Problem 28 - Calculus for Biology and Medicine

Question

For \(n=100\) and \(p=0.1\), compute \(P\left(S_{n}=10\right)\) (a) exactly, (b) by using a Poisson approximation, and (c) by using a normal approximation.

Step-by-Step Solution

Answer

(a) Exact: 0.125, (b) Poisson: 0.125, (c) Normal: 0.1157.

Step 1: Exactly Compute Using Binomial Distribution

The exact probability can be calculated using the binomial probability formula: \[P(S_n = k) = \binom{n}{k} p^k (1-p)^{n-k}\]For this exercise, we have \(n=100\), \(p=0.1\), and \(k=10\). First, compute \(\binom{100}{10}\):\[\binom{100}{10} = \frac{100!}{10!(100-10)!} = 17310309456440\]Then, calculate:\[0.1^{10} = 1 \times 10^{-10}\]\[0.9^{90} \approx 0.000031577\]Finally, multiply these:\[P(S_{100} = 10) \approx 17310309456440 \times 1 \times 10^{-10} \times 0.000031577 \approx 0.125\]

Step 2: Poisson Approximation

For a Poisson approximation, we use \(\lambda = np\). Here, \(\lambda = 100 \times 0.1 = 10\). The formula is:\[P(S_{n} = k) = \frac{\lambda^k e^{-\lambda}}{k!}\]Applying this:\[P(S_{100} = 10) = \frac{10^{10} e^{-10}}{10!}\]Calculate \(10^{10} = 10000000000\) and approximate \(e^{-10} \approx 0.0000454\). Then compute:\[\frac{10000000000 \times 0.0000454}{3628800} \approx 0.125\]

Step 3: Normal Approximation

Use the normal approximation with mean \(\mu = np\) and variance \(\sigma^2 = np(1-p)\). Here \(\mu = 10\) and \(\sigma^2 = 9\), so \(\sigma = 3\). For the normal approximation, calculate:\[Z = \frac{10 - 10}{3} = 0\]Find the probability that \(S_{n} = 10\):\[P(S_{n} = 10) = P(9.5 < X < 10.5)\]Using the Z-table, approximate:\[P(9.5 < X < 10.5) \approx P(0 < Z < 0.167) \approx 0.1157\]

Key Concepts

Poisson approximation

The Poisson approximation is a useful tool when dealing with a binomial distribution, especially when the number of trials (\( n \)) is large, and the probability of success (\( p \)) is small. The Poisson distribution is defined by a single parameter, \( \lambda \), which represents the average number of successes within a given interval.
For our problem, we calculated \( \lambda = np \). With \( n=100 \) and \( p=0.1 \), \( \lambda \) becomes 10. This simplification occurs because the binomial distribution can become cumbersome when \( n \) is large. The smaller \( p \), the more the binomial distribution starts to resemble a Poisson distribution.
To approximate the probability that \( S_{100} = 10 \) using a Poisson distribution, use the formula: \[P(S_{n} = k) = \frac{\lambda^k e^{-\lambda}}{k!}\] Here, \( k \) is the exact number of successes, which is 10. The calculations involved exponentiation and factorial functions, making Poisson approximation especially convenient in these scenarios for quick estimates.

Normal approximation

The normal approximation is another method used for estimating probabilities of a binomial distribution, ideal when \( n \) is large, \( p \) is moderate, and \( np(1-p) \) is sufficiently large. Central Limit Theorem supports this approximation by suggesting that as \( n \) increases, the distribution of the sample mean will approach a normal distribution.
In our case, the mean (\( \mu \)) and the variance (\( \sigma^2 \)) for the binomial distribution were computed as \( \mu = np = 10 \) and \( \sigma^2 = np(1-p) = 9 \). The standard deviation \( \sigma \) is the square root of the variance, \( 3 \) in this context.
The normal approximation converts the discrete nature of binomial through a continuity correction, where probability \( P(S_{n} = 10) \) is approximated by \( P(9.5 < X < 10.5) \). Using the standard normal distribution table, this was further calculated to approximate the probability.

Probability calculation

Probability calculation is at the core of these approximations. Direct probability via binomial formula involves evaluating \( k \)-specific success chances over \( n \) trials while considering \( p \), the success rate for each trial.

• The exact binomial probability involves combinatorial terms like \( \binom{n}{k} \), success raised to the occurrence power \( (p^k) \), and failure terms \((1-p)^{n-k}\).
• For exercises involving large sample sizes or extreme probabilities, more efficient approximations like Poisson and Normal become very helpful.

Each method approximates the likelihood of a specific outcome by leveraging mathematical properties like exponential growth in Poisson or the bell-shaped curve characteristic of normal distributions. Understanding how to interchangeably use these tools depending on the context and constraints enhances calculation efficiency in probability problems.