IB DP Further Mathematics – 3.2 Mean of linear combinations of n random variables HL Paper 2

Question

The independent random variables X and Y are given by X ~ N\(\left( {{\mu _1},\,\sigma _1^2} \right)\) and Y ~ N\(\left( {{\mu _2},\,\sigma _2^2} \right)\).

Two independent random variables X1 and X2 each have a normal distribution with a mean 3 and a variance 9. Four independent random variables Y1, Y2, Y3, Y4 each have a normal distribution with mean 2 and variance 25. Each of the variables Y1Y2Y3Y4 is independent of each of the variables X1, X2. Find

Write down the distribution of aX + bY where a, b \( \in \mathbb{R}\).

[2]
a.

P(X1 + Y1 < 11).

[3]
b.i.

P(3X1 + 4Y1 > 15).

[4]
b.ii.

P(X1X2 + Y1 + Y2 + Y3 + Y4 < 30).

[3]
b.iii.

Given that \({\bar X}\) and \({\bar Y}\) are the respective sample means, find \({\text{P}}\left( {\bar X > \bar Y} \right)\).

[5]
c.
Answer/Explanation

Markscheme

aX + bY ~ N\(\left( {a{\mu _1} + b{\mu _2},\,\,{a^2}\sigma _1^2 + {b^2}\sigma _2^2} \right)\)     A1A1

Note: A1 for N and the mean, A1 for the variance.

[2 marks]

a.

X1 + Y∼ N(5,34)     (A1)(A1)

⇒ P(X1 + Y1 < 11) = 0.848       A1

[3 marks]

b.i.

3X1 + 4Y1 ∼ N(9 + 8, 9 × 9 + 16 × 25)     (A1)(M1)(A1)

Note: Award (A1) for correct expectation, (M1)(A1) for correct variance.

∼ N(17, 481)

⇒ P(3X1 + 4Y1 > 15) = 0.536       A1

[4 marks]

b.ii.

X1 + X2 + Y1 + Y2 + Y3 + Y4 ∼ N(6 + 8, 2 × 9 + 4 × 25)     (A1)(A1)

∼ N(14, 118)

⇒ P(X1 + X2 + Y1 + Y2 + Y3 + Y4 < 30) = 0.930       A1

[3 marks]

b.iii.

consider \(\bar X – \bar Y\)      (M1)

\({\text{E}}\left( {\bar X – \bar Y} \right) = 3 – 2 = 1\)       A1

\({\text{Var}}\left( {\bar X – \bar Y} \right) = \frac{9}{2} + \frac{{25}}{4}\left( { = 10.75} \right)\)      (M1)A1

\( \Rightarrow {\text{P}}\left( {\bar X – \bar Y > 0} \right) = 0.620\)       A1

[5 marks]

c.
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