CAIE S2 2008 November — Question 3 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2008
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeSum versus sum comparison
DifficultyStandard +0.8 This question requires understanding that linear combinations of independent normal variables are themselves normal, then correctly forming the distribution of (2T - 10C), calculating its mean and variance using independence properties, and finally finding a probability. While the individual steps are standard S2 techniques, the multi-step reasoning and careful handling of the '2 tables vs 10 chairs' comparison makes it moderately above average difficulty.
Spec5.04b Linear combinations: of normal distributions
3 Weights of garden tables are normally distributed with mean 36 kg and standard deviation 1.6 kg . Weights of garden chairs are normally distributed with mean 7.3 kg and standard deviation 0.4 kg . Find the probability that the total weight of 2 randomly chosen tables is more than the total weight of 10 randomly chosen chairs.
\(2T \sim N(72, 2 \times 1.6)^2\)
\(10C \sim N(73, 10 \times 0.4)^2\)
\(2T - 10C \sim N(-1, 2 \times 1.6^2 + 10 \times 0.4^2)\)
\(\sim N(-1, 6.72)\)
\(P((2T - 10C) > 0) = 1 - \Phi\left(\frac{0 - (-1)}{\sqrt{6.72}}\right)\)
\(= 1 - \Phi(0.3857)\)
\(= 1 - 0.650\)
AnswerMarks Guidance
\(= 0.350\)B1 Correct mean ±(72–73)
B1\(2 \times 1.6^2 + 10 \times 0.4^2\) or 6.72 seen
M1Consideration of 2T – 10C and standardising, no cc, sq root
M1Correct side (<0.5) – consistent with their working
A1 [5]Correct answer 
$2T \sim N(72, 2 \times 1.6)^2$
$10C \sim N(73, 10 \times 0.4)^2$
$2T - 10C \sim N(-1, 2 \times 1.6^2 + 10 \times 0.4^2)$
$\sim N(-1, 6.72)$

$P((2T - 10C) > 0) = 1 - \Phi\left(\frac{0 - (-1)}{\sqrt{6.72}}\right)$
$= 1 - \Phi(0.3857)$
$= 1 - 0.650$
$= 0.350$ | B1 | Correct mean ±(72–73)
| B1 | $2 \times 1.6^2 + 10 \times 0.4^2$ or 6.72 seen
| M1 | Consideration of 2T – 10C and standardising, no cc, sq root
| M1 | Correct side (<0.5) – consistent with their working
| A1 [5] | Correct answer

---
3 Weights of garden tables are normally distributed with mean 36 kg and standard deviation 1.6 kg . Weights of garden chairs are normally distributed with mean 7.3 kg and standard deviation 0.4 kg . Find the probability that the total weight of 2 randomly chosen tables is more than the total weight of 10 randomly chosen chairs.

\hfill \mbox{\textit{CAIE S2 2008 Q3 [5]}}
This paper (7 questions)
View full paper
Q1 4 Q2 5 Q3 5 Q4 7 Q5 8 Q6 9 Q7 12