Question 1171244
Let's break down this problem step-by-step.

**Understanding the Distribution**

Since we are dealing with the number of events (births) occurring in a fixed interval of time (one minute), and we are given the mean number of events, we will use the **Poisson distribution**.

**Given Information**

* Mean number of births per minute (λ) = 4

**Poisson Distribution Formula**

The probability of observing k events in a given interval is given by:

P(X = k) = (e^(-λ) * λ^k) / k!

Where:

* e is Euler's number (approximately 2.71828)
* λ is the mean number of events
* k is the number of events we are interested in

**Calculations**

**(a) Exactly five births (k = 5)**

P(X = 5) = (e^(-4) * 4^5) / 5!

P(X = 5) = (0.0183156 * 1024) / 120

P(X = 5) = 18.7558 / 120

P(X = 5) ≈ 0.1563

**(b) At least five births (k ≥ 5)**

P(X ≥ 5) = 1 - P(X < 5)

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Let's calculate each of these:

* P(X = 0) = (e^(-4) * 4^0) / 0! ≈ 0.0183
* P(X = 1) = (e^(-4) * 4^1) / 1! ≈ 0.0733
* P(X = 2) = (e^(-4) * 4^2) / 2! ≈ 0.1465
* P(X = 3) = (e^(-4) * 4^3) / 3! ≈ 0.1954
* P(X = 4) = (e^(-4) * 4^4) / 4! ≈ 0.1954

P(X < 5) ≈ 0.0183 + 0.0733 + 0.1465 + 0.1954 + 0.1954 ≈ 0.6289

P(X ≥ 5) = 1 - 0.6289 ≈ 0.3711

**(c) More than five births (k > 5)**

P(X > 5) = 1 - P(X ≤ 5)

P(X ≤ 5) = P(X < 5) + P(X = 5)

P(X ≤ 5) = 0.6289 + 0.1563 = 0.7852

P(X > 5) = 1 - 0.7852 ≈ 0.2148

**Unusual Events**

An event is typically considered unusual if its probability is less than 0.05.

* (a) P(X = 5) ≈ 0.1563 (Not unusual)
* (b) P(X ≥ 5) ≈ 0.3711 (Not unusual)
* (c) P(X > 5) ≈ 0.2148 (Not unusual)

**Summary**

* (a) P(X = 5) ≈ 0.1563
* (b) P(X ≥ 5) ≈ 0.3711
* (c) P(X > 5) ≈ 0.2148
* None of the events are unusual.