Call Center Probabilities

Related Concepts

Poisson Distribution Probability Mass Function (PMF) Discrete Random Variables Lambda (λ) - Average Rate Euler's Number (e) Factorials (k!) Cumulative Probability

Hint

Identify the average rate (λ): The problem states "an average of 5 calls per hour". So, λ = 5.
For "exactly 3 calls":
- This means we are looking for P(X = k) where k = 3.
- Use the Poisson formula: P(X = k) = (e^-λ * λ^k) / k!
- Plug in λ = 5 and k = 3.
- Remember 3! = 3 × 2 × 1 = 6.
For "2 or fewer (≤2) calls":
- This means we could have 0 calls, OR 1 call, OR 2 calls.
- So, you need to calculate: P(X = 0) + P(X = 1) + P(X = 2).
- Calculate each part using the Poisson formula:
  - For P(X = 0): k = 0. Remember 0! = 1 and λ⁰ = 1.
  - For P(X = 1): k = 1. Remember 1! = 1.
  - For P(X = 2): k = 2. Remember 2! = 2.
- Add the three probabilities together.
Calculator Tip: You'll need to calculate e^-5. Many calculators have an e^x button. e^-5 is approximately 0.006738.

Explanation: Call Center - Poisson Distribution

Let's imagine our call center. We know that, on average, 5 calls come in every hour. But "average" doesn't mean exactly 5 calls arrive each hour like clockwork. Some hours might have more, some less. The Poisson distribution helps us figure out the chances of these different numbers happening.

What's the chance of getting exactly 3 calls in one hour? Since the average is 5, getting 3 calls is a bit below average, so it's possible, but maybe not super likely. We'll use our special formula to find out.
What's the chance of getting 2 calls or fewer in one hour? This means we're interested if the call center gets 0 calls, OR 1 call, OR 2 calls. It's like asking, "What's the chance of a really quiet hour?" We'll find the chance for 0 calls, the chance for 1 call, and the chance for 2 calls, and then add these chances together.

The two main ingredients we need for our formula are: 1. The average rate: λ (lambda) = 5 calls per hour. 2. The specific number of calls we're interested in: k.

The number of calls, let's call it X, follows a Poisson distribution. The problem tells us the average rate is λ = 5 calls per hour. The formula to find the probability of getting exactly 'k' calls is: P(X = k) = (e^-λ * λ^k) / k!

For our calculations, we'll use the value of e^-5. Using a calculator, e^-5 ≈ 0.006738. (Remember, 'e' is a special math number, about 2.71828).

1. Probability of receiving exactly 3 calls in an hour (P(X = 3))

Here, we want to find the chance of k=3 calls, and we know λ=5.

Let's plug these into our formula: P(X = 3) = (e^-5 * 5³) / 3!

Now, let's calculate each part:

e^-5 ≈ 0.006738 (as we found out)
5³ means 5 multiplied by itself 3 times: 5 × 5 × 5 = 25 × 5 = 125.
3! (3 factorial) means 3 × 2 × 1 = 6.

Put it all together: P(X = 3) ≈ (0.006738 * 125) / 6 P(X = 3) ≈ 0.84225 / 6 P(X = 3) ≈ 0.140375

So, the probability of receiving exactly 3 calls in a given hour is approximately 0.140 (or about 14.0%).

2. Probability of receiving 2 or fewer (≤2) calls in an hour (P(X ≤ 2))

"2 or fewer calls" means we could have 0 calls, OR 1 call, OR 2 calls. To find the total chance, we calculate the probability for each case and add them up: P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2)

Let's calculate each part using λ=5:

For P(X = 0) (exactly 0 calls): k=0
- λ⁰ = 5⁰ = 1 (any number to the power of 0 is 1)
- 0! = 1 (this is a special rule for factorials)
P(X = 0) = (e^-5 * 5⁰) / 0! = (e^-5 * 1) / 1 = e^-5 P(X = 0) ≈ 0.006738
For P(X = 1) (exactly 1 call): k=1
- λ¹ = 5¹ = 5
- 1! = 1
P(X = 1) = (e^-5 * 5¹) / 1! = (e^-5 * 5) / 1 = 5 * e^-5 P(X = 1) ≈ 5 × 0.006738 ≈ 0.033690
For P(X = 2) (exactly 2 calls): k=2
- λ² = 5² = 5 × 5 = 25
- 2! = 2 × 1 = 2
P(X = 2) = (e^-5 * 5²) / 2! = (e^-5 * 25) / 2 = 12.5 * e^-5 P(X = 2) ≈ 12.5 × 0.006738 ≈ 0.084225

Now, add these probabilities together: P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2) P(X ≤ 2) ≈ 0.006738 + 0.033690 + 0.084225 P(X ≤ 2) ≈ 0.124653

So, the probability of receiving 2 or fewer calls in a given hour is approximately 0.125 (or about 12.5%).

Summary of Results

The probability of receiving exactly 3 calls in an hour is: P(X = 3) ≈ 0.140
The probability of receiving 2 or fewer calls in an hour is: P(X ≤ 2) ≈ 0.125

A Note on Rounding: The value of e^-5 is actually 0.00673794699... If you use a less precise value like e^-5 ≈ 0.0067 (as sometimes used for quick estimates), your intermediate numbers might be slightly different. For example:

P(X=0) using e^-5 ≈ 0.0067 would be 0.0067.
P(X=1) using e^-5 ≈ 0.0067 would be 5 × 0.0067 = 0.0335.
P(X=2) using e^-5 ≈ 0.0067 would be 12.5 × 0.0067 = 0.08375.

Adding these (0.0067 + 0.0335 + 0.08375) gives 0.12395, which rounds to 0.124. Using the more precise e^-5 ≈ 0.006738 (or your calculator's value) and then rounding the final sum (0.124653) leads to 0.125. Small differences in early rounding can change the final decimal places slightly. Generally, it's best to use more precision during intermediate steps and round only at the end. The values like "0.0337" or "0.0842" you might see in some tables or examples often come from rounding each P(X=k) to 4 decimal places from a more precise e^-λ. For instance, P(X=1) = 0.033690 rounds to 0.0337, and P(X=2) = 0.084225 rounds to 0.0842. Summing these (0.0067 + 0.0337 + 0.0842) gives 0.1246. Our final answers of 0.140 and 0.125 are standard results when rounding to three decimal places using sufficient precision for e^-5.

Modeling Counts of Events: The Poisson Distribution

What is it for? Events in Fixed Intervals

The Magic Formula (Poisson PMF)

Call Center - Poisson Distribution