Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is passengers per minute. Required:
a. Compute the probability of no arrivals in a one-minute period (to 6 decimals).
b. Compute the probability that three or fewer passengers arrive in a one-minute period (to 4 decimals).
c. Compute the probability of no arrivals in a -second period (to 4 decimals).
d. Compute the probability of at least one arrival in a -second period (to 4 decimals).

Complete question is;

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute.

a. Compute the probability of no arrivals in a one-minute period. Round your answer to six decimal places.

b. Compute the probability that three or fewer passengers arrive in a one-minute period. Round your answer to four decimal places.

c. Compute the probability of no arrivals in a 15-second period. Round your answer to four decimal places.

d. Compute the probability of at least one arrival in a 15-second period. Round your answer to four decimal places.

A) 0.000045

B) 0.0099

C) 0.0821

D) 0.9179

Step-by-step explanation:

This is a Poisson probability question. We are given; μ = 10

Formula from Poisson probability is;

f(k) = [μ^(k) × e^(-μ)]/k!

A) probability of no arrivals in a one-minute period will be;

f(0) = [10^(0) × e^(-10)]/0!

f(0) = 0.00004539993

To six decimal places gives;

f(0) = 0.000045

B) the probability that three or fewer passengers arrive in a one-minute period is given by;

P(x ≤ 3) = f(0) + f(1) + f(2) + f(3)

We already have f(0) = 0.000045

f(1) = [10^(1) × e^(-10)]/1!

f(1) = 0.000454

f(2) = [10^(2) × e^(-10)]/2!

f(2) = 0.00227

f(3) = [10^(3) × e^(-10)]/3!

f(3) = 0.007567

Thus;

P(x ≤ 3) = 0.000045 + 0.00227 + 0.007567 = 0.009882

To four decimal places gives;

P(x ≤ 3) = 0.0099

C) no arrivals in a 15-second period means that;

μ = 10 × 15/60

μ = 2.5

Thus;

the probability of no arrivals in a 15-second period ia;

f(0) = [2.5^(0) × e^(-2.5)]/0!

f(0) = 0.082085

To four decimal places is;

f(0) = 0.0821

D) the probability of at least one arrival in a 15-second period will be gotten from the complement rule. Thus;

P(x ≥ 1) = 1 - f(0)

P(x ≥ 1) = 1 - 0.0821

P(x ≥ 1) = 0.9179

the red lines denote the population of the bacteria after 35 hours. that would be 2,409.24. the y-intercept is the blue line which is 800 because that is the starting point. the equation used to make the graph is 800(1.032)^h, where h is the number of hours.

step-by-step explanation: