Question

Assume that each soda can arrives at the packaging station according to a Poisson process with = 15 cans per minute, and that a batch of four cans is made instantly (taking ignorable time) by an automated machine. (a) What is the mean time until a batch is formed? (b) What is the standard deviation of the time until a batch is formed? (c) What is the probability that a batch is formed in less than 16 seconds?

Answer 1

We are given the mean arrival rate here as 15. Therefore the distribution here is given as:

X ~ Poisson (15)

a) The mean time until a batch is formed is computed as the mean time to get 4 cans which is computed here as:  

= 4/15 minute = 4*60/15 = 16 seconds. Therefore 16 seconds is the required mean time here.

b) The standard deviation of the time until a batch is formed is same as the mean time because this is an exponential waiting time process and the standard deviation for exponential distribution is same as its mean. Therefore 16 seconds is the required standard deviation here.

c) Probability that a batch is formed in less than 16 seconds is computed here as:

As the mean waiting time here is 16 seconds, therefore the distribution of the waiting time until a batch is formed is modelled here as:

$T \sim exp(1/16)$ as mean is reciprocal of parameter for exponential distribution.

16 P(T 16) = so (1/16)e-t/16 = 1-6-1 = 1-e-1 = 0.6321

Therefore 0.6321 is the required probability here.