Using standard approach to simulating random variables that follow negative exponential distribution with an expected time of 3 minutes, what would be the simulated time (in minutes) if 30 is the two-digit random number picked from the random number table? $\left(T=-\frac{1}{\lambda} \ln (\cdot R N)\right)$
A) 0.060199
B) 0.4013
C) 3.6119
D) 24.079

Question

Quizplus · Accepted Answer

The correct answer is C.The expected time is 3 minutes, so the rate is $\lambda = \frac{1}{3}$. The two-digit random number is 30, so the simulated time is $T=-\frac{1}{\lambda} \ln (\cdot R N)=-\frac{1}{\frac{1}{3}} \ln (0.30) = 3.6119$.

Using Standard Approach to Simulating Random Variables That Follow Negative (T=−1λln⁡(⋅RN))\left(T=-\frac{1}{\lambda} \ln (\cdot R N)\right)(T=−λ1​ln(⋅RN))

Using Standard Approach to Simulating Random Variables That Follow Negative $\left(T=-\frac{1}{\lambda} \ln (\cdot R N)\right)$