In: Statistics and Probability
A credit union is evaluating their staffing schedule to assure they have sufficient staff for their drive-up window during the lunch hour (12:00 pm to 1:00 pm). Assume the number of people who arrive at their drive-up window in a 15-minute time period during the lunch hour has a Poisson distribution with λ = 2.6.
a. What is the probability no customers will arrive between 12:15 and 12:30?
b. What is the probability fewer than 2 people will arrive between 12:15 and 12:45?
X : Number of customers will arrive between12:15 and 12:30 (15 minutes)
Given,
15-minute time period during the lunch hour has a Poisson distribution with = 2.6.
Therefore
X follows a Poisson distirbution with = 2.6.
Probability mass function of X : Probability that 'x' customer arrive between 12:15 and 12:30 is given by
probability no customers will arrive between 12:15 and 12:30 = P(X=0)
probability no customers will arrive between 12:15 and 12:30 = 0.07427
---------------------
Y : Number of customers will arrive between12:15 and 12:45 (30minutes)
Given,
15-minute time period during the lunch hour has a Poisson distribution with = 2.6.
For 30-minute time period Y = 2 x 2.6 = 5.2
Y follows a Poisson distirbution with Y = 5.2
Probability mass function of Y : Probability that 'y' customer arrive between 12:15 and 12:45 is given by
probability fewer than 2 people will arrive between 12:15 and 12:45 = P(Y<2)=P(Y=0)+P(Y=1)
P(Y<2)=P(Y=0)+P(Y=1) = 0.00552+0.02869=0.03421
probability fewer than 2 people will arrive between 12:15 and 12:45 = 0.03421