Question

Farmers' market. A farmer has \(100 \mathrm{lb}\) of apples and \(50 \mathrm{lb}\) of potatoes for sale. The market price for apples (per pound) each day is a random variable with a mean of \(0.5\) dollars and a standard deviation of \(0.2\) dollars. Similarly, for a pound of potatoes, the mean price is \(0.3\) dollars and the standard deviation is \(0.1\) dollars. It also costs him 2 dollars to bring all the apples and potatoes to the market. The market is busy with eager shoppers, so we can assume that he'll be able to sell all of each type of produce at that day's price. a) Define your random variables, and use them to express the farmer's net income. b) Find the mean. c) Find the standard deviation of the net income. d) Do you need to make any assumptions in calculating the mean? How about the standard deviation?

Short answer

Net Income mean: $63; SD: $20.62; Assume price independence for calculations.

Step-by-step solution

Define Random Variables

Let the random variable \(A\) represent the price per pound of apples, and \(P\) represent the price per pound of potatoes. Thus, the total income from selling apples will be \(100A\), and from potatoes \(50P\). The farmer's net income \(I\) is given by the expression: \(I = 100A + 50P - 2\).

Find the Mean of Net Income

To find the mean of the net income, calculate the expected values: \(E(I) = E(100A + 50P - 2) = 100E(A) + 50E(P) - 2\). Given \(E(A) = 0.5\) and \(E(P) = 0.3\). Substituting these values:\(E(I) = 100(0.5) + 50(0.3) - 2 = 50 + 15 - 2 = 63\).

Find the Standard Deviation of Net Income

Since \(A\) and \(P\) are independent random variables, the variance of \(I\) is \(Var(I) = Var(100A + 50P) = 100^2 Var(A) + 50^2 Var(P)\). Given \(SD(A) = 0.2\) and \(SD(P) = 0.1\), so \(Var(A) = 0.2^2 = 0.04\) and \(Var(P) = 0.1^2 = 0.01\). Thus, \(Var(I) = 10000(0.04) + 2500(0.01) = 400 + 25 = 425\). The standard deviation of \(I\) is \(\sqrt{425} \approx 20.62\).

Assumptions in Mean and Standard Deviation Calculations

To calculate the mean, we assume the daily prices of apples and potatoes are independent of each other. For standard deviation, we again assume the independence of these variables so that we can sum variances directly. The independence allows the use of the formula \(Var(X+Y) = Var(X) + Var(Y)\) for the independent variables \(X\) and \(Y\).

Key concepts

Random Variables

Understanding random variables is key when dealing with probability and statistics. A random variable is essentially a variable whose possible values are numerical outcomes of a random phenomenon. In the context of the farmer's market, the price per pound of apples and potatoes fluctuates daily, making it unpredictable. Therefore, we define the price per pound of apples as a random variable, denoted by \(A\), and the price per pound of potatoes as \(P\). These random variables are essential for representing scenarios in probabilistic terms and are characterized by certain attributes such as mean (or expected value) and standard deviation.
In this problem, since we want to calculate the farmer's net income, we model this using the expression \(I = 100A + 50P - 2\). Here, \(I\) is the farmer's income, depending on random variables \(A\) and \(P\). Understanding that \(A\) and \(P\) are random helps us calculate other statistical properties like the expected value or standard deviation with greater ease.

Expected Value

The expected value is a fundamental concept in probability and statistics that represents the average or mean outcome of a random variable if the process were repeated many times. For the farmer's situation, the expected value helps predict his average income from selling apples and potatoes.
To find the expected value of net income \(E(I)\), we utilize the linear property of expectation: \(E(aX + bY + c) = aE(X) + bE(Y) + c\). Applying this to our equation, we have \(E(I) = 100E(A) + 50E(P) - 2\).
From the problem statement, \(E(A) = 0.5\) dollars and \(E(P) = 0.3\) dollars. Substituting these values yields an expected income of \(63\) dollars. This value tells us that on average, the farmer can expect to earn \$63 from selling his apples and potatoes, post transportation cost.

Standard Deviation

Standard deviation is a measure of the amount of variability or spread in a set of data. It indicates how much individual data points deviate from the mean. In this exercise, standard deviation helps the farmer understand how much his daily income can differ from the expected daily income of \$63.
For random variables \(A\) and \(P\), the standard deviation is given. To find the standard deviation of net income \(I\), we first find the variance \(Var(I)\). If the random variables \(A\) and \(P\) are independent, then \(Var(I) = 100^2Var(A) + 50^2Var(P)\).
Substituting \(Var(A) = (0.2)^2 = 0.04\) and \(Var(P) = (0.1)^2 = 0.01\), we find \(Var(I) = 425\), leading to a standard deviation \(SD(I) = \sqrt{425} \approx 20.62\). This shows there is a moderate spread in potential income outcomes around the expected value.

Independence Assumption

The independence assumption simplifies the analysis when dealing with multiple random variables. Two random variables are independent if the occurrence of one does not affect the probability of occurrence of the other. In our market example, we assumed that the daily price of apples \(A\) and potatoes \(P\) are independent.
This assumption allows us to compute the variance of the sum \(100A + 50P\) by simply adding their variances: \(Var(A) + Var(P)\). Without this assumption, we could not directly sum the variances due to potential covariance between the variables.
In real-world applications, independence might be a strong assumption; however, it greatly simplifies computation and provides a reasonable approximation when there's no strong correlation known between variables. It allowed us to calculate both the mean and the standard deviation effectively in this exercise.