Question 1178063
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.**