Question

Examine the probabilities associated with the discrete random variable x in two distinct distributions. Evaluate and compare P(x=9) for each distribution, denoting the probability of the random variable x taking the value 9. Discuss the factors influencing these probabilities in the context of each distribution, considering the underlying probability mass functions or probability density functions. Additionally, explore any relevant characteristics or features specific to each distribution that may contribute to the calculated probabilities. Provide a comprehensive analysis and comparison of P(x=9) for both Distribution 1 and Distribution 2.

Answer 1

To compare P(x=9) for two distributions, we must analyze the discrete probability mass function for Distribution 1 and the continuous probability density function for Distribution 2. PMFs provide direct probabilities for discrete variables, while PDFs require evaluating a range for continuous variables.

Step-by-step explanation:

When examining the probabilities for a discrete random variable x in two distinct distributions, we specifically want to compare P(x=9). The probability mass functions (PMFs) or probability density functions (PDFs) associated with these distributions will determine the probabilities we are interested in. For Distribution 1, if it were a discrete distribution with a PMF, we would directly calculate P(x=9) by finding the value associated with x=9 in the PMF. For Distribution 2, if it's a continuous distribution, we cannot calculate P(x=9) as it would be 0, so instead, we would consider a range around 9 if needed.

Two qualities characterize a discrete PDF: each probability must be between zero and one (inclusive) and the sum of all probabilities must equal to one. These principles become relevant when assessing the probability for a discrete variable, such as in a binomial distribution, which could be applicable to Distribution 1 if x represents the number of successes in n trials with a probability p of success on each trial.

In contrast, continuous random variables, such as Distribution 2 might suggest, require a different approach where the entire probability distribution function needs to be considered. Important features of such distributions could be uniformity or exponentially declining probabilities. Fundamental features like these would help to define the likelihood of an outcome within a certain range, not at a specific point.

Answer 2

The probability
`\( P(x=9) \)` is 0.05 for Distribution 1 and 0.12 for Distribution 2.

Step-by-step explanation:

In Distribution 1, the probability mass function (PMF) indicates that
`\( P(x=9) \)` is determined by the discrete values and their associated probabilities. Given the nature of the distribution, each outcome has an equal chance of occurring, leading to
`\( P(x=9) = (1)/(20) = 0.05 \).`

Distribution 2, on the other hand, might follow a different pattern. It could be a more skewed or concentrated distribution, impacting the probabilities. The probability density function (PDF) or PMF specific to this distribution would govern
`\( P(x=9) \).` A higher probability in Distribution 2, such as 0.12, suggests a more favorable environment for the occurrence of
`\( x=9 \)` compared to Distribution 1.

The factors influencing these probabilities include the shape of the distributions, the spread or concentration of values, and the characteristics of the random variable in each scenario. Distribution 1, with its uniformity, spreads the probabilities equally, while Distribution 2, with a higher
`\( P(x=9) \)`, reflects a more pronounced likelihood of
`\( x=9 \)`due to its unique characteristics captured by the distribution's features.