Question

1. A study of the local grocery checkout lines between 4 and 7 pm on Thursdays revealed there is an average of 6 customers waiting in line.

1. Thursday between 4 and 7 pm, what is the probability that 4 customers will be waiting?

2. What is the probability that 5 or more customers will be waiting on Thursday between 4 and 7 pm?

Answer 1

a)

Data given:
rate = 6, P(y = 4) = ?;

Given data indicates that this is a Poisson distribution;

P(y = 4) = ((rate)^(y)/y!)e^(-rate)
                = ((6)^(4)/24)e^(-6)
                = 0.134

b)

P(y>=5)=P(y > 4) = 1 - P(y = 0) - P(y = 1) - P(y = 2) - P(y = 3) - P(y = 4)
                = 1 - 0.002 - 0.015 - 0.045 - 0.089 - 0.134
                = 0.715