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

10.416 m/s

step-by-step explantion:

distance=100m

time= 9.6s

average speed=100/9.6

=10.416 m/s

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6   = 0.00000180

1.80 x 10-4   = 0.000180 (bigger, or "less small", than the former one)

so

1.8 x 10-2       = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.