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Understanding the probability mass function (pmf) is a key component when dealing with discrete random variables like the number of batches ordered by a customer, denoted by \(X\). In this case, \(X\) is a discrete random variable because it can only take specific values: 1, 2, 3, or 4. The pmf specifies the probability that \(X\) takes on each of these values. Each value of \(X\) has an associated probability, given by: \(p(1) = 0.2\), \(p(2) = 0.4\), \(p(3) = 0.3\), and \(p(4) = 0.1\).

The sum of all probabilities in a pmf must equal 1, ensuring we have a complete distribution for the random variable. This distribution helps us understand the likelihood of various outcomes based on the pmf. It's a critical step in calculating other statistical measures like expected value and variance because it gives each outcome a specific weight or likelihood.

Variance measures the spread of a set of numbers. For a random variable \(X\), the variance \(V(X)\) quantifies how much the values of \(X\) deviate from the expected value \(E(X)\). A higher variance indicates that the values are more spread out.

To find the variance of \(X\), we use the formula \(V(X) = E(X^2) - [E(X)]^2\). Here, \(E(X^2)\) is the expected value of the squares of \(X\), calculated by summing \(x^2 \cdot p(x)\) for each possible value of \(x\). This provides a measure of the typical squared deviations from the mean.

In our exercise, \(E(X^2) = 6.1\) and \(E(X) = 2.3\), so \(V(X) = 6.1 - 5.29 = 0.81\). This tells us about the variability of the number of batches a customer might order. A small variance here indicates that the number of batches typically doesn't deviate much from the average.

Linear transformations involve changing the scale or position of a random variable using mathematical operations like multiplication or addition. In the context of this exercise, we consider the random variable \(Y = 100 - 5X\), where \(Y\) represents the number of pounds of chemical left after a customer's order. This equation is a linear transformation of \(X\).

The expected value and variance of \(Y\) can be calculated using properties of linear transformations. For the expected value, \(E(Y) = E(100 - 5X)\) simplifies to \(100 - 5E(X)\) because the expected value operator is linear. Substituting \(E(X) = 2.3\) into the equation gives \(E(Y) = 88.5\), indicating the average remaining pounds after an order.

The variance of \(Y\) is calculated with \(V(Y) = V(100 - 5X) = 25V(X)\) because variance is affected by the square of the scaling constant (in this case 5) and not by constant shifts like 100. This gives \(V(Y) = 20.25\), illustrating the spread of the number of pounds left. This understanding helps predict how far the actual number left may deviate from the average.