What is the variance of a probability distribution function of the form: #f(x)=ke^(2x)#?
The distribution is an exponential distribution. k = 2 and E (x) = 1/2 ,
E (
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To find the variance of a probability distribution function of the form ( f(x) = ke^{2x} ), you first need to determine the value of ( k ) and then apply the formula for variance.

Determining the value of ( k ): Since ( f(x) ) is a probability distribution function, the total area under the curve must equal 1. Thus, you need to find the value of ( k ) that makes the integral of ( f(x) ) over its entire domain equal to 1.

Integral of ( f(x) ) over its domain: The domain of ( f(x) ) depends on the range of ( x ). Assuming ( x ) ranges from ( 0 ) to ( +\infty ), integrate ( f(x) ) from ( 0 ) to ( +\infty ) and set the result equal to 1 to find ( k ):
[ \int_{0}^{\infty} ke^{2x} dx = 1 ]

Solve for ( k ): [ k\int_{0}^{\infty} e^{2x} dx = 1 ] [ k \left( \frac{1}{2} e^{2x} \right) \Bigg_{0}^{\infty} = 1 ] [ k \left( 0  \frac{1}{2} \right) = 1 ] [ k \left( \frac{1}{2} \right) = 1 ] [ k = 2 ]
Now that you have found the value of ( k ), which is ( 2 ), you can proceed to find the variance.

Formula for variance: The variance ( \sigma^2 ) of a probability distribution function ( f(x) ) is calculated using the formula:
[ \sigma^2 = \int_{\infty}^{\infty} (x  \mu)^2 f(x) dx ]
Where ( \mu ) is the mean of the distribution.

Calculate the mean ( \mu ): To calculate the mean of ( f(x) ), integrate ( xf(x) ) over its domain:
[ \mu = \int_{0}^{\infty} x(2e^{2x}) dx ]

Calculate the variance: Substitute the value of ( k ) into ( f(x) ) and compute the integral to find the mean ( \mu ). Then, use the formula for variance to find ( \sigma^2 ).
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