CAIE S1 2012 November — Question 1 3 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2012
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeFind standard deviation from probability
DifficultyModerate -0.3 This is a straightforward inverse normal distribution problem requiring a single lookup of z = -1.036 from tables (for P(Z < z) = 0.15), then solving 5.6 = 9.3 - 1.036σ. It's slightly easier than average because it's a direct one-step application of the standardization formula with no complications, though it does require understanding the relationship between probability and z-scores.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation

1 In a normal distribution with mean 9.3, the probability of a randomly chosen value being greater than 5.6 is 0.85 . Find the standard deviation.

AnswerMarks Guidance
(i) PE loss = \(0.8g \times (2.5 - 1.8)\) (= 5.6J)B1
Work done is 5.6 JB1 2 marks total
(ii)M1 For using KE gain = PE loss – WD against resistance
\(\frac{1}{2} \times 0.8v^2 = 0.8g \times 2.5 - 0.6 \times 5.6\)A1ft
Speed at \(B\) is 6.45 ms\(^{-1}\)A1 3 marks total
**(i)** PE loss = $0.8g \times (2.5 - 1.8)$ (= 5.6J) | B1 | 

Work done is 5.6 J | B1 | 2 marks total

**(ii)** | M1 | For using KE gain = PE loss – WD against resistance

$\frac{1}{2} \times 0.8v^2 = 0.8g \times 2.5 - 0.6 \times 5.6$ | A1ft |

Speed at $B$ is 6.45 ms$^{-1}$ | A1 | 3 marks total

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1 In a normal distribution with mean 9.3, the probability of a randomly chosen value being greater than 5.6 is 0.85 . Find the standard deviation.

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