Problem 27 - Calculus for Biology and Medicine

Question

\(S_{n}\) is binomially distributed with parameters \(n\) and \(p\). For \(n=100\) and \(p=0.01\), compute \(P\left(S_{n}=0\right)\) (a) exactly, (b) by using a Poisson approximation, and (c) by using a normal approximation.

Step-by-Step Solution

Answer

Exact: 0.366; Poisson: 0.368; Normal: 0.157.

Step 1: Understand the Binomial Probability

The probability of exactly 0 successes in a binomial distribution is given by the formula: \[ P(S_n = 0) = (1-p)^n \] Substitute the given values, where \( n = 100 \) and \( p = 0.01 \), into the formula.

Step 2: Calculate the Exact Binomial Probability

Plug in the values to find the exact probability:\[P(S_{100}=0) = (1-0.01)^{100} = 0.99^{100}\]Now compute \( 0.99^{100} \) using a calculator.

Step 3: Use a Calculator for Exact Value

Using a calculator, compute \( 0.99^{100} \):\[0.99^{100} \approx 0.366\]Thus, the exact probability is approximately 0.366.

Step 4: Apply Poisson Approximation

The Poisson distribution can approximate the binomial distribution when \( n \) is large and \( p \) is small. Use the mean \( \lambda = np = 100 \times 0.01 = 1 \).The Poisson probability is given by:\[P(S_n = 0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-1}\]Calculate \( e^{-1} \) using a calculator.

Step 5: Calculate Poisson Probability

Compute \( e^{-1} \) using a calculator:\[e^{-1} \approx 0.368\]The Poisson approximation gives a probability of approximately 0.368.

Step 6: Apply Normal Approximation

For a normal approximation, check conditions: \( np = 1 \) and \( n(1-p) = 99 \). With the conditions met for approximation, the mean is \( \mu = np = 1 \) and variance \( \sigma^2 = np(1-p) = 0.99 \).Standardize \( X = 0 \):\[ Z = \frac{0 - 1}{\sqrt{0.99}} \approx \frac{-1}{0.995} \approx -1.005 \]Use a Z-table to find \( P(X < -1.005) \).

Step 7: Find Normal Approximation Using Z-Table

Using a Z-table, find the probability:\[P(X < -1.005) \approx 0.157\]Thus, the normal approximation gives a probability of approximately 0.157.

Key Concepts

Understanding the Binomial Distribution

The binomial distribution is a probability distribution that models the number of successes in a fixed number of independent trials or experiments. Each trial can only have two possible outcomes: success or failure. The key parameters are \( n \), the number of trials, and \( p \), the probability of success in a single trial. For a random variable \( S_n \) following a binomial distribution with parameters \( n \) and \( p \), the formula to calculate the probability of exactly \( k \) successes is:

• \( P(S_n = k) = \binom{n}{k} p^k (1-p)^{n-k} \)

However, for the specific case of 0 successes, the formula simplifies since \( k = 0 \):

• \( P(S_n = 0) = (1-p)^n \)

This formula calculates the likelihood of none of the trials resulting in success, which simplifies the computation for our problem at hand with parameters \( n = 100 \) and \( p = 0.01 \). Inserting these values gives us the probability \( P(S_{100} = 0) = 0.99^{100} \), which can be computed using a calculator to find approximately 0.366.

Applying Poisson Approximation

The Poisson distribution can be used as an approximation for the binomial distribution under specific conditions: when \( n \) is large and \( p \) is small, such that the product \( np \) is moderate. In our scenario, \( n = 100 \) and \( p = 0.01 \), making these conditions applicable.
The Poisson distribution is characterized by the parameter \( \lambda \), which is the mean of the distribution. For approximation, \( \lambda = np \). Thus, \( \lambda = 1 \) in our problem. The Poisson probability of observing 0 events (or successes) is given by:

• \( P(S_n = 0) = \frac{e^{-\lambda} \cdot \lambda^0}{0!} = e^{-1} \)

Using a calculator, we find \( e^{-1} \approx 0.368 \). This Poisson approximation offers us a straightforward computational alternative yielding a result very close to the exact binomial probability.

Exploring Normal Approximation

The normal distribution is another alternative to approximate the binomial distribution, especially when \( n \) is very large. The approximation requires that both \( np \) and \( n(1-p) \) are greater than 5, which is partially achieved in our exercise.
However, in cases like this problem where \( np = 1 \), directly using the normal approximation might not meet classical rules of thumb (typically, \( np \) and \( n(1-p) \) both need to be larger than 5).
For demonstration, we use a Z-score calculation. The mean \( \mu \) and variance \( \sigma^2 \) of the approximating normal distribution are calculated as:

• \( \mu = np = 1 \)
• \( \sigma^2 = np(1-p) = 0.99 \)

Translate \( X = 0 \) into a standard normal variable \( Z \):

• \( Z = \frac{0 - 1}{\sqrt{0.99}} \approx -1.005 \)

Using a standard normal distribution table (Z-table), the probability \( P(X < -1.005) \approx 0.157 \) is derived. This figure highlights that the normal approximation may not always be accurate for very skewed binomial distributions where \( p \) is small.

The Art of Probability Calculation

Probability calculation in these contexts involves the strategic selection of methods based on the problem specifics. Each approach, be it exact computation, Poisson, or normal approximation, has a special niche wherein it becomes the most efficient or insightful.
Understanding which conditions justify each approximation method saves time and effort while maintaining accuracy. Key considerations include:

• A binomial distribution is precise when conditions permit straightforward calculations.
• The Poisson approximation is ideal when \( p \) is small, and \( n \) is large enough, simplifying work when decisions are needed quickly.
• The normal approximation suits larger \( n \) with moderate \( p \) values, providing a good balance and intuitive visualization through bell curves.

These calculations form the bedrock of probability assessments, each with distinct mathematical beauty and practical use scenarios to be critically evaluated and applied in varying statistical contexts.