CAIE FP2 2018 November — Question 6 6 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2018
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConfidence intervals
TypeCI from raw data list
DifficultyModerate -0.3 This is a straightforward confidence interval calculation with small sample size requiring t-distribution. Students must calculate sample mean and standard deviation, then apply the standard formula with t₇ critical value. While it involves multiple computational steps (6 marks), it's a direct application of a standard procedure with no conceptual challenges or novel problem-solving required, making it slightly easier than average.
Spec5.05c Hypothesis test: normal distribution for population mean

The heights, in metres, of a random sample of 8 trees of a particular type are as follows. 14.2 11.3 10.8 8.4 12.8 11.5 12.1 9.2 Assuming that heights of trees of this type are normally distributed, calculate a 95% confidence interval for the mean height of trees of this type. [6]

Question 6:
AnswerMarks Guidance
6x = 90.3 / 8 = 11.2875 (to 4 s.f.) B1
s2 = (1043.67 – 90.32 /8) / 7M1 Estimate population variance
= 19 527 / 5600 or 3.487 [or 1.8672 ] (to 4 s.f.)A1 (allow biased here: 3.051 or 1.7472)
90.3 / 8 ± t √(s2/8)M1 Find confidence interval
t = 2.365 (to 4 s.f.)
AnswerMarks Guidance
7, 0.975A1 State or use correct tabular value of t
11.3 ± 1.6 or [9.7, 12.8[5]]A1 Evaluate C.I. (either form)
6
AnswerMarks Guidance
QuestionAnswer Marks
Question 6:
6 | x = 90.3 / 8 = 11.2875 (to 4 s.f.) | B1 | Find sample mean
s2 = (1043.67 – 90.32 /8) / 7 | M1 | Estimate population variance
= 19 527 / 5600 or 3.487 [or 1.8672 ] (to 4 s.f.) | A1 | (allow biased here: 3.051 or 1.7472)
90.3 / 8 ± t √(s2/8) | M1 | Find confidence interval
t = 2.365 (to 4 s.f.)
7, 0.975 | A1 | State or use correct tabular value of t
11.3 ± 1.6 or [9.7, 12.8[5]] | A1 | Evaluate C.I. (either form)
6
Question | Answer | Marks | Guidance
The heights, in metres, of a random sample of 8 trees of a particular type are as follows.

14.2    11.3    10.8    8.4    12.8    11.5    12.1    9.2

Assuming that heights of trees of this type are normally distributed, calculate a 95% confidence interval for the mean height of trees of this type.
[6]

\hfill \mbox{\textit{CAIE FP2 2018 Q6 [6]}}