This guide explains the probability density function (PDF) concept, starting from frequency histograms to probability histograms. Includes solved examples with detailed calculus-based solutions.
We use 2560 time measurements (in hours) for project completion—a continuous random variable. The frequency histogram (class width = 1) shows data distribution.
Convert frequencies to probabilities by dividing by 2560. Each rectangle's area equals the probability for that interval. Total area = 1.
Smaller class widths produce smoother distributions. The red trendline approximates the underlying probability density function.
As class width approaches zero, the histogram approximates a continuous curve—the probability density function.
For a continuous random variable \(X\) with PDF \(f_X(x)\):
A random variable \(X\) has PDF:
\[ f_X(x) = \begin{cases} \dfrac{k}{b-a}, & a \le x \le b \\[1em] 0, & x < a \text{ or } x > b \end{cases} \]where \(k > 0\) is constant.
a) Graph is a rectangle between \(x=a\) and \(x=b\) with height \(\frac{k}{b-a}\):
b) Area under PDF must equal 1. The rectangular area is:
\[ \text{Area} = (b-a) \times \frac{k}{b-a} = k \]Setting \(k = 1\) gives total area = 1. Thus:
\[ f_X(x) = \begin{cases} \dfrac{1}{b-a}, & a \le x \le b \\[1em] 0, & \text{otherwise} \end{cases} \]c) Using integration:
\[ \begin{aligned} \int_{-\infty}^{\infty} f_X(x) \, dx &= \int_a^b \frac{1}{b-a} \, dx \\ &= \left[ \frac{x}{b-a} \right]_a^b \\ &= \frac{b}{b-a} - \frac{a}{b-a} = 1 \end{aligned} \]d) For \(a=3, b=8\):
\[ \begin{aligned} P(4 \le X \le 7) &= \int_4^7 \frac{1}{8-3} \, dx \\ &= \int_4^7 \frac{1}{5} \, dx \\ &= \left[ \frac{x}{5} \right]_4^7 \\ &= \frac{7}{5} - \frac{4}{5} = \frac{3}{5} = 0.6 \end{aligned} \]A random variable \(X\) has PDF:
\[ f_X(x) = \begin{cases} e^{-kx}, & x \ge 0 \\ 0, & x < 0 \end{cases} \]where \(k > 0\).
a) Total area must be 1:
\[ \begin{aligned} \int_{-\infty}^{\infty} f_X(x) \, dx &= \int_0^{\infty} e^{-kx} \, dx \\ &= \lim_{b \to \infty} \left[ -\frac{1}{k} e^{-kx} \right]_0^b \\ &= \lim_{b \to \infty} \left( -\frac{1}{k} e^{-kb} + \frac{1}{k} \right) \\ &= \frac{1}{k} \end{aligned} \]Set \(\frac{1}{k} = 1 \Rightarrow k = 1\).
b) Find \(a\) for \(P(0 \le X \le a) = 0.9\):
\[ \begin{aligned} P(0 \le X \le a) &= \int_0^a e^{-x} \, dx \\ &= \left[ -e^{-x} \right]_0^a \\ &= -e^{-a} + 1 \end{aligned} \]Set equal to 0.9:
\[ \begin{aligned} -e^{-a} + 1 &= 0.9 \\ e^{-a} &= 0.1 \\ -a &= \ln(0.1) \\ a &= -\ln(0.1) \approx 2.3026 \end{aligned} \]