Mean = balance point; mode = the peak: A continuous variable has a mean and a mode, just like a discrete one. Each needs a calculus step.
| Measure | Meaning | How to find it |
|---|---|---|
| Mean E(X) | The balance point of the area. | The discrete sum Σ x·P becomes the integral ∫ x·f(x) dx. |
| Mode | The most likely region: the x where the density f is tallest. | For a smooth pdf, differentiate f and set f'(x) = 0 (then check it's a maximum), or read 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.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
Median cuts the area into two equal halves: The median m is the value with half the area to its left. The variance follows the same pattern as the discrete case.
How to find each
- Median: integrate the pdf from the bottom of the support up to m and set it equal to 0.5, then solve for m.
- Variance: use 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.