OCR S3 2007 January — Question 1 6 marks

Exam BoardOCR
ModuleS3 (Statistics 3)
Year2007
SessionJanuary
Marks6
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Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeLinear transformation to achieve target parameters
DifficultyStandard +0.3 This is a straightforward application of linear combinations of normal distributions. Part (i) requires simple algebra with expectations, part (ii) uses standard formulas for variance of independent variables, and part (iii) asks for a brief conceptual comment. All techniques are routine for S3 level with no novel problem-solving required.
Spec5.04b Linear combinations: of normal distributions

1 The marks obtained by a randomly chosen student in the two papers of an examination are denoted by the random variables \(X\) and \(Y\), where \(X \sim \mathrm {~N} ( 45,81 )\) and \(Y \sim \mathrm {~N} ( 33,63 )\). The student's overall mark for the examination, \(T\), is given by \(T = X + \lambda Y\), where the constant \(\lambda\) is chosen such that \(\mathrm { E } ( T ) = 100\).
  1. Show that \(\lambda = \frac { 5 } { 3 }\).
  2. Assuming that \(X\) and \(Y\) are independent, state the distribution of \(T\), giving the values of its parameters.
  3. Comment on the assumption of independence.

AnswerMarks Guidance
(i) \(E(T) = E(X) + \lambda E(Y) \Rightarrow 100 = 45 + 33\lambda \Rightarrow \lambda = \frac{5}{3}\) AGM1, A1 Use \(E(X + \lambda Y)\); 2 aef
(ii) \(\text{Var}(T) = \text{Var}(X) + (\lambda)^2\text{Var}(Y) = 256\); \(T \sim N(100, 256)\)M1, A1, B1√ 3 ft variance
(iii) Same student for \(X\) and \(Y\) so independence unlikelyB1 1 Sensible reason
**(i)** $E(T) = E(X) + \lambda E(Y) \Rightarrow 100 = 45 + 33\lambda \Rightarrow \lambda = \frac{5}{3}$ AG | M1, A1 | Use $E(X + \lambda Y)$; 2 aef

**(ii)** $\text{Var}(T) = \text{Var}(X) + (\lambda)^2\text{Var}(Y) = 256$; $T \sim N(100, 256)$ | M1, A1, B1√ | 3 ft variance

**(iii)** Same student for $X$ and $Y$ so independence unlikely | B1 | 1 Sensible reason
1 The marks obtained by a randomly chosen student in the two papers of an examination are denoted by the random variables $X$ and $Y$, where $X \sim \mathrm {~N} ( 45,81 )$ and $Y \sim \mathrm {~N} ( 33,63 )$. The student's overall mark for the examination, $T$, is given by $T = X + \lambda Y$, where the constant $\lambda$ is chosen such that $\mathrm { E } ( T ) = 100$.\\
(i) Show that $\lambda = \frac { 5 } { 3 }$.\\
(ii) Assuming that $X$ and $Y$ are independent, state the distribution of $T$, giving the values of its parameters.\\
(iii) Comment on the assumption of independence.

\hfill \mbox{\textit{OCR S3 2007 Q1 [6]}}