customers arrive at the driveup window of a fastfood restaur

customers arrive at the drive-up window of a fast-food restaurant at the rate of 2 per minute during the lunch hour. a.What is the probability that the next customer will arrive within 2 minutes?

b.What is the probability that the next customer will arrive within 4 minutes?

c. during the dinner time oeriod, the arrival rate is 1 per minute. what are your answer to A and B for this period?

Solution

Possion Distribution
PMF of P.D is = f ( k ) = e- x / x!
Where   
= parameter of the distribution.
x = is the number of independent trials

Mean rate is 2 per minute during the lunch hour
a)
For 2 minute mean rate is 4

Probability that the next customer will arrive within 2 minute
P( X < 1) = P(X=0) +   
= e^-4 * 0 ^ 0 / 0! +
= 0.0183
P( X > = 1 ) = 1 - P (X < 1) = 0.9817

b)
For 4 minute mean rate is 8

P( X < 1) = P(X=0)   
= e^-8 * 0 ^ 0 / 0!
= 0.0003
P( X > = 1 ) = 1 - P (X < 1) = 0.9997

c)
For the arrival rate is 1 per minute , in 2 minute, mean rate is 2

P( X < 1) = P(X=0)   
= e^-2 * 0 ^ 0 / 0!
= 0.1353
P( X > = 1 ) = 1 - P (X < 1) = 0.8647

For the arrival rate is 1 per minute , in 4 minute, mean rate is 4
P( X < 1) = P(X=0)   
= e^-8 * 0 ^ 0 / 0!
= 0.0003
P( X > = 1 ) = 1 - P (X < 1) = 0.9997