Question

Find the binomial probability for the given problem, and then compare the normal and the Poisson approximations. Find the probability of \(x\) successes in 100 Bernoulli trials with probability \(p=1 / 5\) of success (a) if \(x=25 ;\) (b) if \(x=21.\)

Short answer

Binomial: For \(x=25\), use \( P(X=25)=\binom{100}{25}(\frac{1}{5})^{25}(\frac{4}{5})^{75} \); for \(x=21\), use \( P(X=21)=\binom{100}{21}(\frac{1}{5})^{21}(\frac{4}{5})^{79} \). Normal and Poisson approximations can simplify these calculations. Compare by evaluating each probability.

Step-by-step solution

Understand the Problem

Given are 100 Bernoulli trials with a success probability of \(p = \frac{1}{5}\). Need to find the probability of \(x\) successes for \(x = 25\) and \(x = 21\). Compare binomial, normal, and Poisson probabilities.

Calculate Binomial Probability for x=25

Use the binomial probability formula: \[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \] Substitute \(n = 100\), \(p = \frac{1}{5}\), \(x = 25\): \[ P(X = 25) = \binom{100}{25} \left( \frac{1}{5} \right)^{25} \left( \frac{4}{5} \right)^{75} \]

Calculate Binomial Probability for x=21

Use the same formula as above with \(x = 21\): \[ P(X = 21) = \binom{100}{21} \left( \frac{1}{5} \right)^{21} \left( \frac{4}{5} \right)^{79} \]

Normal Approximation

For large \(n\), approximate binomial with normal distribution where \(\mu = np\) and \(\sigma = \sqrt{np(1-p)}\). Compute \(\mu\) and \(\sigma\) for \(n = 100\) and \(p = \frac{1}{5}\).\[ \mu = 100 \times \frac{1}{5} = 20 \]\[ \sigma = \sqrt{100 \times \frac{1}{5} \times \frac{4}{5}} = 4 \]For \(x = 25\) and \(x = 21\), convert to z-scores and use standard normal table.

Normal Approximation for x=25 and x=21

Convert \(x\) values to z-scores:\[ z_{25} = \frac{25 - 20}{4} = 1.25 \]\[ z_{21} = \frac{21 - 20}{4} = 0.25 \]Use standard normal distribution to find probabilities.

Poisson Approximation

For Poisson approximation with large \(n\) and small \(p\) such that \(np\) is moderate, use \(\lambda = np\):\[ \lambda = 100 \times \frac{1}{5} = 20 \]Poisson probability formula:\[ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \]Calculate for \(x = 25\) and \(x = 21\).

Compare Results

Compare binomial, normal, and Poisson probabilities for \(x = 25\) and \(x = 21\). Note how close the approximations are.

Key concepts

Bernoulli Trials

Bernoulli trials are named after the Swiss mathematician Jacob Bernoulli. Each trial is a random experiment with exactly two possible outcomes: success or failure. A common example is flipping a coin, where success might be landing on heads, and failure might be landing on tails.
In a Bernoulli trial, the probability of success is denoted by \( p \), and the probability of failure is \( 1 - p \). These trials are the foundation of the binomial distribution. When we have a fixed number of independent Bernoulli trials, the number of successes follows a binomial distribution. For example, if we flip a coin 100 times and are interested in the number of heads, each flip is a Bernoulli trial. The main formula related to Bernoulli trials is:

• \( P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \)

Normal Approximation

The normal approximation to the binomial distribution is useful when the number of trials \( n \) is large and the probability of success \( p \) is not too close to 0 or 1. The central limit theorem tells us that as the number of trials increases, the binomial distribution approaches a normal distribution with mean \( \mu = np \) and standard deviation \( \sigma = \sqrt{np(1-p)} \).
For the binomial distribution with \( n = 100 \) and \( p = \frac{1}{5} \), the mean \( \mu \) and the standard deviation \( \sigma \) are:

• \( \mu = 100 \times \frac{1}{5} = 20 \)
• \( \sigma = \sqrt{100 \times \frac{1}{5} \times \frac{4}{5}} = 4 \)

To use the normal approximation, convert the number of successes \( x \) to a z-score:

• \( z = \frac{x - \mu}{\sigma} \)

Lookup the z-score in the standard normal table to find the corresponding probability. This method simplifies the calculation when \( n \) is large.

Poisson Approximation

The Poisson approximation is another method to approximate the binomial distribution. It is particularly useful when the number of trials \( n \) is large, and the probability of success \( p \) is small, making \( np \) moderate. In such cases, the binomial distribution can be approximated using a Poisson distribution with parameter \( \lambda = np \).
In our example, for \( n = 100 \) and \( p = \frac{1}{5} \), we have:

• \( \lambda = 100 \times \frac{1}{5} = 20 \)

The Poisson probability of getting \( x \) successes is given by:

• \( P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \)

This approximation is handy for calculations that would be otherwise complex using the exact binomial formula, especially when \( x \) is much smaller than \( n \).

Probability of Success

The probability of success, denoted as \( p \), is a fundamental concept in probability and statistics. It represents the likelihood of a success occurring in a single Bernoulli trial. In the context of binomial, normal, and Poisson approximations, \( p \) helps determine the distribution parameters. For example, in our exercise, the probability of success is \( \frac{1}{5} \), or 0.2.

• In binomial distribution: \( p \) directly influences the shape of the distribution and the calculations involved.
• In normal approximation: \( p \) helps calculate the mean \( \mu = np \) and standard deviation \( \sigma = \sqrt{np(1-p)} \).
• In Poisson approximation: \( p \) combined with \( n \) determines the parameter \( \lambda = np \).

The probability of success is crucial in deciding the likelihood of different outcomes and in making statistical inferences about the data.