An installation technician for a specialized communication system is dispatched to a city only when 3 or more orders have been placed. Suppose the orders follow a Poisson distribution with a mean of 0.25 per week for a city with a population of 100,000 and suppose your city contains 800,000. a. What is the probability that a technician is required after a one-week period?
b. If you are the first one in the city to place an order, what is the probability that you have to wait more than two weeks from the time you place your order until a technician is dispatched?

0.3233

0.09

Step-by-step explanation:

Given that :

Mean, λ = 0.25 for a 100,000 per week

For a population of 800,000 :

λ = 800,000 / 100,000 * 0.25 = 8 * 0.25 = 2 orders per week

Probability that technician is required after one week ;

After one week, order is beyond 2 ; hence, order, x ≥ 3

P(x ≥ 3) = 1 - [p(x=0) + p(x= 1) + p(x =2)]

P(x ≥ 3) = 1 - e^-λ(1 + 2¹/1! + 2²/2!)

P(x ≥ 3) = 1 - e^-2(1+2+2) = 1 - e^-2*5 = 1 - e^-10

P(x ≥ 3) = 1 - e^-2 * 10

P(x ≥ 3) = 1 - 0.6766764

P(x ≥ 3) = 0.3233

B.)

Mean, λ for more than 2 weeks = 2 * 2 = 4

P(x < 2) = p(x = 0) + p(x = 1)

P(x < 2) = e^-4(0 + 4^1/1!)

P(x < 2) = e^-4(0 + 4) = e^-4(5)

P(x < 2) = e^-4(5) = 0.0183156 * 5 = 0.0915

P(x < 2) = 0.09

b) 1/3 * 1/5

step-by-step explanation:

each smallest part is 1/15 since the written one is 1/5 and it's divided by 3 equal parts. (1/5 )/3 = 1/15 = 1/3 * 1/5