CAIE S2 2023 November — Question 1 3 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2023
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentral limit theorem
TypeCalculate probabilities using sample mean distribution
DifficultyModerate -0.5 This is a straightforward application of the central limit theorem requiring only substitution into the sampling distribution formula (mean μ, variance σ²/n) followed by a single normal probability calculation. The question is slightly easier than average as it's a direct one-step application with no conceptual complications, though it does require understanding that sample means follow N(410, 400/36).
Spec5.05a Sample mean distribution: central limit theorem
A random variable \(X\) has the distribution N(410, 400). Find the probability that the mean of a random sample of 36 values of \(X\) is less than 405. [3]
Question 1:
AnswerMarks
1405−410
[= –1.5]
20
AnswerMarks Guidance
6M1 For standardising, must have 36.
14580−14760
Allow totals method .
14400
No mixed methods.
AnswerMarks Guidance
ɸ('–1.5') = 1 – ɸ('1.5')M1 For area consistent with their working.
= 0.0668A1
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 1:
1 | 405−410
[= –1.5]
20
6 | M1 | For standardising, must have 36.
14580−14760
Allow totals method .
14400
No mixed methods.
ɸ('–1.5') = 1 – ɸ('1.5') | M1 | For area consistent with their working.
= 0.0668 | A1
3
Question | Answer | Marks | Guidance
A random variable $X$ has the distribution N(410, 400).

Find the probability that the mean of a random sample of 36 values of $X$ is less than 405. [3]

\hfill \mbox{\textit{CAIE S2 2023 Q1 [3]}}
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