SOLUTION: A sales firm receives , on average 3 calls per hour on its toll free number. Let x represent the number of calls received in a given hour. Assuming x has a Poisson distribution.

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Question 1186832: A sales firm receives , on average 3 calls per hour on its toll free number. Let x represent the number of calls received in a given hour. Assuming x has a Poisson distribution.
For any given hour find the probability that it will receive the following
1. At most 3 calls
2. At least 3 calls
3. 5 or more calls
4. E(X)
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Here's how to solve this problem using the Poisson distribution:
**1. Understanding the Poisson Distribution**
The Poisson distribution is used to model the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known average rate and independently of the time since the last event.
**Key Parameter:**
* λ (lambda): The average number of events (calls in this case) per interval (hour). In this problem, λ = 3.
**Formula:**
The probability of *k* events occurring in an interval is given by:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
* e is the base of the natural logarithm (approximately 2.71828)
* k is the number of events (calls)
* k! is the factorial of k
**Calculations:**
**1. At most 3 calls:**
We want to find P(X ≤ 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
* P(X=0) = (e⁻³ * 3⁰) / 0! ≈ 0.0498
* P(X=1) = (e⁻³ * 3¹) / 1! ≈ 0.1494
* P(X=2) = (e⁻³ * 3²) / 2! ≈ 0.2240
* P(X=3) = (e⁻³ * 3³) / 3! ≈ 0.2240
P(X ≤ 3) ≈ 0.0498 + 0.1494 + 0.2240 + 0.2240 ≈ 0.6472
**2. At least 3 calls:**
We want to find P(X ≥ 3) = 1 - P(X < 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]
P(X ≥ 3) = 1 - (0.0498 + 0.1494 + 0.2240) = 1 - 0.4232 ≈ 0.5768
**3. 5 or more calls:**
We want to find P(X ≥ 5) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)]
* P(X=4) = (e⁻³ * 3⁴) / 4! ≈ 0.1680
P(X ≥ 5) = 1 - (0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680) = 1 - 0.8152 ≈ 0.1848
**4. E(X):**
The expected value (mean) of a Poisson distribution is equal to λ.
E(X) = λ = 3
**Answers:**
1. At most 3 calls: ≈ 0.6472
2. At least 3 calls: ≈ 0.5768
3. 5 or more calls: ≈ 0.1848
4. E(X) = 3