An installation technician for a specialized communication system is dispatched to a city only when three or
more orders have been placed. Suppose that orders follow a Poisson distribution with a mean of 0.25 per
week for a city with a population of 100,000, and suppose that your city contains a population of 800,000.
(a) What is the probability that a technician is required after a oneweek
period
To find the probability that a technician is required after a one-week period, we need to determine the probability of having three or more orders in a week.
The orders follow a Poisson distribution with a mean of 0.25 per week for a city with a population of 100,000. However, in this case, the city has a population of 800,000. To adjust for the population difference, we need to multiply the mean by the ratio of the two populations: 800,000 / 100,000 = 8.
Therefore, the mean for the adjusted population is 0.25 * 8 = 2.
Next, we calculate the probability of having three or more orders using the Poisson distribution formula:
P(X >= 3) = 1 - P(X < 3)
Using a Poisson distribution table or a calculator, we can find the probabilities for X = 0, 1, and 2 orders:
P(X = 0) = e^(-2) * (2^0) / 0! = 0.1353
P(X = 1) = e^(-2) * (2^1) / 1! = 0.2707
P(X = 2) = e^(-2) * (2^2) / 2! = 0.2707
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1353 + 0.2707 + 0.2707 = 0.6767
Finally, we calculate the probability of needing a technician:
P(Technician required after one week) = 1 - P(X < 3) = 1 - 0.6767 = 0.3233
Therefore, the probability that a technician is required after a one-week period is approximately 0.3233.
To find the probability that a technician is required after a one-week period, we need to calculate the probability of having three or more orders placed.
Let's start by calculating the average number of orders per week for the given population:
λ = (population / 100,000) * mean orders per week
= (800,000 / 100,000) * 0.25
= 2
Next, we can use the Poisson distribution formula to calculate the probability of having three or more orders in a week:
P(X >= 3) = 1 - P(X < 3)
To find P(X < 3), we can use the cumulative distribution function (CDF) for the Poisson distribution:
P(X < 3) = e^(-λ) * (λ^0 / 0! + λ^1 / 1!)
λ = 2
P(X < 3) = e^(-2) * (2^0 / 0! + 2^1 / 1!)
= e^(-2) * (1 + 2)
= e^(-2) * 3
≈ 0.1804
Finally, we can find P(X >= 3) by subtracting P(X < 3) from 1:
P(X >= 3) = 1 - P(X < 3)
= 1 - 0.1804
≈ 0.8196
Therefore, the probability that a technician is required after a one-week period is approximately 0.8196 or 81.96%.