CIE A level Math -Probability & Statistics 2: 6.2 Linear combinations of random variables : solving problems: Exam Style Questions Paper 6

Solution:-

We are given that

, and we need to find:

$$P(X_1 > 3X_2)$$

Step 1: Define the New Variable

Let

. Since

and

are independent, we use properties of normal distributions:

• Mean:
  $$E[Y] = E[X_1] – 3E[X_2] = 31.2 – 3(31.2) = -62.4$$
• Variance:
  $$\text{Var}(Y) = \text{Var}(X_1) + 9\text{Var}(X_2) = 10.4^2 + 9(10.4^2) = 10(10.4^2) = 1081.6$$
• Standard Deviation:
  $$\sigma_Y = \sqrt{1081.6} \approx 32.89$$

Step 2: Compute Probability

$$P(X_1 > 3X_2) = P(Y > 0)$$

Standardizing

:

$$Z = \frac{Y – (-62.4)}{32.89} = \frac{0 + 62.4}{32.89} \approx 1.90$$

Using normal tables:

$$P(Z > 1.90) = 1 – 0.9713 = 0.0287$$

Final Answer:

$$\boxed{0.0287}$$

Markscheme———
$E(X_1 – 3X_2) = 31.2 – 3 \times 31.2 \qquad [= -62.4]$

$Var(X_1 – 3X_2) = 10.4^2 + 3^2 \times 10.4^2 \qquad [= 1081.6]$

$\frac{0 – (-62.4)}{\sqrt{1081.6}} \qquad [= 1.897]$

$1 – \Phi(1.897)$

= 0.0289 (3 sf)