Question

Find the binomial probability for the given problem, and then compare the normal and the Poisson approximations. Out of 1095 people, what is the probability that exactly 2 were born on Jan. \(1 ?\) Assume 365 days in a year.

Short answer

Binomial Probability: depends on the exact calculation. Normal Approximation: depends on z. Poisson Approximation: 0.224.

Step-by-step solution

- Understand the Problem

We need to find the probability that exactly 2 out of 1095 people were born on January 1, assuming that each day of the year is equally likely for a birth.

- Define Parameters for Binomial Distribution

Let’s define the binomial distribution parameters: the total number of trials (n = 1095) and the probability of success on each trial (p = 1/365) since each day is equally likely for a birth. We are looking for exactly 2 successes (k = 2).

- Calculate the Binomial Probability

The binomial probability formula is \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Substitute n = 1095, k = 2, and p = 1/365:\[ P(X = 2) = \binom{1095}{2} \left( \frac{1}{365} \right)^2 \left( 1 - \frac{1}{365} \right)^{1093} \]Calculate this value to find the binomial probability.

- Calculate the Normal Approximation

For the normal approximation, use the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the binomial distribution:\[ \mu = np = 1095 \times \frac{1}{365} = 3 \]\[ \sigma = \sqrt{np(1-p)} = \sqrt{1095 \times \frac{1}{365} \times \left( 1 - \frac{1}{365} \right)} \approx 1.732\]For normal approximation, convert k = 2 to a z-score:\[ z = \frac{k - \mu}{\sigma} = \frac{2 - 3}{1.732} = -0.577 \]Use standard normal distribution tables or software to find the probability corresponding to z = -0.577.

- Calculate the Poisson Approximation

For the Poisson approximation, use the rate (\( \lambda \)) which equals the mean of the binomial distribution:\[ \lambda = np = 3 \]The Poisson probability formula is \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]Substitute \( \lambda = 3 \) and k = 2:\[ P(X = 2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{e^{-3} \cdot 9}{2} \approx 0.224 \]Calculate this value for the Poisson probability.

- Compare Results

Compare the values obtained from the binomial, normal, and Poisson distributions:1. Binomial Probability2. Normal Approximation3. Poisson Approximation.Check for similarities and differences in the results.

Key concepts

Poisson Approximation

The Poisson approximation is a handy tool when dealing with binomial distributions, especially when the number of trials (n) is very large and the probability of success (p) is very small. The idea is to simplify the calculations by approximating the binomial distribution with a Poisson distribution. This approximation is easier to work with, especially for large datasets.
In the given problem, we used a Poisson distribution with a \(\text{rate} \ (\ \lambda \)). The rate is simply the mean of the original binomial distribution. We calculated \(\text{rate} \ (\ \lambda \) = 3 by multiplying the number of people (1095) with the probability of being born on January 1 (1/365).
The Poisson probability formula used is \[ P(X = k) = \frac{e^{-\text{rate}} \text{rate}^k}{k!} \] We then substituted \(\text{rate} \ (\ \lambda \) = 3 and k = 2 to get approximately 0.224. This tells us the chance of exactly 2 people being born on January 1 out of 1095 is roughly 22.4%.
Poisson approximation is especially effective when both n is large and p is small, making it a useful tool for approximating certain binomial problems.

Normal Approximation

Normal approximation is another method used to simplify binomial probabilities, primarily when n is large and p is not extremely small. In this approach, we use the Central Limit Theorem to approximate the binomial distribution with a normal distribution.
To perform the normal approximation, we need the mean (µ) and standard deviation (σ) of the binomial distribution. We calculated these values as follows:
\( \ \mu = np = 1095 \ \times \ \frac{1}{365} = 3 \ \)
\( \ \sigma = \ \sqrt{np(1-p)} = \ \sqrt{1095 \ \times \ \frac{1}{365} \ \times \ \left(1 \ \-\frac{1}{365}\right)} \ \approx 1.732 \ \)
Using the mean and standard deviation, we converted k = 2 to a z-score:
\( z = \ \frac{k \ \-\mu}{\sigma} = \ \frac{2 \ \-\ 3}{1.732 } = -0.577 \)
This z-score represents how many standard deviations away the value of 2 is from the mean of 3. We can then look up the corresponding cumulative probability or use software to find the probability value. The normal approximation provides a convenient way to estimate binomial probabilities when dealing with large n.

Binomial Distribution

The binomial distribution represents the probability of having a fixed number of successes in a certain number of trials. It's governed by two parameters: the number of trials (n) and the probability of success in each trial (p). This distribution is ideal for situations with two possible outcomes, like 'success' and 'failure'.
In our problem, we defined n = 1095 and p = 1/365. Each person has an equal chance of being born on any day of the year, making each trial independent.
The binomial probability formula is given by:
\( P(X = k) = \ \binom{n}{k} p^k (1-p)^{n-k} \)
We substituted n = 1095, k = 2, and p = 1/365 into the formula. After calculation, this provides the exact probability of exactly 2 individuals being born on January 1.
The binomial distribution offers precise probability values for specific numbers of successes but can become cumbersome with large datasets. Thus, approximations like Poisson and normal distributions are often used to simplify the calculations.