Variance formulae; continuous random variables (pdf) practice questions.

A continuous random variable X has probability density function f(x) = k(4 − x) for 0 ≤ x ≤ 4 and f(x) = 0 otherwise. Determine the value of k that makes f a valid pdf.

• 1/8
• 1/4
• 1/16
• 8

Answer: 1/8

∫₀⁴ k(4 − x) dx = k(16 − 8) = 8k must equal 1, so k = 1/8. The total area under a pdf is always 1.

A random variable X satisfies E(X) = 3 and E(X²) = 13. Calculate Var(X).

• 10
• 4
• 2
• 13

Answer: 4

Var(X) = E(X²) − [E(X)]² = 13 − 9 = 4. Subtracting E(X) instead of its square (13 − 3 = 10) is a classic error.

X is a random variable with Var(X) = 5. Determine Var(3X − 2).

• 15
• 13
• 45
• 43

Answer: 45

Var(aX + b) = a²Var(X), so Var(3X − 2) = 9 × 5 = 45. Multiplying by 3 instead of 3² gives the distractor 15, and the constant −2 changes nothing.

A continuous random variable X has pdf f(x) = x/2 for 0 ≤ x ≤ 2 and 0 otherwise. Determine the median of X.

• 4/3
• √2
• 1
• √3

Answer: √2

The median m satisfies ∫₀^m (x/2) dx = m²/4 = 1/2, so m² = 2 and m = √2. The distractor 4/3 is the mean E(X), not the median.

A continuous random variable X has pdf f(x) = (3/32)x(4 − x) for 0 ≤ x ≤ 4 and 0 otherwise. State the mode of X.

• 2
• 4
• 16/5
• 3/32

Answer: 2

The mode is the value of x where f is greatest; f is a downward parabola with zeros at 0 and 4, so its maximum is at the vertex x = 2.

Which pair of conditions is necessary and sufficient for a function f to be a probability density function on ℝ?

• f(x) ≤ 1 for all x, and the total integral of f equals 1
• f is continuous, and f(x) ≥ 0 for all x
• The total integral of f equals 1
• f(x) ≥ 0 for all x, and the total integral of f equals 1

Answer: f(x) ≥ 0 for all x, and the total integral of f equals 1

A pdf must be non-negative and enclose total area 1; it need not be continuous, and its values may exceed 1.