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The binomial distribution is a fundamental concept in statistics used to model the number of successes in a fixed number of trials. It's particularly useful when each trial is independent and has the same probability of success. In the problem provided, we are looking at whether employees are part of a self-insurance health plan. Here, the probability of success (an employee participating) is 0.40, and each employee is a trial.

Basic properties include:

• Number of trials (n): This is fixed at 100 employees.
• Probability of success (Ï€): Fixed at 0.40, which is the chance an employee participates in the self-insurance plan.
• Random variable (X): The number of successes, or employees, participating in this context.

The binomial distribution provides a discrete probability distribution, meaning it deals with distinct, separate outcomes rather than a continuous range. When calculations with binomial formulas are difficult, techniques like the Central Limit Theorem can help approximate probabilities.

The Central Limit Theorem (CLT) is a crucial statistical concept. It explains that when you have a sufficiently large sample size, the sampling distribution of the sample means will be approximately normally distributed, regardless of the original population's distribution. For our exercise, we use CLT to approximate the binomial distribution.

This allows us to calculate probabilities more easily because the shape of the distribution of outcomes fits into the well-known normal distribution curve. In the exercise about employee health plans:

• The sample mean (μ) is calculated as 40 (since 100 employees × 0.40).
• The standard deviation (σ) is calculated using the formula: \( \sigma = \sqrt{n \pi (1-\pi)} = 5 \).

Using CLT and these calculated values, we transform the binomial distribution into a normal distribution to estimate probabilities more straightforwardly. This transformation helps us then calculate the Z-score for specific number of successes, like 50.

A Z-score is a statistical measurement that describes a value's position in relation to the mean of a group of values. It’s expressed in terms of standard deviations from the mean, allowing us to understand how far away a particular data point is from the average. In the context of our health plan probability exercise:

• We needed to find the Z-score for 50 successes to solve part a. The formula is \( Z = (X - \mu)/\sigma \), where X is the number of successes we are testing. For example, at 50 successes, \( Z = (50 - 40)/5 = 2 \).
• For part b, we calculate a Z-score for 60 employees, so \( Z = (60 - 40)/5 = 4 \).

These Z-scores allow us to use standard normal distribution tables (or calculators) to find the probability of achieving such outcomes. For instance, a higher Z-score signifies a less common occurrence under our initial assumption, like having such a high number of participating employees.

Probability calculation allows us to measure the likelihood of various possible outcomes. In our exercise, we are specifically interested in the chances of at least half or more employees opting into a self-insurance health plan within a given sample.

To estimate this probability:

• First calculate the mean and standard deviation using the binomial approximation and Central Limit Theorem.
• Use the calculated Z-scores to determine probabilities. For a Z-score of 2 (50 successes), consult a Z-table or employ software to find the corresponding probability value. This indicates the cumulative probability up to that Z-score, requiring subtraction from 1 to get the probability of results being 50 or higher.

For the second part of the problem, seeing a Z-score of 4 suggests the probability of having 60 employees participate is very low if the true participation rate were indeed 40%. This low probability may lead statisticians to question the initial participation rate used as the basis for the assumption.