91Ó°ÊÓ

To understand probability distributions, we must first discuss the normalization constant. This constant ensures that the sum of probabilities in a distribution equals one. Consider a situation where the number of forms, denoted by \(y\), required by a contractor is proportional to \(y\) itself: \(p(y) = k y\). Here, \(k\) is the normalization constant.

Finding the correct \(k\) is crucial for forming a valid probability distribution. In our problem, the range of \(y\) is from 1 to 5. Therefore, the sum of \(p(y)\) over these values must be 1. We achieve this by solving the equation:

\[\sum_{y=1}^{5} k y = 1\]

Calculating gives us:
\[k (1 + 2 + 3 + 4 + 5) = 1\]
or \(k = \frac{1}{15}\).

Thus, \(k\) is the normalization constant that makes \(p(y) = ky\) a valid probability distribution.

The concept of summation of probabilities is central to confirming a valid probability distribution. It means that the total probability over all possible outcomes equals one.

For example, consider the probability \(p(y) = k y\), where \(k\) is the normalization constant \(\frac{1}{15}\). We calculated previously that the sum over \(y = 1\) to \(5\) equals 1:

• \(p(1) = \frac{1}{15} \cdot 1 \)
• \(p(2) = \frac{1}{15} \cdot 2 \)
• \(p(3) = \frac{1}{15} \cdot 3 \)
• \(p(4) = \frac{1}{15} \cdot 4 \)
• \(p(5) = \frac{1}{15} \cdot 5 \)

The sum of these gives us \(1\), satisfying the condition; thus confirming a valid probability distribution.

A probability density function (PDF) represents probabilities in continuous scenarios, but here, in this discrete context, it can be seen as analogous to describing probability distributions. It characterizes how the total probability is distributed across possible outcomes.

In our case, \(p(y) = ky\) with \(k = \frac{1}{15}\) serves as this function, mapping each form count \(y\) to its corresponding probability. The probability that at most three forms are needed is found by using the PDF and calculating:

\[P(y \leq 3) = p(1) + p(2) + p(3) = 6k = 0.4\]

Similarly, to find the probability that between two and four forms are needed, we compile:

\[P(2 \leq y \leq 4) = p(2) + p(3) + p(4) = 9k = 0.6\]

A valid probability distribution must meet specific criteria, primarily non-negativity and total sum equaling one. If a proposed distribution fails in either aspect, it's not a valid probability distribution.

In the exercise above, we considered whether \(p(y) = \frac{y^2}{50}\) is a valid distribution. Performing the summation from \(y = 1\) to \(5\) yields:
\[\sum_{y=1}^{5} \frac{y^2}{50} = \frac{55}{50} = 1.1\]

This result is greater than 1, indicating that \(p(y) = \frac{y^2}{50}\) violates the condition for a probability sum. Thus, it's not valid, as the total probability exceeds one.

Understanding these properties helps ensure that proposed models for data behave correctly within probabilistic frameworks.