Question

Set up sample spaces for Problems 1 to 7 and list next to each sample point the value of the indicated random variable \(x,\) and the probability associated with the sample point. Make a table of the different values \(x_{i}\) of \(x\) and the corresponding probabilities \(p_{i}=f\left(x_{i}\right)\) Compute the mean, the variance, and the standard deviation for \(x\). Find and plot the cumulative distribution function \(F(x)\). A random variable \(x\) takes the values \(0,1,2,3,\) with probabilities \(\frac{5}{12}, \frac{1}{3}, \frac{1}{12}, \frac{1}{6}\).

Short answer

Mean = 1, Variance = \(\frac{7}{6}\), Standard Deviation = \(\sqrt{\frac{7}{6}}\). CDF values: \(F(0) = \frac{5}{12}\), \(F(1) = \frac{3}{4}\), \(F(2) = \frac{5}{6}\), \(F(3) = 1\).

Step-by-step solution

Define the sample space

The sample space is the set of all possible outcomes of the random variable. For the random variable x, the possible values are 0, 1, 2, 3.

List the sample points with their probabilities

Assign probabilities to each value of the random variable x: - For x = 0, probability = \(\frac{5}{12}\) - For x = 1, probability = \(\frac{1}{3}\) - For x = 2, probability = \(\frac{1}{12}\) - For x = 3, probability = \(\frac{1}{6}\)

Create a probability table

Construct a table showing the different values of x and their corresponding probabilities:\[\begin{array}{|c|c|}\hlinex_i & p_i \ \hline0 & \frac{5}{12} \ 1 & \frac{1}{3} \ 2 & \frac{1}{12} \ 3 & \frac{1}{6} \ \hline\end{array}\]

Compute the mean (expected value)

The expected value or mean \(E(x)\) of the random variable x is given by the formula: \[ E(x) = \sum_{i} x_i p_i \] Substitute the values: \[ E(x) = 0 \times \frac{5}{12} + 1 \times \frac{1}{3} + 2 \times \frac{1}{12} + 3 \times \frac{1}{6} \] \[ E(x) = 0 + \frac{1}{3} + \frac{2}{12} + \frac{3}{6} \] \[ E(x) = \frac{4}{12} + \frac{2}{12} + \frac{6}{12} \] \[ E(x) = \frac{12}{12} = 1 \]

Compute the variance

The variance \(Var(x)\) is given by the formula: \[ Var(x) = \sum_{i} (x_i - E(x))^2 p_i \] First, find \((x_i - E(x))^2\) for each x_i: \[ 0: (0 - 1)^2 = 1 \]\[ 1: (1 - 1)^2 = 0 \]\[ 2: (2 - 1)^2 = 1 \]\[ 3: (3 - 1)^2 = 4 \] Next, multiply each by its corresponding probability and sum them up: \[ Var(x) = 1 \times \frac{5}{12} + 0 \times \frac{1}{3} + 1 \times \frac{1}{12} + 4 \times \frac{1}{6} \] \[ Var(x) = \frac{5}{12} + 0 + \frac{1}{12} + \frac{4}{6} \] \[ Var(x) = \frac{5}{12} + \frac{1}{12} + \frac{8}{12} \] \[ Var(x) = \frac{14}{12} = \frac{7}{6} \]

Compute the standard deviation

The standard deviation \(\sigma\) is the square root of the variance: \[ \sigma = \sqrt{Var(x)} = \sqrt{\frac{7}{6}} \]

Find and plot the cumulative distribution function (CDF)

The CDF \(F(x)\) is the probability that the random variable is less than or equal to x. \[ F(0) = P(x \leq 0) = \frac{5}{12} \]\[ F(1) = P(x \leq 1) = \frac{5}{12} + \frac{1}{3} = \frac{5}{12} + \frac{4}{12} = \frac{9}{12} = \frac{3}{4} \]\[ F(2) = P(x \leq 2) = \frac{3}{4} + \frac{1}{12} = \frac{9}{12} + \frac{1}{12} = \frac{10}{12} = \frac{5}{6} \]\[ F(3) = P(x \leq 3) = \frac{5}{6} + \frac{1}{6} = 1 \]In a table: \[ \begin{array}{|c|c|}\hlinex & F(x) \ \hline0 & \frac{5}{12} \ 1 & \frac{3}{4} \ 2 & \frac{5}{6} \ 3 & 1 \ \hline\end{array} \] Plot the CDF on a graph, with x-axis as the values of x (0, 1, 2, 3) and y-axis as the corresponding probabilities.

Key concepts

Random Variable

A random variable is a numerical outcome of a random process. It's like a box that holds potential values based on the outcomes of an experiment. For example, consider rolling a fair die. Here, the box (random variable) could hold any value from 1 to 6. In the provided problem, our random variable, denoted as \(x\), can take on the values 0, 1, 2, or 3. Each value comes with a specific probability. These probabilities are the chances of each outcome happening. The list of outcomes and their probabilities helps us understand the behavior of \(x\).

Expected Value

The expected value, often denoted as \(E(x)\) or \(\mu\), is essentially the mean or average value that you would expect from the random variable if you repeated the experiment many times. Think of it as a weighted average where each possible value of the random variable is multiplied by its probability, and then all these products are summed up. In the given exercise, we calculated the expected value of \(x\) to be 1 using the formula: \[ E(x) = \sum_{i} x_i p_i \]This tells us that if we were to repeat our experiment many times, on average, we'd expect the value of our random variable \(x\) to be 1.

Variance

Variance, denoted as \(Var(x)\), measures how much the values of the random variable differ from the expected value (mean). In simpler terms, it tells us the spread or dispersion of the data. The variance is calculated as the average of the squared differences between each value and the mean, weighted by their probabilities. In our problem, the variance of \(x\) was found to be: \[ Var(x) = \sum_{i} (x_i - E(x))^2 p_i = \frac{7}{6} \]This indicates how much the values of \(x\) are spread out around the mean.

Standard Deviation

The standard deviation is simply the square root of the variance. It gives us a measure of how spread out the values of the random variable are but in the same units as the variable itself (unlike variance which is in squared units). For our example, the standard deviation \(\sigma\) is calculated as: \[ \sigma = \sqrt{Var(x)} = \sqrt{\frac{7}{6}} \]This value is useful because it helps us more easily understand the extent of variability or dispersion in our random variable \(x\).

Cumulative Distribution Function

The cumulative distribution function (CDF), denoted as \(F(x)\), is a function that tells us the probability that the random variable is less than or equal to a certain value. It adds up the probabilities of all outcomes up to the given point. The CDF for our problem was calculated as follows: For \(x = 0\), \[ F(0) = P(x \leq 0) = \frac{5}{12} \]For \(x = 1\), \[ F(1) = P(x \leq 1) = \frac{5}{12} + \frac{1}{3} = \frac{3}{4} \]For \(x = 2\), \[ F(2) = P(x \leq 2) = \frac{3}{4} + \frac{1}{12} = \frac{5}{6} \]For \(x = 3\), \[ F(3) = P(x \leq 3) = 1 \]The CDF is useful for understanding the overall probability distribution and making predictions about the random variable.