Question

A discrete probability distribution for a random variable \(X\) is given. Use the given distribution to find \((a)\) \(P(X \geq 2)\) and \((b) E(X)\). $$ \begin{array}{l|lllll} x_{i} & -2 & -1 & 0 & 1 & 2 \\ \hline p_{i} & 0.2 & 0.2 & 0.2 & 0.2 & 0.2 \end{array} $$

Short answer

(a) 0.2, (b) 0.

Step-by-step solution

Identify Probabilities for P(X ≥ 2)

To find \(P(X \geq 2)\), focus on the probability of the specific outcome satisfying the condition. From the distribution, \(X = 2\) with a probability of \(p_{i} = 0.2\). Since 2 is the only value greater than or equal to 2, \(P(X \geq 2) = p(2) = 0.2\).

Calculate Expected Value (E(X))

The expected value \(E(X)\) is the sum of each possible value of \(X\) multiplied by its corresponding probability. Calculate it as follows:\[ E(X) = (-2) \cdot 0.2 + (-1) \cdot 0.2 + 0 \cdot 0.2 + 1 \cdot 0.2 + 2 \cdot 0.2 \] Simplifying this:\[ E(X) = (-0.4) + (-0.2) + 0 + 0.2 + 0.4 = 0 \] The expected value is 0.

Key concepts

Understanding Random Variables

A random variable is a fundamental concept in probability and statistics. It represents a numerical outcome of a random phenomenon. For example, consider rolling a fair die. The possible outcomes, which are numbers 1 through 6, represent the random variable for this situation. In the context of the given exercise, the random variable is denoted by \(X\), which can take specific discrete values like -2, -1, 0, 1, and 2.
It's essential to recognize that each value of a random variable is associated with a probability. This probability indicates how likely it is for the random variable to land on a specific value during an experiment. Here, each value of \(X\) has an equal probability of 0.2. This uniform distribution makes our calculations straightforward, as each outcome is equally likely.
In practice, random variables come in two types: discrete and continuous. The challenge often lies in identifying and working with these kinds of variables, where discrete variables, like in our exercise, take on specific and countable values.

Diving into Expected Value

The expected value, denoted as \(E(X)\), is a crucial element for understanding the average outcome of a random variable over many trials. In simpler terms, it's the weighted average, where each possible outcome is scaled by its probability.
In the exercise at hand, we find the expected value by multiplying each possible outcome of our random variable \(X\) by its corresponding probability and then summing all these products. This can be expressed mathematically as:
\[ E(X) = (-2) imes 0.2 + (-1) imes 0.2 + 0 imes 0.2 + 1 imes 0.2 + 2 imes 0.2 \]
The simplification of this calculation gives: \[ E(X) = 0 \]
This result tells us that if we could observe \(X\) many times, the long-term average of these outcomes would be 0. Having an expected value of 0 in this distribution indicates complete symmetry around zero, which is consistent given the equal probabilities.

Interpreting Probability Distribution

A probability distribution is a statistical function that describes all the possible values and likelihoods that a random variable can take within a given range. In simpler terms, it tells us all the potential outcomes of the random variable and their respective chances of occurring.
In this exercise, we are dealing with a discrete probability distribution for the random variable \(X\). The distribution is characterized by a set of values \(x_i\) and their associated probabilities \(p_i\). The sum of these probabilities should always equal 1 (i.e., total certainty), which holds true here as \(0.2 + 0.2 + 0.2 + 0.2 + 0.2 = 1\).
The role of a probability distribution is vital not only because it summarizes the likelihood of outcomes but also because it helps in other calculations like finding expectation, variance, and more. Analyzing and understanding the given distribution allows us to predict and comprehend the behavior of the random variable \(X\), thus providing insights into the potential results of associated random experiments.