CAIE S2 2006 June — Question 4 6 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2006
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeDifferent variables, one observation each
DifficultyStandard +0.3 This is a straightforward application of linear combinations of normal distributions (W - D ~ N(μ_W - μ_D, σ_W² + σ_D²)) followed by a standard normal probability calculation. The 'differs by more than 3 minutes' requires recognizing this means |W - D| > 3, which translates to a two-tailed probability calculation. While it requires understanding of variance addition and absolute value interpretation, these are standard S2 techniques with no novel problem-solving required.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)

4 A certain make of washing machine has a wash-time with mean 56.9 minutes and standard deviation 4.8 minutes. A certain make of tumble dryer has a drying-time with mean 61.1 minutes and standard deviation 6.3 minutes. Both times are normally distributed and are independent of each other. Find the probability that a randomly chosen wash-time differs by more than 3 minutes from a randomly chosen drying-time.

\(D - W \sim N(4.2, 6.3^2 + 4.8^2)\)
\(P(D - W > 3) = 1 - \Phi\left(\frac{3-4.2}{\sqrt{62.73}}\right) = \Phi(0.152) = 0.560\)
\(P(D - W < -3) = \Phi\left(\frac{-3-4.2}{\sqrt{62.73}}\right) = 1 - \Phi(0.909) = 1 - 0.8182 = 0.182\)
AnswerMarks Guidance
total prob \(= 0.742\)B1 \(6.3^2 + 4.8^2\) or \(62.7\) seen
M1*considering \(P(D - W > 3)\) or considering \(P(W - D > 3)\), or \(P(D - W < -3)\) and standardising
M1 dep*correct prob area for their prob considered above ie \(> 0.5\) or \(< 0.5\) depending
M1Considering the other version not considered above
A1\(0.560\) or \(0.182\) seen
A1 (6 marks)correct answer
$D - W \sim N(4.2, 6.3^2 + 4.8^2)$

$P(D - W > 3) = 1 - \Phi\left(\frac{3-4.2}{\sqrt{62.73}}\right) = \Phi(0.152) = 0.560$

$P(D - W < -3) = \Phi\left(\frac{-3-4.2}{\sqrt{62.73}}\right) = 1 - \Phi(0.909) = 1 - 0.8182 = 0.182$

total prob $= 0.742$ | B1 | $6.3^2 + 4.8^2$ or $62.7$ seen

| M1* | considering $P(D - W > 3)$ or considering $P(W - D > 3)$, or $P(D - W < -3)$ and standardising

| M1 dep* | correct prob area for their prob considered above ie $> 0.5$ or $< 0.5$ depending

| M1 | Considering the other version not considered above

| A1 | $0.560$ or $0.182$ seen

| A1 (6 marks) | correct answer
4 A certain make of washing machine has a wash-time with mean 56.9 minutes and standard deviation 4.8 minutes. A certain make of tumble dryer has a drying-time with mean 61.1 minutes and standard deviation 6.3 minutes. Both times are normally distributed and are independent of each other. Find the probability that a randomly chosen wash-time differs by more than 3 minutes from a randomly chosen drying-time.

\hfill \mbox{\textit{CAIE S2 2006 Q4 [6]}}