91Ó°ÊÓ

When dealing with the hypergeometric distribution, the probability mass function (PMF) is a crucial concept. The PMF gives us the probability of selecting a particular number of granite specimens out of a total selection made without replacement. This is exactly what occurs in our problem where a laboratory assistant randomly selects specimens for analysis.

In mathematical terms, the PMF for the number of granite specimens, say \( X \), is calculated as follows: \[ P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{\binom{N}{n}} \] Where:

• \( K \) is the total number of granite specimens.
• \( N \) is the total number of specimens.
• \( n \) is the number of specimens selected.
• \( k \) is the number of granite specimens you want to find the probability for.

To apply this to our exercise, we substitute \( K = 10 \), \( N = 20 \), and \( n = 15 \). Then, we evaluate the formula for various permissible values of \( k \) (in this scenario from 5 to 10). This method helps us to understand how the probabilities are distributed over the possible outcomes of granite specimen counts selected.

The mean and standard deviation are statistical measures that help describe characteristics of our dataset. For the hypergeometric distribution, these measures are especially helpful to analyze the expected number of granite specimens being selected.

The mean of the distribution, denoted by \( \mu \), is calculated by multiplying the proportion of granite specimens by the sample size:\[ \mu = n \cdot \frac{K}{N} \] With \( n = 15 \) and the proportion \( \frac{K}{N} = \frac{10}{20} = 0.5 \), the mean is:\[ \mu = 15 \times 0.5 = 7.5 \] The mean value suggests that, on average, you can expect to select 7.5 granite specimens out of 15.Standard deviation, represented by \( \sigma \), indicates the extent of variation or dispersion from the mean:\[ \sigma = \sqrt{n \cdot \frac{K}{N} \cdot \left(1-\frac{K}{N}\right) \cdot \frac{N-n}{N-1}} \] This calculates how spread out the number of selected granite rocks are expected to be. Once computed, \( \sigma \) offers a range (mean ± standard deviation) within which most outcomes should fall. In our problem, this range helps determine the probability in one standard deviation, emphasizing the central peak of the distribution.

A discrete probability distribution characterizes outcomes that can take on distinct and separate values. This is contrasting with continuous distributions, where variables represent data that can assume any value within a continuous range.

The hypergeometric distribution is a perfect example of a discrete probability distribution. Here, the possible outcomes are the counts of granite specimens selected. Instead of having an infinite continuum of values, the granite specimens count is restricted to integers from 5 to 10, based on the construction of the problem.

• Each possible value of the count of granite specimens corresponds to a probability, defined by our PMF.
• The total of all these probabilities adds up to 1, ensuring it is a valid discrete distribution.

These narrow, defined outcomes allow us to calculate not just central tendencies like mean and variance, but also the specific probabilities of scenarios such as selecting all specimens of one type. In the broader context, discrete distributions are crucial to solving a multitude of problems where outcomes are fixed and distinct.