Problem 29 - Calculus for Biology and Medicine

Question

\(S_{n}\) is binomially distributed with parameters \(n\) and \(p\). For \(n=50\) and \(p=0.1\), compute \(P\left(S_{n}=5\right)\) (a) exactly, (b) by using a Poisson approximation, and (c) by using a normal approximation.

Step-by-Step Solution

Answer

Exact: 0.1746; Poisson Approx.: 0.1755; Normal Approx.: 0.152.

Step 1: Compute Exactly using Binomial Formula

The probability of exactly k successes in a binomial distribution is given by the formula: \( P(S_n = k) = \binom{n}{k} p^k (1-p)^{n-k} \). For \( n=50 \), \( p=0.1 \), and \( k=5 \), compute \( P(S_{50} = 5) \): \[P(S_{50} = 5) = \binom{50}{5} (0.1)^5 (0.9)^{45}.\] First, calculate \( \binom{50}{5} = \frac{50!}{5!(45!)} = 2118760 \). Thus, the probability is \[P(S_{50} = 5) = 2118760 \times (0.1)^5 \times (0.9)^{45} \]Calculating the values, we find \( P(S_{50} = 5) \approx 0.1746 \).

Step 2: Use Poisson Approximation

When \( n \) is large and \( p \) is small, the binomial distribution can be approximated by a Poisson distribution with parameter \( \lambda = np \). Here, \( \lambda = 50 \times 0.1 = 5 \). For a Poisson distribution, we calculate \[P(S_n = k) = \frac{\lambda^k e^{-\lambda}}{k!}\] for \( k=5 \):\[P(S_n = 5) = \frac{5^5 e^{-5}}{5!}.\]Computing this gives us \( P(S_n = 5) \approx 0.1755 \).

Step 3: Use Normal Approximation

For larger \( n \), if both \( np \) and \( n(1-p) \) are sufficiently large (greater than 5), the binomial can be approximated using a normal distribution with mean \( \mu = np = 5 \) and variance \( \sigma^2 = np(1-p) = 4.5 \). Thus, \( \sigma = \sqrt{4.5} \approx 2.12 \).Normalize by computing the z-score with continuity correction for \( P(4.5 \leq S_n \leq 5.5) \):\[z_1 = \frac{4.5 - 5}{2.12} \approx -0.24, \quad z_2 = \frac{5.5 - 5}{2.12} \approx 0.24.\]Use \( P(z_1 \leq Z \leq z_2) \) from standard normal tables,\[P(-0.24 \leq Z \leq 0.24) \approx 0.1894 - 0.4052 \approx 0.1524,\]which is the approximate probability \( P(S_{50} = 5) \approx 0.152 \).

Key Concepts

Understanding Binomial Distribution

The binomial distribution is crucial in probability theory and statistics. It models the number of successes in a fixed number of trials, where each trial has only two possible outcomes: "success" and "failure." Each outcome must have consistent probabilities in every trial. The formula for binomial probabilities is given by:\[P(S_n = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where:

• \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successes out of \( n \) trials.
• \( p \) is the probability of success in each trial.
• \( k \) is the number of successful outcomes you want to find the probability for.

In our exercise, for instance, with \( n=50 \), \( p=0.1 \), and \( k=5 \), the probability \( P(S_{50} = 5) \) was calculated using this formula, yielding approximately \( 0.1746 \). This method remains exact and is the foundational approach for computing probabilities in binomial settings.

Exploring Poisson Approximation

When dealing with a large number of trials and a small probability of success, using the binomial distribution can be complex. Here, the Poisson approximation offers a simplified alternative. If the product \( np \) is small, the binomial distribution for number of successes can be closely approximated by a Poisson distribution with parameter \( \lambda = np \).The Poisson probability mass function is written as:\[P(S_n = k) = \frac{\lambda^k e^{-\lambda}}{k!}\]For this approximation, you expect the mean number of occurrences, \( \lambda \), to provide an accurate estimate of the distribution.In our example with \( n=50 \) and \( p=0.1 \), \( \lambda = 5 \). Calculating the probability \( P(S_n = 5) \) using the Poisson formula gives us approximately \( 0.1755 \). This technique is efficient and simplifies the computations, making it a valuable tool for probabilistic estimates in suitable scenarios.

Utilizing Normal Approximation

The normal approximation helps to estimate binomial probabilities when both \( np \) and \( n(1-p) \) are large, which makes the distribution symmetric and bell-shaped, similar to a normal distribution. The approximation uses a normal distribution with the same mean and variance as the binomial distribution: mean \( \mu = np \) and variance \( \sigma^2 = np(1-p) \).We calculate the z-scores to understand the range of values:\[z = \frac{X - \mu}{\sigma}\]where \( X \) is the value of interest, \( X = k \pm 0.5 \) for continuity correction.In our given problem, with \( \mu = 5 \) and \( \sigma \approx 2.12 \), we calculated approximate z-scores to find:\[z_1 = \frac{4.5 - 5}{2.12}, \quad z_2 = \frac{5.5 - 5}{2.12}\]The approximate probability found was \( 0.1524 \). While less precise than exact calculations, this method gives a good estimation if conditions of symmetry and large sample size are met, particularly useful for quick, rough estimates.