Question 1171244: Find the indicated probabilities using the geometric distribution, the Poisson distribution, or the binomial distribution. Then determine if the events are unusual. If convenient, use the appropriate probability table or technology to find the probabilities.
The mean number of births per minute in a country in a recent year was about
four.
Find the probability that the number of births in any given minute is (a) exactly
five,
(b) at least
five,
and (c) more than
five.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.
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