Question 5 - A-Level Maths

CAIE FP2 2015 November — Question 5 5 marks

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2015
Session	November
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Confidence intervals
Type	Unbiased estimates then CI
Difficulty	Standard +0.3 This is a straightforward confidence interval calculation requiring standard formulas for sample mean and standard deviation, then applying the t-distribution with 9 degrees of freedom. While it's a Further Maths topic, the question is entirely procedural with no conceptual challenges—students simply substitute values into learned formulas, making it slightly easier than average overall.
Spec	5.05d Confidence intervals: using normal distribution

A random sample of 10 observations of a normal variable $X$ gave the following summarised data, where $\bar{x}$ is the sample mean. $$\Sigma x = 222.8 \qquad \Sigma(x - \bar{x})^2 = 4.12$$ Find a 95% confidence interval for the population mean. [5]

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks
5	Find or use sample mean x = 222⋅8 / 10 = 22⋅28

and estimate population variance: s 2 = 4⋅12 / 9

(allow biased here: 0⋅412 or 0⋅642 2 ) = 0⋅458 or 103/225 or 0⋅677 2 M1

Find confidence interval (allow z in place of t) e.g.:

22⋅28 ± t √(0⋅458 / 10) M1 A1

Use of correct tabular value: t9, 0.975 = 2⋅26[2] A1

Answer	Marks	Guidance
Evaluate C.I. correct to 3 s.f.: 22⋅3 ± 0⋅48[4] or [21⋅8, 22⋅8] A1	5	5
Page 5	Mark Scheme	Syllabus
Cambridge International A Level – October/November 2015	9231	21

6 (i)

(ii)

Answer	Marks
(iii)	Find prob. p of head from mean = 2 × variance: 1/p = 2 × (1 – p)/p 2 , p = ⅔ A.G. M1 A1

Find P(X = 4) (denoting 1 – p by q [= ⅓]): P(X = 4) = q 3 × p

= 2/81 or 0⋅0247 B1

Find or state P(X > 4): P(X > 4) [= 1 – (1 + q + q 2 + q 3 ) × p

4 4

= 1 – (1 – q ) ] = q

= 1/81 or 0⋅0123 M1 A1

Formulate condition for N: 1 – q N > 0⋅999, [ (⅓) N < 0⋅001 ] M1

Take logs (any base) to give bound for N: N > log 0⋅001 / log ⅓ M1

Find Nmin: N > 6⋅29, Nmin = 7 A1

Answer	Marks
(N < 6⋅29 or N = 6⋅29 earns M2 A0)	2

1

2

Answer	Marks
3	8

Question 5:
5 | Find or use sample mean x = 222⋅8 / 10 = 22⋅28
and estimate population variance: s 2 = 4⋅12 / 9
(allow biased here: 0⋅412 or 0⋅642 2 ) = 0⋅458 or 103/225 or 0⋅677 2 M1
Find confidence interval (allow z in place of t) e.g.:
22⋅28 ± t √(0⋅458 / 10) M1 A1
Use of correct tabular value: t9, 0.975 = 2⋅26[2] A1
Evaluate C.I. correct to 3 s.f.: 22⋅3 ± 0⋅48[4] or [21⋅8, 22⋅8] A1 | 5 | 5
Page 5 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2015 | 9231 | 21
6
(i)
(ii)
(iii) | Find prob. p of head from mean = 2 × variance: 1/p = 2 × (1 – p)/p 2 , p = ⅔ A.G. M1 A1
Find P(X = 4) (denoting 1 – p by q [= ⅓]): P(X = 4) = q 3 × p
= 2/81 or 0⋅0247 B1
Find or state P(X > 4): P(X > 4) [= 1 – (1 + q + q 2 + q 3 ) × p
4 4
= 1 – (1 – q ) ] = q
= 1/81 or 0⋅0123 M1 A1
Formulate condition for N: 1 – q N > 0⋅999, [ (⅓) N < 0⋅001 ] M1
Take logs (any base) to give bound for N: N > log 0⋅001 / log ⅓ M1
Find Nmin: N > 6⋅29, Nmin = 7 A1
(N < 6⋅29 or N = 6⋅29 earns M2 A0) | 2
1
2
3 | 8

Show LaTeX source

A random sample of 10 observations of a normal variable $X$ gave the following summarised data, where $\bar{x}$ is the sample mean.

$$\Sigma x = 222.8 \qquad \Sigma(x - \bar{x})^2 = 4.12$$

Find a 95% confidence interval for the population mean. [5]

\hfill \mbox{\textit{CAIE FP2 2015 Q5 [5]}}

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Q1 9 Q2 10 Q3 11 Q4 13 Q5 5 Q6 8 Q7 9 Q8 10 Q9 11 Q10 28