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The hypergeometric distribution is a probability distribution that describes scenarios where objects are selected without replacement from a finite population. This means that each selection affects subsequent selections because the overall population size decreases. It is particularly useful for understanding the probability of obtaining a specific number of successes in a sample drawn from a population with a known number of successes and failures.

In our example, we have 50 industrial firms, with 15 violating regulations. We select 10 firms without replacement. To find the probability of selecting a certain number of violating firms (successes), we use the probability mass function (PMF):

\[ P(X = k) = \frac{{\binom{15}{k} \binom{35}{10-k}}}{\binom{50}{10}} \]

Here, \(k\) is the number of firms in violation among the 10 selected. This formula helps us calculate the probability for each possible value of \(k\) taking into account the constraints of selection without replacement.

The binomial distribution applies when there are fixed numbers of independent trials, each with the same probability of a certain outcome. It's a perfect fit for situations where you perform an action multiple times, under the same conditions, and are interested in the number of successes.

In the exercise, for part (b), we have 500 firms with 150 violators. Although the population size is still finite, it's large enough compared to the sample size to use the binomial approximation. We sum up the approximation with its probability mass function (PMF):

\[ P(X = k) \approx \binom{10}{k} (0.3)^k (0.7)^{10-k} \]

Here, \(0.3\) is the approximate probability of an individual firm being a violator, calculated as \(\frac{150}{500}\). This approximation simplifies calculations when the population size is much larger than the sample size.

Expected value is a key concept in probability and statistics that provides the average outcome of a random process if it were repeated many times. It gives us a sense of the central tendency of a distribution.

For the hypergeometric distribution in our example, the expected value \(E(X)\) is found by:

\[ E(X) = \frac{nK}{N} \]

Substituting the known values, \(n = 10\), \(K = 15\), and \(N = 50\), we find that \(E(X) = 3\).

For the binomial approximation, the expected value is similarly calculated using \(E(X) = np\). With \(n = 10\) and \(p = 0.3\), the result is also \(E(X) = 3\), indicating that, on average, the inspector will find 3 violations.

Variance measures how much the values in a distribution are spread out around the expected value. It provides insight into the variability of the data.

For the hypergeometric distribution, variance \(V(X)\) is calculated with the formula:

\[ V(X) = n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1} \]

In our case, substituting the values results in a variance that accounts for the natural decrease in population as items are selected without replacement.

In contrast, for the binomial distribution, variance is calculated as \(V(X) = np(1-p)\), which sums up the expected variability based on independent trials. With \(n = 10\) and \(p = 0.3\), we find \(V(X) = 2.1\). This variance reflects the expected spread of finding violators in independent selections with a probability of \(0.3\).