Problem 1 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 1 and 2, the outcomes and corresponding probability assignments for a discrete random variable \(X\) are listed. Draw the histogram for \(X\). Then find the expected value \(E(X)\), the variance \(\operatorname{Var}(X)\), and the standard deviation \(\sigma(X)\). $$ \begin{array}{l|c|c|c|c|c} \hline \text { Outcomes for } X & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Probability } & \frac{1}{9} & \frac{2}{9} & \frac{1}{3} & \frac{1}{9} & \frac{2}{9} \\ \hline \end{array} $$

Step-by-Step Solution

Answer

Expected value \(E(X) = \frac{28}{9}\), variance \(\operatorname{Var}(X) = \frac{160}{81}\), standard deviation \(\sigma(X) = \frac{4\sqrt{10}}{9}\).

Step 1: List Outcomes and Probabilities

List the outcomes and their corresponding probabilities:- Outcomes: 1, 2, 3, 4, 5- Probabilities: \( \frac{1}{9} \), \( \frac{2}{9} \), \( \frac{1}{3} \), \( \frac{1}{9} \), \( \frac{2}{9} \)

Step 2: Draw the Histogram

Using the listed probabilities and outcomes, draw a bar for each outcome indicating its probability. The x-axis will represent the outcomes (1, 2, 3, 4, 5) and the y-axis will represent the probabilities.

Step 3: Calculate the Expected Value

The expected value \(E(X)\) is calculated using the formula: \[ E(X) = \sum_{i=1}^{n} x_i p_i \] where \( x_i \) are the outcomes and \( p_i \) are the probabilities:\[ E(X) = 1\left(\frac{1}{9}\right) + 2\left(\frac{2}{9}\right) + 3\left(\frac{1}{3}\right) + 4\left(\frac{1}{9}\right) + 5\left(\frac{2}{9}\right) \]\[ E(X) = \frac{1}{9} + \frac{4}{9} + \frac{9}{9} + \frac{4}{9} + \frac{10}{9} \]\[ E(X) = \frac{28}{9} \]

Step 4: Calculate the Variance

The variance \(\operatorname{Var}(X)\) is calculated using the formula: \[ \operatorname{Var}(X) = \sum_{i=1}^{n} (x_i - E(X))^2 p_i \]Where \(E(X) = \frac{28}{9}\):\[ \operatorname{Var}(X) = \left(1 - \frac{28}{9}\right)^2\left(\frac{1}{9}\right) + \left(2 - \frac{28}{9}\right)^2\left(\frac{2}{9}\right) + \left(3 - \frac{28}{9}\right)^2\left(\frac{1}{3}\right) + \left(4 - \frac{28}{9}\right)^2\left(\frac{1}{9}\right) + \left(5 - \frac{28}{9}\right)^2\left(\frac{2}{9}\right) \]Calculate each term separately:\[ \left(1 - \frac{28}{9}\right)^2\left(\frac{1}{9}\right) = \left(-\frac{19}{9}\right)^2 \left(\frac{1}{9} \right) = \frac{361}{81} \left(\frac{1}{9} \right) = \frac{361}{729} \]\[ \left(2 - \frac{28}{9}\right)^2\left(\frac{2}{9}\right) = \left(-\frac{10}{9}\right)^2 \left(\frac{2}{9} \right) = \frac{100}{81} \left(\frac{2}{9} \right) = \frac{200}{729} \]\[ \left(3 - \frac{28}{9}\right)^2\left(\frac{1}{3}\right) = \left(\frac{1}{9}\right)^2 \left(\frac{1}{3}\right) = \frac{1}{81} \left(\frac{1}{3} \right) = \frac{1}{243} \]\[ \left(4 - \frac{28}{9}\right)^2\left(\frac{1}{9}\right) = \left(\frac{8}{9}\right)^2 \left(\frac{1}{9} \right) = \frac{64}{81} \left(\frac{1}{9} \right) = \frac{64}{729} \]\[ \left(5 - \frac{28}{9}\right)^2\left(\frac{2}{9}\right) = \left(\frac{17}{9}\right)^2 \left(\frac{2}{9} \right) = \frac{289}{81} \left(\frac{2}{9} \right) = \frac{578}{729} \]Summing these, we get:\[ \operatorname{Var}(X) = \frac{361}{729} + \frac{200}{729} + \frac{1}{243} + \frac{64}{729} + \frac{578}{729} \]\[ \operatorname{Var}(X) = \frac{1440}{729} = \frac{160}{81} \]

Step 5: Calculate the Standard Deviation

The standard deviation \(\sigma(X)\) is the square root of the variance:\[ \sigma(X) = \sqrt{\operatorname{Var}(X)} = \sqrt{\frac{160}{81}} = \frac{4\sqrt{10}}{9} \]

Key Concepts

Expected Value

The expected value, or mean, of a discrete random variable gives us a measure of the center of its probability distribution. It essentially tells us the average value we would expect if the experiment were repeated many times. It is calculated using the formula: \[ E(X) = \frac{1}{9} + \frac{4}{9} + \frac{9}{9} + \frac{4}{9} + \frac{10}{9} = \frac{28}{9} \ \ E(X) = \frac{28}{9} ~\approx~ 3.11 \]. Here, you can see that each outcome is multiplied by its probability, then summed up to give the expected value.

Variance

Variance measures how spread out the values of a random variable are around the mean. A higher variance indicates a wider spread. Calculating variance involves the formula: \[ \operatorname{Var}(X) = \frac{361}{729} + \frac{200}{729} + \frac{1}{243} + \frac{64}{729} + \frac{578}{729} = \frac{1440}{729} = \frac{160}{81} \]. Each outcome’s deviation from the mean is squared, multiplied by its probability, and then summed up. This aggregated value tells us about the overall dispersion of the outcomes.

Standard Deviation

Standard deviation provides a measure of the dispersion or spread in the same units as the original data by taking the square root of the variance. It is useful as it makes interpretation easier. Using the formula: \[ \sigma(X) = \ \sqrt{\operatorname{Var}(X)} = \sqrt{\frac{160}{81}} = \frac{4\sqrt{10}}{9} \], we find that the standard deviation calculates the average amount by which the values differ from the mean. It’s a more intuitive measure of variability than variance.

Probability Distribution

A probability distribution shows how the probabilities are distributed over the possible values of a discrete random variable. In this exercise, the probability distribution is given as:

• 1 with a probability of \( \frac{1}{9} \)
• 2 with a probability of \( \frac{2}{9} \)
• 3 with a probability of \( \frac{1}{3} \) (or \( \frac{3}{9} \))
• 4 with a probability of \( \frac{1}{9} \)
• 5 with a probability of \( \frac{2}{9} \)

. This distribution helps us quantify the likelihood of each outcome. It is an essential concept because it allows us to make meaningful predictions and calculations regarding the behavior of the random variable.

Histogram

A histogram is a graphical representation of the probability distribution of a discrete random variable. To create a histogram:

• Draw the x-axis representing the outcomes (1, 2, 3, 4, 5).
• Draw the y-axis representing the probabilities (\( \frac{1}{9} \), \( \frac{2}{9} \), \( \frac{1}{3} \), etc.).
• For each outcome, draw a bar that reaches up to its probability.

. A histogram provides a visual summary of how the probabilities are distributed and makes it easier to see the relative likelihood of different outcomes.