Mean = balance point; mode = the peak: For a continuous variable the mean E(X) is the balance point of the area — the discrete sum Σ x·P becomes the integral ∫ x·f(x) dx.
The mode is the most likely region: the x where the density f is tallest. For a smooth pdf, find it by differentiating f and setting f'(x) = 0 (then checking it's a maximum), or by reading off the highest point on the curve.
IB-style question — find the mean
A continuous random variable X has pdf f(x) = (3/8)x² for 0 ≤ x ≤ 2, and 0 elsewhere.
Find E(X).
Step by step
- Multiply f by x and integrate over the support.
- Integrate: antiderivative of (3/8)x³ is (3/8)(x⁴/4) = (3/32)x⁴.
- Evaluate: 2⁴ = 16.
Final answer
E(X) = 3/2 = 1.5.
IB-style question — find the mode
A continuous random variable X has pdf f(x) = (3/4)x(2 − x) for 0 ≤ x ≤ 2, and 0 elsewhere.
Find the mode.
Step by step
- Mode = where f is tallest. Expand, then differentiate.
- Set f'(x) = 0.
- f' goes + then − across x = 1, so it's a maximum (and 1 is inside [0, 2]).
Final answer
Mode = 1.
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Median cuts the area into two equal halves: The median m is the value with half the area to its left: integrate the pdf from the bottom of the support up to m and set it equal to 0.5, then solve for m.
The variance uses the same E(X²)−[E(X)]² idea as the discrete case, but E(X²) becomes an integral: ∫ x²·f(x) dx.
IB-style question — find the median
A continuous random variable X has pdf f(x) = (3/8)x² for 0 ≤ x ≤ 2, and 0 elsewhere.
Find the median m.
Step by step
- Set the area from 0 to m equal to 0.5.
- Integrate: antiderivative is x³/8.
- Solve for m³, then take the cube root.
Final answer
Median m = ∛4 ≈ 1.59.
IB-style question — find the variance
For the same pdf f(x) = (3/8)x² on [0, 2], with E(X) = 3/2 from before.
Find Var(X).
Step by step
- First find E(X²): integrate x²·f(x).
- Integrate: antiderivative of (3/8)x⁴ is (3/40)x⁵.
- Subtract the square of the mean.
- Common denominator 20: 48/20 − 45/20.
Final answer
Var(X) = 3/20 = 0.15.