# 5.1 Continuous probability functions

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This module introduces the continuous probability function and explores the relationship between the probability of X and the area under the curve of f(X).

We begin by defining a continuous probability density function. We use the function notation $f(x)$ . Intermediate algebra may have been your first formal introduction to functions. In the study of probability, the functions we study are special. We define the function $f(x)$ so that the area between it and the x-axis is equal to a probability. Since the maximum probability is one,the maximum area is also one.

For continuous probability distributions, PROBABILITY = AREA.

Consider the function $f(x)=\frac{1}{20}$ for $((0, x), \mathrm{20})$ . $x$ = a real number. The graph of $f(x)=\frac{1}{20}$ is a horizontal line. However, since $((0, x), \mathrm{20})$ , $f(x)$ is restricted to the portion between $(x, 0)$ and $(x, 20)$ , inclusive .

$f(x)=\frac{1}{20}$ for $((0, x), 20)$ .

The graph of $f(x)=\frac{1}{20}$ isa horizontal line segment when $((0, x), \mathrm{20})$ .

The area between $f(x)=\frac{1}{20}$ where $((0, x), \mathrm{20})$ and the x-axis is the area of a rectangle with base = $20$ and height = $\frac{1}{20}$ .

$((\text{AREA}, 20\cdot \frac{1}{20}), 1)$

This particular function, where we have restricted $x$ so that the area between the function and the x-axis is 1, is an example of a continuousprobability density function. It is used as a tool to calculate probabilities.

Suppose we want to find the area between $f(x)=\frac{1}{20}$ and the x-axis where $((0, x), 2)$ .

$((\text{AREA}, \left(2-0\right)\cdot \frac{1}{20}), 0.1)$

$((\left(2-0\right), 2), \text{base of a rectangle})$

$\frac{1}{20}$ = the height.

The area corresponds to a probability. The probability that $x$ is between 0 and 2 is 0.1, which can be written mathematically as $\text{P(0 .

Suppose we want to find the area between $f(x)=\frac{1}{20}$ and the x-axis where $((4, x), 15)$ .

$((\text{AREA}, \left(15-4\right)\cdot \frac{1}{20}), 0.55)$

$\left(15-4\right)=11=\text{the base of a rectangle}$

$\frac{1}{20}$ = the height.

The area corresponds to the probability $(((P\left(4, x), 15\right)), 0.55)$ .

Suppose we want to find $P\left(x=\mathrm{15}\right)$ . On an x-y graph, $x=\mathrm{15}$ is a vertical line. A vertical line has no width (or 0 width). Therefore, $P\left(x=\mathrm{15}\right)=\text{(base)}\text{(height)}=\left(0\right)\left(\frac{1}{20}\right)=0$ .

$(P\left(X, x\right))$ (can be written as $(P\left(X, x\right))$ for continuous distributions) is called the cumulative distribution function or $\mathrm{CDF}$ . Notice the "less than or equal to" symbol. We can use the $\mathrm{CDF}$ to calculate $(P\left(X, x\right))$ . The $\mathrm{CDF}$ gives "area to the left" and $(P\left(X, x\right))$ gives "area to the right." We calculate $(P\left(X, x\right))$ for continuous distributions as follows: $(((P\left(X, x\right)), 1-P\left(X), x\right))$ .

Label the graph with $\mathrm{f\left(x\right)}$ and $x$ . Scale the x and y axes with the maximum $x$ and $y$ values. $(f(x), \frac{1}{20})$ , $((0, x), \mathrm{20})$ .

$(((((P\left(2.3, x), 12.7\right)), \left(\text{base}\right)\left(\text{height}\right)), \left(12.7-2.3\right)\left(\frac{1}{20}\right)), 0.52)$

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