Pre-U Pre-U 9795/2 2011 June — Question 1 3 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2011
SessionJune
Marks3
TopicLinear combinations of normal random variables
TypeComparison involving sums or multiples
DifficultyStandard +0.3 This is a straightforward application of standard results for linear combinations of independent normal random variables. Part (i) requires recalling that sums of normals are normal and applying variance rules (variance of sum = sum of variances for independent variables, variance of difference = sum of variances). Part (ii) converts to a single normal distribution and finds a probability using tables. While it involves multiple random variables, the method is entirely routine for Further Maths statistics with no problem-solving insight required beyond recognizing the standard technique.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions

1 The independent random variables \(X\) and \(Y\) have distributions \(\mathrm { N } ( 30,9 )\) and \(\mathrm { N } ( 20,4 )\) respectively.
  1. Give the distribution of $$\left( X _ { 1 } + X _ { 2 } + X _ { 3 } \right) - \left( Y _ { 1 } + Y _ { 2 } + Y _ { 3 } + Y _ { 4 } \right)$$ where \(X _ { i } , i = 1,2,3\), and \(Y _ { j } , j = 1,2,3,4\), are independent observations of \(X\) and \(Y\) respectively. The time for female cadets to complete an assault course is \(X\) minutes and the time for male cadets to complete the same assault course is \(Y\) minutes.
  2. Find the probability that the total time for three randomly chosen female cadets to complete the assault course is greater than the total time for four randomly chosen male cadets to complete the assault course.

(i) \((X_1 + X_2 + X_3) - (Y_1 + Y_2 + Y_3 + Y_4) \sim \text{N}((3 \times 30 - 4 \times 20), (3 \times 9 + 4 \times 4))\)
i.e. N(10, 43) B1, B1 [2]
(ii) \(z = \dfrac{0 - 10}{\sqrt{43}} = -1.52(5)\) (Accept 1.52 to 1.53) M1A1
\(\text{P}(X_1 + X_2 + X_3 > Y_1 + Y_2 + Y_3 + Y_4) = 0.936\) A1 [3]
(i) $(X_1 + X_2 + X_3) - (Y_1 + Y_2 + Y_3 + Y_4) \sim \text{N}((3 \times 30 - 4 \times 20), (3 \times 9 + 4 \times 4))$

i.e. N(10, 43) **B1, B1** [2]

(ii) $z = \dfrac{0 - 10}{\sqrt{43}} = -1.52(5)$ (Accept 1.52 to 1.53) **M1A1**

$\text{P}(X_1 + X_2 + X_3 > Y_1 + Y_2 + Y_3 + Y_4) = 0.936$ **A1** [3]
1 The independent random variables $X$ and $Y$ have distributions $\mathrm { N } ( 30,9 )$ and $\mathrm { N } ( 20,4 )$ respectively.\\
(i) Give the distribution of

$$\left( X _ { 1 } + X _ { 2 } + X _ { 3 } \right) - \left( Y _ { 1 } + Y _ { 2 } + Y _ { 3 } + Y _ { 4 } \right)$$

where $X _ { i } , i = 1,2,3$, and $Y _ { j } , j = 1,2,3,4$, are independent observations of $X$ and $Y$ respectively.

The time for female cadets to complete an assault course is $X$ minutes and the time for male cadets to complete the same assault course is $Y$ minutes.\\
(ii) Find the probability that the total time for three randomly chosen female cadets to complete the assault course is greater than the total time for four randomly chosen male cadets to complete the assault course.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2011 Q1 [3]}}