Question

The probability density function of a continuous random variable is \[ f(x)= \begin{cases} \dfrac35 e^{-3x/5}, & x>0 \\ 0, & x\le0 \end{cases} \] The mean of the distribution is

• A. \(1\)
• B. 5/3
• C. \(2\)
• D. \(3\)

Hint:

For exponential distribution \[ f(x)=\lambda e^{-\lambda x}, \] remember instantly: \[ \text{Mean}=\frac1\lambda, \qquad \text{Variance}=\frac1{\lambda^2}. \]

Answer 1

The Correct Option is B

Concept:& nbsp;
For a continuous random variable with probability density function \(f(x)\), the mean (expected value) is given by \[ E(X)=\int_{-\infty}^{\infty} x\,f(x)\,dx. \] Given \[ f(x)= \begin{cases} \dfrac{3}{5}e^{-3x/5}, & x>0,\\[6pt] 0, & x\le 0, \end{cases} \] we need to evaluate \[ E(X)=\int_{0}^{\infty} x\left(\frac{3}{5}e^{-3x/5}\right)dx. \]& nbsp;

Step 1: Substitute the density function into the mean formula.
\[ E(X)=\frac{3}{5}\int_{0}^{\infty} xe^{-3x/5}\,dx. \] Let \[ a=\frac{3}{5}. \] Then \[ E(X)=a\int_{0}^{\infty} xe^{-ax}\,dx. \]& nbsp;

Step 2: Evaluate the integral using integration by parts.
Let \[ u=x, \qquad dv=e^{-ax}dx. \] Then \[ du=dx, \qquad v=-\frac{1}{a}e^{-ax}. \] Applying integration by parts, \[ \int xe^{-ax}dx = -\frac{x}{a}e^{-ax} +\frac{1}{a}\int e^{-ax}dx. \] Since \[ \int e^{-ax}dx = -\frac{1}{a}e^{-ax}, \] we get \[ \int xe^{-ax}dx = -\frac{x}{a}e^{-ax} -\frac{1}{a^2}e^{-ax}. \] Evaluating from \(0\) to \(\infty\), \[ \int_{0}^{\infty} xe^{-ax}dx = \left[ -\frac{x}{a}e^{-ax} -\frac{1}{a^2}e^{-ax} \right]_{0}^{\infty}. \] As \(x\to\infty\), \[ e^{-ax}\to 0, \] so the upper limit becomes \(0\). At \(x=0\), \[ -\frac{0}{a}e^0-\frac{1}{a^2}e^0 = -\frac{1}{a^2}. \] Hence, \[ \int_{0}^{\infty} xe^{-ax}dx = 0-\left(-\frac{1}{a^2}\right) = \frac{1}{a^2}. \]& nbsp;

Step 3: Substitute \(a=\frac{3}{5}\).
\[ E(X) = a\left(\frac{1}{a^2}\right) = \frac{1}{a}. \] Therefore, \[ E(X) = \frac{1}{\frac{3}{5}} = \frac{5}{3}. \]& nbsp;

Alternative Observation.
The given density function is of the exponential form \[ f(x)=\lambda e^{-\lambda x}, \] where \[ \lambda=\frac{3}{5}. \] For an exponential distribution, \[ E(X)=\frac{1}{\lambda}. \] Thus, \[ E(X)=\frac{1}{3/5}=\frac{5}{3}. \]& nbsp;

Final Answer: \[ \boxed{\frac{5}{3}} \]