A continuous random variable X has PDF f(x) = 3x² for 0 ≤ x ≤ 1. What is f(0.9)?
AThis is impossible — a probability value can never exceed 1
B2.43, which is the probability of X equaling 0.9
C2.43, which is a valid density value even though it exceeds 1
D0.81, after normalizing f(0.9) to ensure it stays below 1
f(0.9) = 3(0.81) = 2.43. This is perfectly valid — f(x) is a *density*, not a probability, so it can exceed 1. The constraint on a PDF is not that f(x) ≤ 1; it is that f(x) ≥ 0 everywhere and that the total integral ∫f(x)dx = 1. Option A is the classic misconception: confusing a density value with a probability. A density measures probability per unit length, and there is no reason that rate must stay below 1.
Question 2 Multiple Choice
A classmate computes P(X = 2.7) for a continuous random variable by evaluating f(2.7). What is the correct answer, and why is f(2.7) wrong?
Af(2.7) is correct; the PDF gives the probability at each point for continuous variables
BP(X = 2.7) = 0; probability at a single point equals the integral over zero width, which is zero
CP(X = 2.7) = F(2.7), the CDF evaluated at 2.7
DP(X = 2.7) = f(2.7)·Δx for some small Δx around 2.7
For any continuous random variable, P(X = c) = 0 for every specific value c. Probability is obtained by integrating the PDF over an interval: P(a ≤ X ≤ b) = ∫ₐᵇ f(x)dx. The integral over a single point has zero width and therefore zero area — giving zero probability. f(x) is a density, measuring how rapidly probability accumulates near a point, not the probability itself. Evaluating f(2.7) gives a density value that has no direct interpretation as a probability.
Question 3 True / False
For a continuous random variable X, P(X ≤ 3) = P(X < 3).
TTrue
FFalse
Answer: True
For a continuous distribution, P(X = 3) = 0, so including or excluding the endpoint makes no difference: P(X ≤ 3) = P(X < 3) + P(X = 3) = P(X < 3) + 0 = P(X < 3). This contrasts with discrete random variables, where P(X = 3) can be positive and the distinction between ≤ and < matters significantly.
Question 4 True / False
If f(x) is the PDF of a continuous random variable X, then f(x) represents the probability that X equals x.
TTrue
FFalse
Answer: False
f(x) is a probability *density*, not a probability. Probabilities are obtained by integrating the PDF over an interval: P(a ≤ X ≤ b) = ∫ₐᵇ f(x)dx. At any single point, f(x) represents probability per unit length in the neighborhood of x — a rate, not an amount. Consequently, f(x) can take any non-negative value, including values greater than 1, as long as the total integral equals 1.
Question 5 Short Answer
Explain why the probability that a continuous random variable X takes any specific value c is exactly zero, and what this means for computing probabilities in practice.
Think about your answer, then reveal below.
Model answer: P(X = c) = 0 because probability comes from integrating the PDF over an interval. The integral from c to c has zero width, giving zero area and therefore zero probability. This doesn't mean c is impossible — it means a continuous variable has uncountably many possible values and probability is spread continuously rather than concentrated at points. In practice, we always compute P(a ≤ X ≤ b) = ∫ₐᵇ f(x)dx; asking for the probability at a single point is the wrong question for a continuous distribution.
This is the fundamental distinction between continuous and discrete distributions. A discrete variable concentrates probability mass at specific values. A continuous variable spreads probability across intervals, and any single point gets zero. The PDF encodes density — how rapidly probability accumulates near a point — not probability itself. This is why computing P(X = 2.7) using f(2.7) is a category error: you need an integral, not a function evaluation.