   Chapter 9.3, Problem 1CP ### 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
18 views

# Find the expected value of the probability density function f ( x ) = 1 32 ( 3 x ) ( 4 − x ) on the interval [ 0 , 4 ] .

To determine

To calculate: The expected value of the probability density function f(x)=132(3x)(4x) on the interval [0,4].

Explanation

Given information:

The function is,

f(x)=132(3x)(4x)

And the interval is [0,4].

Formula used:

To find the expected value or mean of x,

μ=E(x)=abxf(x)dx

Where, f is a probability density function with a continuous random variable x in the interval [a,b]

Calculation:

Consider the provided function is,

f(x)=132(3x)(4x)

Now apply the formula μ=E(x)=abxf(x)dx to find the expected value,

μ=E(x)=04x[132(3x)(4x)]dx=13204x[(3x)(4x)]dx=13204(12x23x3)dx

Now further solve,

μ=132[(12)(x33)3(x44)]04=132[

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