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Grasping the concept of the mean of a random variable is important because it tells us the average outcome we can expect if we were to repeat a random experiment many times. The mean, also referred to as the expected value, is calculated by multiplying each possible outcome by the probability of that outcome occurring and summing those products.

For the problem given, we found the mean by using the formula: \(10 \times 0.7 + 20 \times 0.2 + 30 \times 0.1\). In this example, we have a set of outcomes \(x\) which are 10, 20, and 30, with their respective probabilities \(p(x)\) of 0.7, 0.2, and 0.1. By multiplying each outcome by its probability and adding them together, we sum up to 14, which is the mean of the given discrete probability distribution. This calculation provides a weighted average that is skewed toward the more likely outcomes.

Understanding standard deviation in the context of a random variable can help us gauge the level of dispersion or spread in the possible outcomes around the mean. The further the outcomes are from the mean, the higher the standard deviation, indicating a greater variability in outcomes.

To calculate the standard deviation, as seen in the step by step solution, we start by determining the deviation of each outcome from the mean, squaring it, and then multiplying by the corresponding probability. Sum these products to get the variance, and then take the square root to find the standard deviation:
\[\sqrt{(10-14)^2 \times 0.7 + (20-14)^2 \times 0.2 + (30-14)^2 \times 0.1} = \sqrt{11.2 + 7.2 + 25.6} = \sqrt{44}\].
By following this method, we calculate that the standard deviation of the random variable is the square root of 44. This quantifies the expected deviation of the outcomes from the mean, providing a measure of uncertainty in the random variable.

A discrete probability distribution is a key concept in understanding how probabilities are assigned to specific outcomes of a discrete random variable, which can take on a finite or countable number of outcomes.

In the case of the exercise, we are provided with a simple discrete probability distribution given in a table format. The table presents a set of possible outcomes \(x\), along with their corresponding probabilities \(p(x)\). These probabilities must sum to 1, ensuring that one of the outcomes must occur. For instance, in this example, the probabilities of 10, 20, and 30 occurring are 0.7, 0.2, and 0.1, respectively, and \(0.7 + 0.2 + 0.1 = 1\), satisfying the fundamental property of a probability distribution.

Understanding discrete probability distributions is essential, as they form the foundation for calculating important statistical measures, such as the mean and standard deviation, which in turn, provide insight into the nature of the random variable in question.