   Chapter 9.3, Problem 16E ### 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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# Using two Methods In Exercises 13–16, find the variance of the probability density function using two methods. f ( x ) = 2 3 − 2 9 x ,   [ 0 ,   3 ]

To determine

To calculate:

The variance of the probability density function f(x)=2329x by the use of two methods in the interval [0,3].

Explanation

Given Information:

The provided probability density function is f(x)=2329x in the interval [0,3].

Formula used:

For any probability density function f of a continuous random variable x for the interval [a,b], the variance of x is given as,

V(x)=ab(xu)2f(x)dx

Where, μ is the mean of x.

The alternative formula used to find the variance is given as,

V(x)=abx2f(x)dxu2

The formula used for expected value or mean is given as,

μ=abxf(x)dx

Calculation:

Consider the probability density function, f(x)=2329x

Now, first calculate the mean μ of x by the use of formula,

μ=abxf(x)dx

Substitute f(x)=2329x and a=0, b=3 in the above formula,

μ=03x[2329x]dx=03(23x29x2)dx=[23×2x229×3x3]03=[2×962×2727]0

Simplify the above equation,

μ=32=1

Thus, the mean is 6527.

Consider the formula used for finding the variance,

V(x)=ab(xu)2f(x)dx

Substitute f(x)=2329x, mean μ=1 and a=0, b=3 in the above equation,

V(x)=03(x1)2(2329x)dx=03[(x2+12x)(2329x

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