CAIE S1 2015 June — Question 1 4 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2015
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeFind mean from probability statement
DifficultyStandard +0.3 This is a straightforward normal distribution problem requiring understanding of symmetry and use of standard tables. Students must recognize that P(3.2 < X < μ) = 0.475 implies 3.2 is positioned symmetrically below μ, then use inverse normal tables to find the z-score and solve for μ. It's slightly above average difficulty due to the need to interpret the probability statement correctly, but the calculation itself is routine once set up.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation

1 The lengths, in metres, of cars in a city are normally distributed with mean \(\mu\) and standard deviation 0.714 . The probability that a randomly chosen car has a length more than 3.2 metres and less than \(\mu\) metres is 0.475 . Find \(\mu\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(x < 3.273) = 0.5 - 0.475 = 0.025\)M1 Attempt to find \(z\)-value using tables in reverse
\(z = -1.96\)A1 \(\pm 1.96\) seen
\(\frac{3.2 - \mu}{0.714} = -1.96\)M1 Solving their standardised equation; \(z\)-value not necessary
\(\mu = 4.60\) sA1 [4] Correct answer; accept 4.6
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(x < 3.273) = 0.5 - 0.475 = 0.025$ | M1 | Attempt to find $z$-value using tables in reverse |
| $z = -1.96$ | A1 | $\pm 1.96$ seen |
| $\frac{3.2 - \mu}{0.714} = -1.96$ | M1 | Solving their standardised equation; $z$-value not necessary |
| $\mu = 4.60$ s | A1 [4] | Correct answer; accept 4.6 |

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1 The lengths, in metres, of cars in a city are normally distributed with mean $\mu$ and standard deviation 0.714 . The probability that a randomly chosen car has a length more than 3.2 metres and less than $\mu$ metres is 0.475 . Find $\mu$.

\hfill \mbox{\textit{CAIE S1 2015 Q1 [4]}}