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In statistical terms, a random variable is a variable whose possible values are numerical outcomes of a random phenomenon. In this context, the random variable is denoted by \(X\), which represents the number of radios selected that have two slots.
To further understand this, consider the 20 radios being the population. Among these, 12 have two slots, considered a 'success' in this context. When selecting 6 radios without replacement, \(X\) becomes a hypergeometric random variable, counting the number of successes in the sample.
Unlike other distributions such as the binomial, the hypergeometric distribution does not assume that each draw is independent. This is a crucial aspect and makes it particularly useful in scenarios where the sampling is done without replacement.

Calculating probability involves determining the likelihood of various outcomes of a random variable. For a hypergeometric distribution like ours, this requires specific calculations.
In our scenario, if you want to find \(P(X=2)\), you use the hypergeometric probability formula:
\[P(X = k) = \frac{\binom{K}{k} \cdot \binom{N-K}{n-k}}{\binom{N}{n}}\] where \(\binom{a}{b}\) denotes a binomial coefficient. This notation represents the number of ways to choose \(b\) items from a total of \(a\) without regard to order.
Using the formula, you compute \(P(X=2)\) by calculating separate binomial coefficients for different components and then computing their product.
This approach extends to finding cumulative probabilities such as \(P(X \leq 2)\) and \(P(X \geq 2)\), which involve summing individual probabilities or employing complementary probabilities.

The mean and standard deviation of a distribution provide measures of central tendency and variability, respectively. For the hypergeometric distribution, they are calculated based on the parameters of the population and the sample.
The mean \(\mu\) reflects the expected number of two-slot radios among the selected ones:
\[\mu = n \left(\frac{K}{N}\right)\]This formula utilizes the total number of draws \(n\), the number of available two-slot radios \(K\), and the total radios in the shipment \(N\).
The standard deviation \(\sigma\) indicates how much variability or spread exists around the mean:
\[\sigma = \sqrt{n \left(\frac{K}{N}\right) \left(\frac{N-K}{N}\right) \left(\frac{N-n}{N-1}\right)}\]Both calculated values, a mean of 3.6 and a standard deviation of about 1.063, give us insights into what we can typically expect from such selections.