Question

If the mean of a random variable which follows exponential distribution is \( \frac{1}{3} \), then the probability that the random variable \( X \) takes the values more than \( r \) is greater than \( \frac{1}{e^5} \), then

• A. \( r>\frac{5}{3} \)
• B. \( r<\frac{5}{3} \)
• C. \( r>\frac{1}{6} \)
• D. \( r<\frac{1}{6} \)

Hint:

For exponential distribution, use \( P(X>r) = e^{-\lambda r} \) and relate it directly with the given probability to find bounds on \( r \).

Answer 1

The Correct Option is B

The exponential distribution is characterized by its mean, which is given by \( \frac{1}{\lambda} \). Here, the mean is \( \frac{1}{3} \), so we have \( \frac{1}{\lambda} = \frac{1}{3} \), giving us \( \lambda = 3 \).

The probability that the random variable \( X \) exceeds a certain value \( r \) is given by the survival function: \[ P(X > r) = 1 - F(r) = e^{-\lambda r} \] where \( F(r) \) is the cumulative distribution function. Therefore, \[ P(X > r) = e^{-3r} \]

We are given that this probability is greater than \( \frac{1}{e^5} \): 
\[ e^{-3r} > \frac{1}{e^5} \]

Taking the natural logarithm on both sides: \[ -3r > -5 \]

Dividing through by \(-3\), we reverse the inequality: \[ r < \frac{5}{3} \]

Thus, the correct condition is \( r < \frac{5}{3} \).