A continuous random variable X has probability density given by
f(x) = \begin{cases} 2e^{-2x}, & x > 0 \\ 0, & x \le 0 \end{cases}
The variance is:
(A) Var(X) = \frac{1}{4}
(B) Var(X) = \frac{1}{2}
(C) Var(X) = \frac{1}{8}
(D) Var(X) = \frac{1}{16}
Answer (1)
To find the variance of this continuous random variable X, which is defined by the probability density function (pdf) 0 \\ 0, & x \le 0 \end{cases}"> f ( x ) = { 2 e − 2 x , 0 , x > 0 x ≤ 0 , we follow these steps:
Understand the Probability Density Function:
The given pdf indicates that X is an exponentially distributed random variable with the parameter λ = 2 . This type of distribution is defined only for 0"> x > 0 , and the exponential distribution is often used to model the time between events in a Poisson process.
Identify the Mean μ :
For an exponential distribution with parameter λ , the mean is given by μ = λ 1 .
Thus, μ = 2 1 .
Determine the Variance Var ( X ) :
The variance of an exponentially distributed random variable with parameter λ is Var ( X ) = λ 2 1 .
For this function, Var ( X ) = 2 2 1 = 4 1 .
Conclusion:
The variance of the random variable X is 4 1 .
Therefore, the correct multiple choice answer is (A) Var(X) = \frac{1}{4}.
This solution involves identifying the type of distribution and utilizing the known formulas for mean and variance of an exponential distribution.
