CAIE S1 2012 June — Question 1 3 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2012
SessionJune
Marks3
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TopicNormal Distribution
TypeInterval probability P(a < X < b)
DifficultyEasy -1.2 This is a straightforward normal distribution probability calculation requiring only standardization and table lookup. Given mean and variance explicitly, students simply compute two z-scores and find the difference between table values—a routine procedural task with no conceptual challenges or multi-step reasoning.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation
1 It is given that \(X \sim \mathrm {~N} ( 28.3,4.5 )\). Find the probability that a randomly chosen value of \(X\) lies between 25 and 30 .
AnswerMarks Guidance
\(z_1 = \frac{30 - 28.3}{\sqrt{4.5}} = 0.8014\)M1 Standardising at least one value, sq rt.ess; no cc
\(z_2 = \frac{25 - 28.3}{\sqrt{4.5}} = -1.5556\)M1 \(\Phi_1 + \Phi_2 - 1\) oe
\(\Phi_1 - (1 - \Phi_2) = 0.7884 + 0.9401 - 1 = 0.729\)A1 [3] Correct answer 
$z_1 = \frac{30 - 28.3}{\sqrt{4.5}} = 0.8014$ | M1 | Standardising at least one value, sq rt.ess; no cc

$z_2 = \frac{25 - 28.3}{\sqrt{4.5}} = -1.5556$ | M1 | $\Phi_1 + \Phi_2 - 1$ oe

$\Phi_1 - (1 - \Phi_2) = 0.7884 + 0.9401 - 1 = 0.729$ | A1 [3] | Correct answer
1 It is given that $X \sim \mathrm {~N} ( 28.3,4.5 )$. Find the probability that a randomly chosen value of $X$ lies between 25 and 30 .

\hfill \mbox{\textit{CAIE S1 2012 Q1 [3]}}
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