Final answer:
The frame size to achieve a 0.1 probability of a new job arriving should be approximately 3.183 seconds based on the exponential distribution with an arrival rate of 2 jobs per minute.
Step-by-step explanation:
The student question revolves around the probability and time between events in a Poisson process, where jobs are sent to a mainframe computer at a constant rate.
Given that the jobs arrive at an average rate of 2 jobs per minute, and we want to find a frame size in which the probability of a new job arriving during each frame is 0.1.
To model this situation using exponential distribution, we can apply the formula P(X < t) = 1 - e^{-λt}, where λ represents the average rate of arrival and t is the time frame in question.
Given λ = 2 jobs/minute, we want to find the time frame t such that P(X < t) = 0.1. Solving this, we'll have:
0.1 = 1 - e^{-2t}
e^{-2t} = 0.9
-2t = ln(0.9)
t = -ln(0.9)/2
t ≈ 0.05305 minutes
Converting this time frame to seconds (since it's less than one minute), we have:
t ≈ 3.183 seconds
Thus, the frame size (or time frame) in which there is a 0.1 probability of a new job arriving is approximately 3.183 seconds.