   Chapter 9.3, Problem 2CP ### 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
1 views

# Find the variance and standard deviation of the probability density function in Checkpoint 1. 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 variance and standard deviation of the probability density function f(x)=132(3x)(4x) on the interval [0,4].

Explanation

Given information:

The function is,

f(x)=6343x(7x)

And the interval is [0,7].

Formula used:

To find the expected value or mean of x,

μ=E(x)=abxf(x)dx

The variance of x is,

V(x)=ab(xμ)2f(x)dx

The standard deviation of x is,

σ=V(x)

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

Calculation:

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[(12)(x33)3(x44)]04=132[4x33x44]04=132[4(40)33(40)44]

Further solve,

μ=132[4(40)33(40)44]=132=2

Now draw the function f(x)=132(3x)(4x) by using ti-83 calculator,

Step1: Press ON key

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