| Date | May 2018 | Marks available | 3 | Reference code | 18M.2.hl.TZ0.1 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 1 | Adapted from | N/A |
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 Y1, Y2, Y3, Y4 is independent of each of the variables X1, X2. Find
Write down the distribution of aX + bY where a, b \( \in \mathbb{R}\).
P(X1 + Y1 < 11).
P(3X1 + 4Y1 > 15).
P(X1 + X2 + Y1 + Y2 + Y3 + Y4 < 30).
Given that \({\bar X}\) and \({\bar Y}\) are the respective sample means, find \({\text{P}}\left( {\bar X > \bar Y} \right)\).
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]
X1 + Y1 ∼ N(5,34) (A1)(A1)
⇒ P(X1 + Y1 < 11) = 0.848 A1
[3 marks]
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]
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]
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]
