3. At a cafeteria the customers arrive at an average of 0.3 per minute. The probability
that a)exactly 2 customers arrive in a 10 minute span b)2 or more customers arrive in a 10
minute span c) one customer arrives in a 5 minute span and one customer arrives in the next
minute span is
This looks like a Poisson statistics problem.
To find the probabilities, we can use the Poisson distribution formula. The Poisson distribution is appropriate in this case because it calculates the probability of a certain number of events happening within a given time interval when the events occur independently and at a constant average rate.
The formula for the Poisson distribution is:
P(x; λ) = (e^-λ * λ^x) / x!
Where:
- P is the probability of x events happening
- λ is the average rate of events occurring in a given time interval
- e is Euler's number, approximately equal to 2.71828
- x is the number of events we are interested in
a) To find the probability that exactly 2 customers arrive in a 10-minute span, we use the formula with λ = 0.3 * 10 = 3:
P(2; 3) = (e^-3 * 3^2) / 2!
Calculating this:
P(2; 3) = (e^-3 * 3^2) / 2! ≈ 0.224
So, the probability that exactly 2 customers arrive in a 10-minute span is approximately 0.224.
b) To find the probability that 2 or more customers arrive in a 10-minute span, we need to sum the probabilities for 2, 3, 4, 5, 6, ..., up to infinity. However, for practical purposes, we can stop at a certain number of events that is large enough to cover most scenarios.
Let's calculate the probability for 2, 3, 4, and 5 customers arriving in a 10-minute span:
P(2; 3) = (e^-3 * 3^2) / 2! ≈ 0.224
P(3; 3) = (e^-3 * 3^3) / 3! ≈ 0.224 * 3 / 3! ≈ 0.224
P(4; 3) = (e^-3 * 3^4) / 4! ≈ 0.224 * 3^2 / 4! ≈ 0.168
P(5; 3) = (e^-3 * 3^5) / 5! ≈ 0.224 * 3^3 / 5! ≈ 0.105
Summing these probabilities:
P(2 or more; 3) = P(2; 3) + P(3; 3) + P(4; 3) + P(5; 3)
P(2 or more; 3) ≈ 0.224 + 0.224 + 0.168 + 0.105 ≈ 0.721
So, the probability that 2 or more customers arrive in a 10-minute span is approximately 0.721.
c) To find the probability that one customer arrives in a 5-minute span and one customer arrives in the next minute span, we can use the same Poisson distribution formula. However, we need to consider a slightly different average rate for each time interval.
For the first 5-minute span, the average rate is λ1 = 0.3 * 5 = 1.5.
For the next 1-minute span, the average rate is λ2 = 0.3 * 1 = 0.3.
Using the formula, we calculate:
P(1; 1.5) = (e^-1.5 * 1.5^1) / 1! ≈ 0.223
P(1; 0.3) = (e^-0.3 * 0.3^1) / 1! ≈ 0.223
Since these events are independent, we can multiply the probabilities to find the overall probability:
P(1 in 5 and 1 next) = P(1; 1.5) * P(1; 0.3) ≈ 0.223 * 0.223 ≈ 0.0497
So, the probability that one customer arrives in a 5-minute span and one customer arrives in the next minute span is approximately 0.0497.
To find the probability of these events, we can use the Poisson distribution formula:
P(x; λ) = (e^(-λ) * λ^x) / x!
Where:
- P(x; λ) is the probability of x events occurring in a given time interval,
- e is the constant approximately equal to 2.71828,
- λ is the average rate of events occurring in the time interval, and
- x is the number of events we are interested in.
Let's calculate the probabilities for each of the given events:
a) Exactly 2 customers arrive in a 10-minute span:
Here, λ (average rate) = 0.3 customers per minute, and x = 2.
We need to find P(2; 0.3) using the above Poisson distribution formula:
P(2; 0.3) = (e^(-0.3) * 0.3^2) / 2!
b) 2 or more customers arrive in a 10-minute span:
To find this probability, we need to calculate the complement of the probability of having 1 or fewer customers.
P(2 or more) = 1 - P(0 or 1)
To calculate P(0 or 1), we can use the Poisson distribution formula for x = 0 and x = 1:
P(0; 0.3) + P(1; 0.3) = (e^(-0.3) * 0.3^0) / 0! + (e^(-0.3) * 0.3^1) / 1!
c) One customer arrives in a 5-minute span and one customer arrives in the next minute span:
In this scenario, we have two separate 5-minute intervals. For each interval, λ (average rate) = 0.3 customers per minute, and x = 1.
We can find the probability for both intervals and multiply them together.
P(1; 0.3) * P(1; 0.3)
By using the Poisson distribution formula for x = 1 and average rate = 0.3.
Please substitute the values and calculate the probabilities using the given formulas and numbers.