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The Probability Mass Function (pmf) describes the probability that a discrete random variable takes on a specific value. It is essential for understanding distributions that arise from sampling without replacement, like the hypergeometric distribution. In our example concerning firms in violation, we want to calculate the probability of selecting a certain number of violating firms when visiting 10 firms from a total of 50, where 15 firms are known violators.

The formula for the hypergeometric pmf involves the combination function, denoted by \(\binom{n}{k}\), which determines the number of ways to choose \(k\) successes (violating firms) out of \(n\) possible ones. The hypergeometric pmf is calculated using:

• \(\binom{15}{k}\): Ways to choose \(k\) violating firms from 15.
• \(\binom{35}{10-k}\): Ways to choose non-violating firms from the remaining 35.
• \(\binom{50}{10}\): All possible ways to select any 10 firms from 50.

Thus, the pmf is given by:
\[P(X = k) = \frac{\binom{15}{k} \binom{35}{10-k}}{\binom{50}{10}} \] This applies to any integer \(k\) between 0 and 10 inclusive, although the realistic range is determined by logic constraints.

The Binomial Distribution is often used as an approximation of the Hypergeometric Distribution when the sample size is much smaller than the population size. This simplification is crucial when we deal with large populations, like approximating the scenario with 500 firms total. We aim to use a binomial model to mimic the hypergeometric distribution with fewer calculations.

Here's how the approximation works:

• The total population is 500 firms, and 150 are violators.
• We need to find the probability of choosing any number of violators out of the 10 firms inspected.
• The probability of choosing a violator (success) is \(p = \frac{150}{500}\) or 0.3.

In the binomial pmf, \(n=10\) (sample size) and \(p=0.3\) (probability of choosing a violator), leading to:
\[P(X = k) = \binom{10}{k} (0.3)^k (0.7)^{10-k} \]
This formula is simpler and works under the assumption that the sample size is small relative to the population, which is often acceptable in practical scenarios.

The Expected Value and Variance of a random variable give us essential insights into its behavior. These metrics tell us what to expect on average and how much variability that expectation has. Let's explore these concepts for both hypergeometric and binomial distributions.

**Expected Value (E(X)):**
- For a hypergeometric distribution, expected value \(E(X)\) is calculated:
\[E(X) = \frac{sample\ size \times successes}{total\ population} = \frac{10 \times 15}{50} = 3\]- The binomial approximation gives the same result:
\[E(X) = n \times p = 10 \times 0.3 = 3\]
In both cases, on average, 3 firms are expected to be violators.

**Variance (V(X)):**
- Hypergeometric distribution variance is calculated using:
\[V(X) = \frac{sample\ size \times successes}{total\ population} \times \left(1 - \frac{successes}{total\ population}\right) \times \frac{total\ population - sample\ size}{total\ population - 1} = 1.7143\]- For the binomial distribution, the formula simplifies:
\[V(X) = n \times p \times (1-p) = 10 \times 0.3 \times 0.7 = 2.1\]The binomial approximation slightly overestimates variance compared to the hypergeometric due to its nature of replacement assumption. Understanding these calculations helps you predict and account for variation in samples.