A student calculates a PDF for a continuous random variable and finds f(1.5) = 2.5. The student concludes the probability P(X = 1.5) = 2.5, which is impossible since probabilities can't exceed 1. What went wrong in their reasoning?
AThe student made an arithmetic error; a valid PDF value can never exceed 1
Bf(x) is a density, not a probability; it can legitimately exceed 1, and probability requires integration over an interval
CThe variable must be discrete, not continuous, since the PDF exceeds 1
DThe student should have used the CDF instead, since PDFs only apply to symmetric distributions
f(x) is a probability density, not a probability. The PDF can exceed 1 — for example, a Uniform distribution on [0, 0.5] has f(x) = 2. What matters is that the PDF integrates to 1 over its entire domain. To find any probability, you must integrate: P(a ≤ X ≤ b) = ∫ₐᵇ f(x) dx. Reading probability directly off the y-axis is the defining misconception when moving from discrete to continuous distributions.
Question 2 Multiple Choice
For a continuous random variable X with PDF f(x), what is the exact value of P(X = 3.0)?
Af(3.0) — the PDF value at that point
BF(3.0) — the CDF evaluated at 3.0
C0, because the probability of any single exact value is zero for a continuous distribution
DUndefined, because continuous distributions have no probability assigned to individual points
For any continuous random variable, P(X = c) = ∫_c^c f(x) dx = 0, regardless of how high f(c) is. This is not undefined — it is exactly zero. There are infinitely many possible values, so no single point can hold positive probability. This is why option D is wrong: the probability exists, it's just 0. This also explains why P(a < X < b) = P(a ≤ X ≤ b) for continuous variables — the endpoints contribute nothing.
Question 3 True / False
If f(5) > f(3) for some PDF, then the probability of getting exactly X = 5 is greater than the probability of getting exactly X = 3.
TTrue
FFalse
Answer: False
Both P(X = 5) and P(X = 3) are exactly 0 for any continuous random variable, regardless of the PDF values. A higher PDF value at x = 5 means probability is more densely concentrated near 5 — you are more likely to fall in a small interval around 5 than an equally-sized interval around 3 — but you cannot read probability from individual PDF values. Probability requires integration over an interval.
Question 4 True / False
A valid probability density function must be non-negative everywhere and must integrate to 1 over its entire domain.
TTrue
FFalse
Answer: True
These are the two necessary and sufficient conditions for a valid PDF. Non-negativity ensures that no region has 'negative probability.' Integrating to 1 ensures that the total probability across all possible outcomes equals 1, consistent with certainty that the random variable takes some value. Note that f(x) can exceed 1 at individual points — only the total area is constrained to equal 1.
Question 5 Short Answer
Explain why you cannot calculate the probability that a continuous random variable equals exactly 3.7 by reading the value f(3.7) from the PDF.
Think about your answer, then reveal below.
Model answer: f(3.7) is a probability density, not a probability. For continuous variables, probability is defined as area under the curve, not height. Since a single point has zero width, the area above any single point is zero — so P(X = 3.7) = 0, regardless of how large f(3.7) is. To find probability, you must integrate f(x) over an interval.
The density analogy helps here: f(x) describes how thick the pile of probability is at each point, like sand on a table. The height of the pile at one spot tells you relative concentration, but the amount of sand in a single geometric point (no width) is zero. Probability requires a region with positive width. This is the fundamental shift from discrete to continuous probability — from summing function values to integrating them.