   Chapter 9.2, Problem 4E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

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

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Verifying a Probability Density Function In Exercises 1-6, show that the function is a probability density function over the given interval. See Examples 1 and 2. f ( x ) = 4 27 x ( 3 − x ) 2 , [ 0 , 2 ]

To determine

To prove: The function, f(x)=427x(3x)2 is a probability density function over the interval [0,3].

Explanation

Given Information:

The function, f(x)=427x(3x)2 over the interval [0,3].

Formula used:

Consider a function f of a continuous random variables x whose set of values is the interval [a,b] The function is a probability density function when it is non-negative and continuous on the interval [a,b] and when

abf(x)dx=1

Proof:

Consider the function,

f(x)=427x(3x)2=427x[(3)22(3)(x)+(x)2]=427x[96x+x2]=427[9x6x2+x3]

Further simplify the above equation:

f(x)=4279x4276x2+427x3=43x89x2+427x3

f(x) is expressible as the sum or difference of different polynomial functions.

All polynomial functions are non-negative and continuous over a bounded interval.

Thus, f(x) is non-negative and continuous over the interval [0,3]

A function f of a continuous random variables x whose set of values is the interval [a,b]. The function is a probability density function when it is non-negative and continuous on the interval [a,b] and when

abf(x)dx=1

Now, evaluate abf(x)dx as,

03f(x)dx=0343xdx0389

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