Probability Density Functions

This lesson deepens your understanding of probability density functions (PDFs) for continuous random variables, covering how to find the mean, variance, mode, and median from a PDF, and how to apply these concepts in exam-style problems.

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Mean (Expected Value) of a Continuous Distribution

The expected value of a continuous random variable $X$X with PDF $f(x)$f(x) is:

$E(X) = \mu = \int_{-\infty}^{\infty} x \cdot f(x) \, dx$E(X)=μ=∫−∞∞​x⋅f(x)dx

More generally, for any function $g(X)$g(X):

$E(g(X)) = \int_{-\infty}^{\infty} g(x) \cdot f(x) \, dx$E(g(X))=∫−∞∞​g(x)⋅f(x)dx

Worked Example

Let $f(x) = \frac{3}{8}x^2$f(x)=83​x2 for $0 \leq x \leq 2$0≤x≤2.

$E(X) = \int_0^2 x \cdot \frac{3}{8}x^2 \, dx = \frac{3}{8}\int_0^2 x^3 \, dx = \frac{3}{8}\left[\frac{x^4}{4}\right]_0^2 = \frac{3}{8} \times 4 = \frac{3}{2} = 1.5$E(X)=∫02​x⋅83​x2dx=83​∫02​x3dx=83​[4x4​]02​=83​×4=23​=1.5

$E(X^2) = \int_0^2 x^2 \cdot \frac{3}{8}x^2 \, dx = \frac{3}{8}\int_0^2 x^4 \, dx = \frac{3}{8}\left[\frac{x^5}{5}\right]_0^2 = \frac{3}{8} \times \frac{32}{5} = \frac{12}{5} = 2.4$E(X2)=∫02​x2⋅83​x2dx=83​∫02​x4dx=83​[5x5​]02​=83​×532​=512​=2.4

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Variance of a Continuous Distribution

$\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2

Using the values from above:

$\text{Var}(X) = 2.4 - 1.5^2 = 2.4 - 2.25 = 0.15$Var(X)=2.4−1.52=2.4−2.25=0.15

$\text{SD}(X) = \sqrt{0.15} = 0.387$SD(X)=0.15​=0.387