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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

The standard deviation for a random variable with probability density function f and mean μ is defined by

σ = [ ( x μ ) 2 f ( x ) d x ] 1 / 2

Find the standard deviation for an exponential density function with mean μ.

To determine

To find: The standard deviation for an exponential density function with mean μ .

Explanation

Given information:

The standard deviation for a random variable with probability density function f and mean μ is σ=[(xμ)2f(x)dx]12 .

Calculation:

Show the exponential density function as follows.

f(x)={0        ifx<01μexμ ifx0 (1)

Standard deviation σ=[(xμ)2f(x)dx]12 .

Modify the above Equation.

σ2=(xμ)2f(x)dx (2)

Consider

b=1μ=c

Substitute b for 1μ in Equation (1).

f(x)={0        ifx<0beb ifx0

Find the standard deviation for an exponential density function with mean μ as shown below.

Substitute bebx for f(x) in Equation (2) and apply the limits.

σ2=0+0(xμ)2(bebx)dx=b0(x2+μ22xμ)(ebx)dx=2bμlimt0txebxdxblimt0tx2ebxdxbμ2limt0tebxdx (3)

Show the method of integration by parts as shown below:

udv=uvvdu (4)

Apply method of integration by parts for the integral xebxdx in Equation (3).

Consider u=x

Differentiate both sides of the Equation.

du=dx

Consider dv=ebxdx

Integrate both sides of the Equation.

v=1bebx

Apply method of integration by parts as shown below.

Substitute x for u, ebxdx for dv, dx for du, and 1bebx for v in Equation (4).

xebxdx=x(1bebx)(1bebx)dx=1bxebx1b2ebx=ebxb2(bx1)

Apply method of integration by parts for the integral x2ebxdx in Equation (3).

Consider u=x2

Differentiate both sides of the Equation.

du=2xdx

Consider dv=ebxdx

Integrate both sides of the Equation.

v=1bebx

Apply method of integration by parts as shown below.

Substitute x2 for u, ebxdx for dv, 2x dx for du, and 1bebx for v in Equation (4).

x2ebxdx=x2(1bebx)(1bebx)2xdx=x2bexμ2bxebxdx

Substitute ebxb2(bx1) for xebxdx in the above Equation

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