Each year (365 days) the fire department receives 270 emergency calls on average. In a give 7-day week, what is the probability of receiving
A. no calls
b. more than one call
c. one call or less
d. at least two calls
please help and show me some work or how to do it so i can learn from this.
Thanks!
To find the probabilities, we'll use the concept of a Poisson distribution. In a given time period, the Poisson distribution is often used to model the number of events that occur. The Poisson distribution can be described by a single parameter, which in this case is the average number of events.
The average number of calls per day is given as 270 divided by 365:
λ = 270 / 365
Now, let's answer each part of the question using this information.
A. Probability of receiving no calls in a 7-day week:
The Poisson distribution can be used to calculate the probability of receiving a specific number of calls. The probability of receiving a specific number of calls in a given time period is given by the Poisson formula:
P(x, λ) = (e^-λ * λ^x) / x!
For the case of zero calls, we just substitute x = 0 into the formula:
P(0, λ) = (e^-λ * λ^0) / 0!
Since 0! = 1, the formula simplifies to:
P(0, λ) = e^-λ
Plugging the average value of λ into the formula, we can calculate the probability of no calls in a 7-day week.
B. Probability of receiving more than one call:
To find the probability of receiving more than one call, we need to find the cumulative probability of zero and one call and subtract that from 1.
C. Probability of receiving one call or less:
This is the cumulative probability of zero and one call.
D. Probability of receiving at least two calls:
To find the probability of receiving at least two calls, we need to subtract the cumulative probability of zero and one call from 1.
Now, let's go step by step in calculating these probabilities.
A. Probability of receiving no calls:
P(0, λ) = e^-λ
B. Probability of receiving more than one call:
P(x ≤ 1, λ) = 1 - (P(0, λ) + P(1, λ))
C. Probability of receiving one call or less:
P(x ≤ 1, λ) = P(0, λ) + P(1, λ)
D. Probability of receiving at least two calls:
P(x ≥ 2, λ) = 1 - P(x ≤ 1, λ)
Now, let's calculate the probabilities using the given average number of calls per day.
We first need to calculate the value of λ:
λ = 270 / 365
Then, we can use this value to calculate the probabilities for each case.