   # Identifying Probability Density Functions In Exercises 25-30, use a graphing utility to graph the function. Then determine whether the function f represents a probability density function over the given interval. If f is not a probability density function, identify the condition(s) that is (are) not satisfied. f ( x ) = 1 12 , [ 0 , 12 ] ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
Publisher: Cengage Learning
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
Publisher: Cengage Learning
ISBN: 9781305860919
Chapter 9, Problem 25RE
Textbook Problem
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## Identifying Probability Density Functions In Exercises 25-30, use a graphing utility to graph the function. Then determine whether the function f represents a probability density function over the given interval. If f is not a probability density function, identify the condition(s) that is (are) not satisfied. f ( x ) = 1 12 ,        [ 0 , 12 ]

To determine

To graph: The function f(x)=112 using a graphing utility and determine whether the function f(x)=112 represents a probability density function over the interval [0,12].

### Explanation of Solution

Given Information:

The function f(x)=112 over the interval [0,12].

Graph:

Now, consider the provided function f(x)=112.

Use the ti-83 graphing calculator to construct the graph of the function.

Step 1: Open the ti-83 graphing calculator.

Step 2: Press [y=] and enter the function Y1=112.

Step 3: Press the [window] key and adjust the scale.

Xmin=0Xmax=12Xscl=1

And,

Ymin=0Ymax=212Yscl=0.01

Step 4: Press the [graph] key.

The graph of the function f(x)=112 is obtained as,

Consider the provided function f(x)=112 over the interval [0,12].

A function f of a continuous random variable x in the interval [a,b] is a probability density function if it is nonnegative and continuous on the interval [a,b].

And,

abf(x) dx=1

The provided function is a constant function and so the function f is continuous and nonnegative for all values on the interval [0,12].

Now, show that abf(x) dx=1

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