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The expected value, often denoted as \(E(X)\), is a fundamental concept in probability that gives us the average or mean outcome of a random variable over a large number of trials. It provides a measure of the 'center' of the probability distribution. For discrete random variables, this is computed by multiplying each possible value of the random variable by its probability and summing all these products.
In our exercise, \(X\) represents the number of 5-pound batches ordered by a customer. We can calculate \(E(X)\) using the probabilities given:

• When \(x = 1\), the contribution to \(E(X)\) is \(1 \times 0.2 = 0.2\).
• When \(x = 2\), the contribution is \(2 \times 0.4 = 0.8\).
• When \(x = 3\), it is \(3 \times 0.3 = 0.9\).
• Finally, when \(x = 4\), it's \(4 \times 0.1 = 0.4\).

Adding these up, the expected value \(E(X)\) is 2.3. This means that, on average, a customer orders 2.3 batches. Notably, this is not necessarily a number one would ever observe in a single instance, but rather describes the central tendency of many orders.

Variance is another important concept in probability, often denoted by \(V(X)\). Variance measures how much the values of a random variable differ from the expected value, providing a sense of the spread or variability within the distribution. To compute variance, we need to calculate each value's deviation from the expected value, square it, multiply by its probability, and sum these for all possible values.
For the problem at hand:

• Calculate \((1 - 2.3)^2 \times 0.2 = 0.338\)
• \((2 - 2.3)^2 \times 0.4 = 0.036\)
• \((3 - 2.3)^2 \times 0.3 = 0.147\)
• \((4 - 2.3)^2 \times 0.1 = 0.289\)

Summing these values gives us \(V(X) = 0.81\).
Variance tells us how much variability or risk there is in the orders that customers make. A higher variance would indicate that batch sizes vary widely between orders, while a lower variance would suggest they are more consistent.

A discrete random variable is a type of random variable that can take on a finite or countably infinite set of values. These are typically counts or whole numbers, and they differ from continuous random variables, which can take any value in a continuum. Probability mass functions (pmfs) are used to specify the probability that a discrete random variable is exactly equal to some value.
In this exercise, \(X\) is a discrete random variable representing the number of 5-lb batches that a customer orders. The possible values are 1, 2, 3, and 4, given in the form of a pmf. Each value has a corresponding probability as follows:

• 1 batch: 0.2
• 2 batches: 0.4
• 3 batches: 0.3
• 4 batches: 0.1

Understanding \(X\) as a discrete random variable helps in calculating both the expected value and the variance. This offers insights into not only the average behavior of orders but also the spread, or variability, of these orders as described by the variance.