Probability is the area under the curve: For a continuous variable (like a waiting time or a length), single values have probability 0 — instead probability lives in areas under a curve f(x).
Two rules make f(x) a genuine probability density function:
• it is never negative (an area below the axis would be a negative probability), and
• the total area under it is exactly 1 (something is certain to happen).
The probability that X lands in an interval is just the area of that strip.
IB-style question — probability as an area
A continuous random variable X has pdf f(x) = (3/8)x² for 0 ≤ x ≤ 2, and f(x) = 0 elsewhere.
Find P(0 ≤ X ≤ 1).
Step by step
- The probability is the area, so integrate f from 0 to 1.
- Integrate: the antiderivative of x² is x³/3, so (3/8)(x³/3) = x³/8.
- Evaluate at the limits.
Final answer
P(0 ≤ X ≤ 1) = 1/8 = 0.125.
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Integrate over the whole support, set it equal to 1: Many exam pdfs come with an unknown constant, e.g. f(x) = kx² on some interval. To find k:
integrate f over the interval where it's non-zero, set the answer equal to 1, and solve. The total area must be 1, so this always gives one equation in one unknown.
This is the single most common opening part of a continuous-distribution question.
IB-style question — find k
A continuous random variable X has pdf f(x) = kx for 0 ≤ x ≤ 4, and f(x) = 0 elsewhere.
Find the value of k.
Step by step
- The total area must be 1: integrate kx from 0 to 4 and set equal to 1.
- Integrate: the antiderivative of kx is k·x²/2.
- Evaluate at the limits.
- Solve for k.
Final answer
k = 1/8 = 0.125.
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IB-style question — k with a quadratic pdf
A continuous random variable X has pdf f(x) = kx² for 0 ≤ x ≤ 3, and 0 elsewhere.
Find k.
Step by step
- Set the total area to 1.
- Integrate: antiderivative of kx² is k·x³/3.
- Evaluate.
- Solve.
Final answer
k = 1/9.