A woman is seated beside a conveyer belt and her job is to r

A woman is seated beside a conveyer belt, and her job is to remove items from belt. She has a narrow line of vision and can get these items only when they are right in front of her. Items arrive on the belt according to a Poisson process with Lambda 2 per minute. The woman and convery belt both begin working at 12:00:00 noon.

A) What is the probability that exactly 3 items arrive in the woman’s first 2 minutes of work?

B) If the woman takes a 5 minute break, what is the expected value of the number of items that will arrive on the belt in her absence?

Solution

a)

Note that the probability of x successes out of n trials is          
          
P(x) = u^x e^(-u) / x!          
          
where          
          
u = the mean number of successes = 2 items/min * 2min =    4      
          
x = the number of successes =    3      
          
Thus, the probability is          
          
P (    3   ) =    0.195366815 [ANSWER]

************************

b)

E(x) = 2 per minute * 5 minutes = 10 items [ANSWER]