Linear transformation and combinations of random variables; unbiased estimates practice questions.

A random variable X has E(X) = 12 and Var(X) = 5. Let Y = 3X − 4. Determine Var(Y).

• 11
• 15
• 45
• 41

Answer: 45

Var(aX + b) = a²Var(X), and the constant −4 has no effect, so Var(Y) = 3² × 5 = 45.

A random variable X has E(X) = 20 and Var(X) = 9. Let Y = −2X + 7. Calculate E(Y).

• −33
• 47
• −40
• 33

Answer: −33

E(aX + b) = aE(X) + b = −2 × 20 + 7 = −33.

X and Y are independent with Var(X) = 4 and Var(Y) = 9. Determine Var(X − Y).

• 5
• 13
• −5
• 1

Answer: 13

For independent variables Var(X − Y) = Var(X) + Var(Y) = 4 + 9 = 13; variances always add, never subtract.

X and Y are independent with E(X) = 5, E(Y) = 8, Var(X) = 2, Var(Y) = 3. Let W = 2X + 3Y. Calculate Var(W).

• 13
• 35
• 5
• 34

Answer: 35

Var(2X + 3Y) = 2²Var(X) + 3²Var(Y) = 4 × 2 + 9 × 3 = 8 + 27 = 35 (independence needed).

A sample of size n = 10 gives Σx = 250 and Σx² = 6400. Determine the unbiased estimate of the population variance, to 3 s.f.

• 15.0
• 16.7
• 150
• 13.5

Answer: 16.7

Sₙ₋₁² = (Σx² − (Σx)²/n)/(n−1) = (6400 − 250²/10)/9 = (6400 − 6250)/9 = 150/9 = 16.7 (3 s.f.).

X₁ and X₂ are two independent observations of a variable X with mean μ and variance σ². Determine E(X₁ + X₂) and Var(X₁ + X₂).

• 2μ and σ²
• μ and 2σ²
• 2μ and 2σ²
• 2μ and 4σ²

Answer: 2μ and 2σ²

Means add: E = μ + μ = 2μ; for independent variables variances add: Var = σ² + σ² = 2σ².