The passengers of the area lines arrive at random and independently to the docum
ID: 3220505 • Letter: T
Question
The passengers of the area lines arrive at random and independently to the documentation section in the average frequency of arrivals is 2.0 passengers per minute.A. What is the probability of non-arrivals in a 0.90 minute interval? B. What is the probability that three or less passengers arrive in a 0.6 minute interval? C.what is the non-arrival probability in an interval of 2.25 minutes? D.what is the probability that three or less passengers will arrive in an interval of 1.85 minutes? The passengers of the area lines arrive at random and independently to the documentation section in the average frequency of arrivals is 2.0 passengers per minute.
A. What is the probability of non-arrivals in a 0.90 minute interval? B. What is the probability that three or less passengers arrive in a 0.6 minute interval? C.what is the non-arrival probability in an interval of 2.25 minutes? D.what is the probability that three or less passengers will arrive in an interval of 1.85 minutes?
A. What is the probability of non-arrivals in a 0.90 minute interval? B. What is the probability that three or less passengers arrive in a 0.6 minute interval? C.what is the non-arrival probability in an interval of 2.25 minutes? D.what is the probability that three or less passengers will arrive in an interval of 1.85 minutes? A. What is the probability of non-arrivals in a 0.90 minute interval? B. What is the probability that three or less passengers arrive in a 0.6 minute interval? C.what is the non-arrival probability in an interval of 2.25 minutes? D.what is the probability that three or less passengers will arrive in an interval of 1.85 minutes?
Explanation / Answer
given that average no of arrivals are 2 per miniute
formula of posson distribution is P(X=n) = e-n/n!
A)
for 0.9 minute = 2*0.9 = 1.8
probability of non-arrivals in a 0.90 minute interval i.e P(arriavals=0 for 0.90 minute) = e-1.81.80/0! = 0.1653
B)
for 0.6 minute = 2*0.6 = 1.2
probability that three or less passengers arrive in a 0.6 minute interva i.e P(arriavals<=3 for 0.60 minute)
=P(X=0) +P(X=1)+P(X=2)+P(X=3)
= e-1.21.20/0! +e-1.21.21/1!+e-1.21.22/2!+e-1.21.23/3!
= 0.9662
C)
for 2.25 minute = 2*2.25 = 4.5
probability of non-arrivals in a 2.25 minutes interval i.e P(arriavals=0 for 2.25 minute) = e-4.54.50/0! = 0.0111
D)
for 1.85 minute = 2*1.85 = 3.7
probability that three or less passengers arrive in a 1.85 minute interva i.e P(arriavals<=3 for 1.85 minutes)
=P(X=0) +P(X=1)+P(X=2)+P(X=3)
= e-3.73.70/0! +e-3.73.71/1!+e-3.73.72/2!+e-3.73.73/3!
= 0.4942
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