Problem 2 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Records kept by the chief dietitian at the university cafeteria over a 30-wk period show the following weekly consumption of milk (in gallons). $$\begin{array}{lccccc}\hline \text { Milk } & 200 & 205 & 210 & 215 & 220 \\\\\hline \text { Weeks } & 3 & 4 & 6 & 5 & 4 \\\\\hline\end{array}$$ $$\begin{array}{lcccc}\hline \text { Milk } & 225 & 230 & 235 & 240 \\\\\hline \text { Weeks } & 3 & 2 & 2 & 1 \\ \hline\end{array}$$ a. Find the average number of gallons of milk consumed per week in the cafeteria. b. Let the random variable \(X\) denote the number of gallons of milk consumed in a week at the cafeteria. Find the probability distribution of the random variable \(X\) and compute \(E(X)\), the expected value of \(X\).

Step-by-Step Solution

Answer

The average number of gallons of milk consumed per week in the cafeteria is: $$ \frac{200 \cdot 3 + 205 \cdot 4 + 210 \cdot 6 + 215 \cdot 5 + 220 \cdot 4 + 225 \cdot 3 + 230 \cdot 2 + 235 \cdot 2 + 240 \cdot 1}{30} $$ The probability distribution of the random variable \(X\) is: $$ \begin{array}{ lccccccccc}\hline \text { Milk } & 200 & 205 & 210 & 215 & 220 & 225 & 230 & 235 & 240 \\\\\hline\text { Probability } & \frac{3}{30} & \frac{4}{30} & \frac{6}{30} & \frac{5}{30} & \frac{4}{30} & \frac{3}{30} & \frac{2}{30} & \frac{2}{30} & \frac{1}{30} \\ \hline\end{array} $$ The expected value of \(X\) is: $$ E(X) = 200 \cdot \frac{3}{30} + 205 \cdot \frac{4}{30} + 210 \cdot \frac{6}{30} + 215 \cdot \frac{5}{30} + 220 \cdot \frac{4}{30} + 225 \cdot \frac{3}{30} + 230 \cdot \frac{2}{30} + 235 \cdot \frac{2}{30} + 240 \cdot \frac{1}{30} $$

Step 1: Find the total number of gallons of milk consumed.

To find the total number of gallons of milk consumed, we have to multiply the number of gallons of milk by the number of weeks it was consumed and then add everything up. Here is the calculation: $$ \begin{aligned} 200 \cdot 3 + 205 \cdot 4 + 210 \cdot 6 + 215 \cdot 5 + 220 \cdot 4 + 225 \cdot 3 + 230 \cdot 2 + 235 \cdot 2 + 240 \cdot 1 \end{aligned} $$

Step 2: Calculate the total number of weeks.

Add the number of weeks for each category of milk consumption to find the total number of weeks: $$ 3 + 4 + 6 + 5 + 4 + 3 + 2 + 2 + 1 = 30 $$

Step 3: Compute the average number of gallons consumed per week.

To find the average number of gallons consumed per week, divide the total number of gallons consumed by the total number of weeks: $$ \frac{200 \cdot 3 + 205 \cdot 4 + 210 \cdot 6 + 215 \cdot 5 + 220 \cdot 4 + 225 \cdot 3 + 230 \cdot 2 + 235 \cdot 2 + 240 \cdot 1}{30} $$ Now, moving on to part (b).

Step 4: Calculate the probability distribution.

To find the probability distribution of the random variable \(X\), we have to find the probability for each category of milk consumption. To do this, divide the number of weeks for each category by the total number of weeks (30). Here is the probability distribution table: $$ \begin{array}{lccccccccc}\hline \text { Milk } & 200 & 205 & 210 & 215 & 220 & 225 & 230 & 235 & 240 \\\\\hline\text { Probability } & \frac{3}{30} & \frac{4}{30} & \frac{6}{30} & \frac{5}{30} & \frac{4}{30} & \frac{3}{30} & \frac{2}{30} & \frac{2}{30} & \frac{1}{30} \\ \hline\end{array} $$

Step 5: Compute the expected value of \(X\).

To find the expected value \(E(X)\) of \(X\), we have to multiply each value of \(X\) by its probability and then sum up the products. Here is the calculation: $$ E(X) = 200 \cdot \frac{3}{30} + 205 \cdot \frac{4}{30} + 210 \cdot \frac{6}{30} + 215 \cdot \frac{5}{30} + 220 \cdot \frac{4}{30} + 225 \cdot \frac{3}{30} + 230 \cdot \frac{2}{30} + 235 \cdot \frac{2}{30} + 240 \cdot \frac{1}{30} $$

Key Concepts

Expected Value

The expected value, often symbolized as \(E(X)\), is a fundamental concept in probability and statistics used to determine the center of a probability distribution. It represents the average outcome one can anticipate from a random experiment if it is repeated many times. To calculate the expected value, you multiply each potential outcome, which are values of the random variable \(X\), by the probability of that outcome occurring and then sum these products.

For our exercise, this involves taking the amounts of milk consumed in gallons and their respective probabilities of being consumed during a certain week. For instance, if 200 gallons of milk were consumed in 3 out of 30 weeks, the probability of consuming 200 gallons of milk in a week is \(\frac{3}{30}\) or \(\frac{1}{10}\). Doing this calculation for every possible amount and adding them together gives us the expected value of milk consumption per week. This figure would provide a useful estimate for the cafeteria’s planning and purchasing decisions, reducing wastage and ensuring sufficient supply.

To give an example, if we wanted to calculate the expected value of two outcomes, where outcome A is 200 gallons with a probability of 0.1 and outcome B is 210 gallons with a probability of 0.2, the expected value \(E(X)\) would be calculated as:\(E(X) = (200 \times 0.1) + (210 \times 0.2)\). The principle is the same regardless of how many different outcomes and probabilities there are.

Average Consumption

Average consumption in this context refers to the mean quantity of a resource, such as milk, consumed over a specific period. It provides a practical measure for assessing use patterns, which is essential for inventory management, forecasting, and budgetary considerations in a setting like a university cafeteria.

By computing the average consumption, as shown in the exercise, we can derive the typical amount of milk needed each week. This value is obtained by totaling the consumption for each category and dividing by the number of observations, which in this scenario, is the 30-week period. To improve comprehension, we can visualize the average as the balance point of consumption: in a perfectly balanced distribution, every week would see the exact same amount of milk consumed as our calculated average.

An illustration of how we find the average might include adding the total gallons of milk consumed across all weeks and then dividing by the total number of weeks. For example, if the total gallons of milk consumed over 30 weeks was 6300 gallons, the average consumption per week would be \(\frac{6300}{30} = 210\) gallons. It's like figuring out what each week would look like if every week was perfectly average.

Random Variable

A random variable, denoted generally as \(X\), is an essential element of probability theory. It is a variable whose possible values are numerical outcomes of a random phenomenon. Essentially, a random variable maps outcomes of a random process to numbers. It can be discrete, as in our milk consumption example, where the consumption is recorded at regular intervals (200, 205, 210 gallons, etc.), or continuous, covering a range of values without gaps.

The concept of a random variable is crucial when we try to model real-world situations mathematically. In our case, \(X\) represents the amount of milk consumed in a week – an event with various possible outcomes. By defining \(X\), we lay the groundwork for building a probability distribution which tells us how likely each consumption level is within a certain timeframe and for calculating the expected value. Understanding random variables helps students to grasp how we can use probability to make predictions and inform decisions based on seemingly unpredictable events.

It’s important to differentiate random variables from deterministic variables, which are predictable and have a known value, such as the radius of a circle given its circumference. In contrast, random variables embody uncertainty and require using probabilities to describe their potential values.