Question 1178063: Suppose that 2500 customers subscribe to a telephone exchange. There are 80 trunk lines available. Any one customer has the probability of 0.03 of needing a trunk line on a given call. Consider the situation as 2500 trials with probability of “success” p = 0.03. What is the approximate probability that the 2500 customers will “tie up” the 80 trunk lines at any given time?
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Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**Understanding the Problem**
* **Number of Customers (n):** 2500
* **Number of Trunk Lines:** 80
* **Probability of Needing a Trunk Line (p):** 0.03
* **Goal:** Find the probability that at least 81 customers need a trunk line (since there are only 80 available).
**Approximation Using Normal Distribution**
Since n is large and p is relatively small, we can approximate the binomial distribution with a normal distribution.
1. **Calculate the Mean (μ) and Variance (σ²) of the Binomial Distribution:**
* μ = np = 2500 * 0.03 = 75
* σ² = np(1 - p) = 2500 * 0.03 * 0.97 = 72.75
* σ = √72.75 ≈ 8.529
2. **Continuity Correction:**
* We want to find P(X ≥ 81), where X is the number of customers needing a trunk line.
* Since we are approximating a discrete distribution with a continuous one, we need to apply a continuity correction.
* We will use P(X > 80.5) to approximate P(X ≥ 81).
3. **Calculate the Z-score:**
* Z = (X - μ) / σ
* Z = (80.5 - 75) / 8.529 = 5.5 / 8.529 ≈ 0.6448
4. **Find the Probability Using the Standard Normal Table:**
* We want to find P(Z > 0.6448).
* From the standard normal table, P(Z ≤ 0.6448) ≈ 0.7397.
* Therefore, P(Z > 0.6448) = 1 - P(Z ≤ 0.6448) = 1 - 0.7397 ≈ 0.2603.
**Therefore, the approximate probability that the 2500 customers will "tie up" the 80 trunk lines at any given time is approximately 0.2603.**
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