnswerMarks
also –0.392: M1A1A0B1
M1A1
M1A1 [5]
M1
A1√
A1 [3]
3 (i)
(ii)
AnswerMarks
(iii)400 20
Let N be the estimated number; = ⇒ N = 8000
N 400
98% confidence: z = ±2.326
p s = 0.05
0.05×0.95
0.05±2.326
400
= (0.0246(5), 0.0753(5))
AnswerMarks
400 ÷ 0.07535 = 5309 and 400 ÷ 0.02465 = 16228pq
Need
n
0.05 and variance correct
Both, correct to 3 SF
400 ÷ CV
Integer answers, a.r.t. 5310,
AnswerMarks
16200B1 [1]
B1
B1
M1A1
A1 [5]
M1
A1 [2]
AnswerMarks Guidance
Page 3Mark Scheme Syllabus
Pre-U – May/June 20149795 02
4 (i)
(ii)
(iii)
AnswerMarks
(iv)[ ]k 2
−3e−x =1 ⇒ –3e –k + 3 = 1 ⇒ e –k = [AG]
0 3
M X ( t ) =∫ 0 k 3e (t−1)xdx=   3 e ( t (t − −1 1 )x )   k
0
e (t−1)k 3 3 ( )
=3 − = 1−e−kekt
( ) ( ) ( )
t−1 t−1 t−1
3  2 
= 1− ekt [AG]
( )
t−1  3 
( ) 2 1 
M (t)=31+t+t2  1− 1+kt+ k2t2 
X  3   2   
= 1 + (1 – 2k)t + (1 – 2k – k 2 )t 2 [AG]
3
E(X) = 1 – 2k = 1–21n  [AG]
AnswerMarks
2Both series correct
2
Deals with
3
AnswerMarks
Allow longer methods if correctM1A1 [2]
M1
A1
M1
A1 [4]
B1
B1
B1 [3]
B1 [1]
5 (i)
(ii)
AnswerMarks
(iii)λ (t – 1)
G(t) = e
G′(t) = λe λ (t – 1)
⇒ µ = G′(1) = λ
G′′(t) = λ 2 e λ (t – 1)
⇒ G′(1) = λ 2
∴σ 2 = E(X 2 ) + µ – µ 2 = λ 2 + λ – λ 2 = λ
229.5−250
z = =−1.297
1
250
260.5−250
z = =0.664
2
250
P(230 ≤ X ≤ 260) = 0.649
We are approximating to B (250, 0.1).
Using N (25,22.5) or Using Po (25)
30.5−25 30.5−25
( ) ( )
z= =1.1595 z= =1.1
22.5 25
AnswerMarks
P(>30) = 0.123 P(>30) = 0.136In tables
One cc correct
Both ccs, √250
Correct handling of tails
Answer in range [0.649, 0.650]
Distribution stated or implied
Correct CC and √
Exact binomial 0.125: M0
AnswerMarks
Exact Poisson 0.136691: 3M1
A1
M1
A1
A1 [5]
M1A1
A1
M1
A1 [5]
M1
A1
A1 [3]
AnswerMarks Guidance
Page 4Mark Scheme Syllabus
Pre-U – May/June 20149795 02
6 (i)
(ii)
(iii)
AnswerMarks
(iv)4[ ] 4[ ]
0.41 0.41
tan−1x =0.495K, tan−1x =0.506K
π 0 π 0
⇒ 0.41 < median < 0.42 [AG]
4[ ]m π
Or: tan−1x =0.5; m = = 0.414…
π 0 8
4 ∫1 x
E(X) = dx
π 01+x2
2 [ ] 1 2
= ln 1+x2 = ln2 [AG]
π 0 π
4 ∫1 x2 4 ∫1 1
E(X 2 ) = dx 1− dx
π 01+x2 π 0 1+x2
4[ ] 1 4  π  4
= x−tan−1x =  1−  = −1
π 0 π  4 π
 4   2  2
Var (X) =  −1− 1n2 =0.0785
π  π 
4[ ] 1 1 1
tan−1x =1− =
π 2−1 2 2
4[ ] 1 2 1
tan−1x =1− =
π 3/3 3 3
π π
−
1 1 2 2
4 6
AnswerMarks
P(X > 1/3 3X > 2 – 1) = ÷ = or =
3 2 3 π π 3
−
AnswerMarks
4 8Correct integral
One correct output
Conclusion fully justified (need
some comment if 0.41 and 0.42
used)
∫ x f(x) dx attempted
∫ x 2 f(x) dx seen
Method for integration
 4 
 −1 or 0.2732
π 
Answers from calculator can
get full marks
Quotient of relevant
probabilities
AnswerMarks
One correctM1
A1
A1 [3]
M1
A1 [2]
M1
M1
A1
M1
A1 [5]
M1
A1
A1 [3]
Question 2:
--- 2 (i)
(ii) ---
2 (i)
(ii) | σ2
X ~ N(µ, ) stated or implied
n
µ+0.5σ −µ
>1.96
σ / n
⇒ 0.5 n > 1.96 ⇒ n > 15.355 ⇒ Least n is 16
µ−0.1σ −µ
z= =−0.4
σ /4
⇒ P( X > µ – 0.1σ) = 0.655 | 1.96
FT on their integer n, and
also –0.392: M1A1A0 | B1
M1A1
M1A1 [5]
M1
A1√
A1 [3]
3 (i)
(ii)
(iii) | 400 20
Let N be the estimated number; = ⇒ N = 8000
N 400
98% confidence: z = ±2.326
p s = 0.05
0.05×0.95
0.05±2.326
400
= (0.0246(5), 0.0753(5))
400 ÷ 0.07535 = 5309 and 400 ÷ 0.02465 = 16228 | pq
Need
n
0.05 and variance correct
Both, correct to 3 SF
400 ÷ CV
Integer answers, a.r.t. 5310,
16200 | B1 [1]
B1
B1
M1A1
A1 [5]
M1
A1 [2]
Page 3 | Mark Scheme | Syllabus | Paper
Pre-U – May/June 2014 | 9795 | 02
4 (i)
(ii)
(iii)
(iv) | [ ]k 2
−3e−x =1 ⇒ –3e –k + 3 = 1 ⇒ e –k = [AG]
0 3
M X ( t ) =∫ 0 k 3e (t−1)xdx=   3 e ( t (t − −1 1 )x )   k
0
e (t−1)k 3 3 ( )
=3 − = 1−e−kekt
( ) ( ) ( )
t−1 t−1 t−1
3  2 
= 1− ekt [AG]
( )
t−1  3 
( ) 2 1 
M (t)=31+t+t2  1− 1+kt+ k2t2 
X  3   2   
= 1 + (1 – 2k)t + (1 – 2k – k 2 )t 2 [AG]
3
E(X) = 1 – 2k = 1–21n  [AG]
2 | Both series correct
2
Deals with
3
Allow longer methods if correct | M1A1 [2]
M1
A1
M1
A1 [4]
B1
B1
B1 [3]
B1 [1]
5 (i)
(ii)
(iii) | λ (t – 1)
G(t) = e
G′(t) = λe λ (t – 1)
⇒ µ = G′(1) = λ
G′′(t) = λ 2 e λ (t – 1)
⇒ G′(1) = λ 2
∴σ 2 = E(X 2 ) + µ – µ 2 = λ 2 + λ – λ 2 = λ
229.5−250
z = =−1.297
1
250
260.5−250
z = =0.664
2
250
P(230 ≤ X ≤ 260) = 0.649
We are approximating to B (250, 0.1).
Using N (25,22.5) or Using Po (25)
30.5−25 30.5−25
( ) ( )
z= =1.1595 z= =1.1
22.5 25
P(>30) = 0.123 P(>30) = 0.136 | In tables
One cc correct
Both ccs, √250
Correct handling of tails
Answer in range [0.649, 0.650]
Distribution stated or implied
Correct CC and √
Exact binomial 0.125: M0
Exact Poisson 0.136691: 3 | M1
A1
M1
A1
A1 [5]
M1A1
A1
M1
A1 [5]
M1
A1
A1 [3]
Page 4 | Mark Scheme | Syllabus | Paper
Pre-U – May/June 2014 | 9795 | 02
6 (i)
(ii)
(iii)
(iv) | 4[ ] 4[ ]
0.41 0.41
tan−1x =0.495K, tan−1x =0.506K
π 0 π 0
⇒ 0.41 < median < 0.42 [AG]
4[ ]m π
Or: tan−1x =0.5; m = = 0.414…
π 0 8
4 ∫1 x
E(X) = dx
π 01+x2
2 [ ] 1 2
= ln 1+x2 = ln2 [AG]
π 0 π
4 ∫1 x2 4 ∫1 1
E(X 2 ) = dx 1− dx
π 01+x2 π 0 1+x2
4[ ] 1 4  π  4
= x−tan−1x =  1−  = −1
π 0 π  4 π
 4   2  2
Var (X) =  −1− 1n2 =0.0785
π  π 
4[ ] 1 1 1
tan−1x =1− =
π 2−1 2 2
4[ ] 1 2 1
tan−1x =1− =
π 3/3 3 3
π π
−
1 1 2 2
4 6
P(X > 1/3 3 | X > 2 – 1) = ÷ = or =
3 2 3 π π 3
−
4 8 | Correct integral
One correct output
Conclusion fully justified (need
some comment if 0.41 and 0.42
used)
∫ x f(x) dx attempted
∫ x 2 f(x) dx seen
Method for integration
 4 
 −1 or 0.2732
π 
Answers from calculator can
get full marks
Quotient of relevant
probabilities
One correct | M1
A1
A1 [3]
M1
A1 [2]
M1
M1
A1
M1
A1 [5]
M1
A1
A1 [3]
The mean of a random sample of $n$ observations drawn from a normal distribution with mean $\mu$ and variance $\sigma^2$ is denoted by $\bar{X}$. It is given that P($\mu - 0.5\sigma < \bar{X} < \mu + 0.5\sigma$) > 0.95.

\begin{enumerate}[label=(\roman*)]
\item Find the smallest possible value of $n$. [5]

\item With this value of $n$, find P($\bar{X} > \mu - 0.1\sigma$). [3]
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2014 Q2 [8]}}
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