CAIE S2 2002 June — Question 3 6 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2002
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeMixed sum threshold probability
DifficultyStandard +0.3 This is a straightforward application of linear combinations of independent normal variables requiring students to form the combined distribution (3×500 + 5×200 + 350 for mean, √(3×20² + 5×12²) for SD) and calculate a single probability using tables. While it involves multiple variables, the method is standard and computational rather than requiring insight.
Spec5.04b Linear combinations: of normal distributions

3 Mary buys 3 packets of sugar and 5 packets of coffee and puts them in her shopping basket, together with her purse which weighs 350 g . Weights of packets of sugar are normally distributed with mean 500 g and standard deviation 20 g . Weights of packets of coffee are normally distributed with mean 200 g and standard deviation 12 g . Find the probability that the total weight in the shopping basket is less than 2900 g .

3 sugar \(\sim N(1500, 1200)\)
5 coffee \(\sim N(1000, 720)\)
AnswerMarks Guidance
Total weight \(\sim N(2850, 1920)\) or \(\sim N(2500, 1920)\)B1, B1, M1, A1 (4 marks) For (normal dist with) correct means for both; For (normal dist with) correct variance for both; For adding their variances and means(+ purse)for coffee and sugar; For correct mean and variance for their total weight ie with or without the purse
\(P(W < 2900) = \Phi\left(\frac{2900 - 2850}{\sqrt{1920}}\right)\)
\(\Phi\left(\frac{2550 - 2500}{\sqrt{1920}}\right) = 0.873\)
AnswerMarks Guidance
or \(P(W < 2550) = \)M1, A1 (6 marks) For standardising and use of tables (consistent inclusion/exclusion of purse); For correct answer
3 sugar $\sim N(1500, 1200)$

5 coffee $\sim N(1000, 720)$

Total weight $\sim N(2850, 1920)$ or $\sim N(2500, 1920)$ | B1, B1, M1, A1 (4 marks) | For (normal dist with) correct means for both; For (normal dist with) correct variance for both; For adding their variances and means(+ purse)for coffee and sugar; For correct mean and variance for their total weight ie with or without the purse

$P(W < 2900) = \Phi\left(\frac{2900 - 2850}{\sqrt{1920}}\right)$

$\Phi\left(\frac{2550 - 2500}{\sqrt{1920}}\right) = 0.873$

or $P(W < 2550) = $ | M1, A1 (6 marks) | For standardising and use of tables (consistent inclusion/exclusion of purse); For correct answer

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3 Mary buys 3 packets of sugar and 5 packets of coffee and puts them in her shopping basket, together with her purse which weighs 350 g . Weights of packets of sugar are normally distributed with mean 500 g and standard deviation 20 g . Weights of packets of coffee are normally distributed with mean 200 g and standard deviation 12 g . Find the probability that the total weight in the shopping basket is less than 2900 g .

\hfill \mbox{\textit{CAIE S2 2002 Q3 [6]}}