Problem 30 - Basic Statistics for Business and Economics

Question

A tube of Listerine Tartar Control toothpaste contains 4.2 ounces. As people use the toothpaste, the amount remaining in any tube is random. Assume the: amount of toothpaste left in the tube follows a uniform distribution. From this information, we can determine the following information about the amount remaining in a toothpaste tube without invading anyone's privacy. a. How much toothpaste would you expect to be remaining in the tube? b. What is the standard deviation of the amount remaining in the tube? c. What is the likelihood there is less than 3.0 ounces remaining in the tube? d. What is the probability there is more than 1.5 ounces remaining in the tube?

Step-by-Step Solution

Answer

a. 2.1 ounces; b. 1.21 ounces; c. 0.714; d. 0.643.

Step 1: Understanding the Problem

We are given that the amount of toothpaste in the tube follows a uniform distribution. The total amount is 4.2 ounces. We need to find the expected amount remaining, the standard deviation, and certain probabilities related to the remaining amount.

Step 1: Calculate the Expectation (Mean)

For a uniform distribution from 0 to 4.2 ounces, the expected value (mean) is the midpoint of the interval. Thus, the expected value is:\[E[X] = \frac{0 + 4.2}{2} = 2.1\ ext{ ounces}\]

Step 2: Compute the Standard Deviation

The formula for the standard deviation \(\sigma\) for a uniform distribution \([a,b]\) is:\[\sigma = \frac{b-a}{\sqrt{12}}\]Substituting \(a = 0\) and \(b = 4.2\),\[\sigma = \frac{4.2 - 0}{\sqrt{12}} \approx 1.21 \ ext{ ounces}\]

Step 3: Calculate the Probability of Less Than 3.0 Ounces

For a uniform distribution, the probability that the amount is less than 3.0 ounces is:\[P(X < 3.0) = \frac{3.0 - 0}{4.2 - 0} = \frac{3.0}{4.2} \approx 0.714\]

Step 4: Determine the Probability of More Than 1.5 Ounces

Similarly, the probability that the amount is more than 1.5 ounces is:\[P(X > 1.5) = 1 - P(X \leq 1.5) = 1 - \frac{1.5 - 0}{4.2 - 0} = 1 - \frac{1.5}{4.2} \approx 0.643\]

Key Concepts

Expected Value

In probability and statistics, the expected value is the mean or average of a random variable. It represents the center or balancing point of the distribution. When dealing with a uniform distribution, the expected value can be easily calculated because each value is equally likely.

For our toothpaste tube scenario, the amount of toothpaste left follows a uniform distribution between 0 and 4.2 ounces. The expected value of toothpaste left is the midpoint of this interval. Using the formula \[E[X] = \frac{a + b}{2}\]where \(a\) and \(b\) are the lower and upper bounds (0 and 4.2 ounces, respectively), the expected value would be 2.1 ounces.

Therefore, on average, you would expect 2.1 ounces of toothpaste to remain in a tube, given random usage within the limits of 0 to 4.2 ounces.

Standard Deviation

The standard deviation is a measure of how spread out the values in a distribution are. It tells us how much variance exists from the expected value (mean). In a uniform distribution, the standard deviation provides insight into the average amount that values deviate from the mean.

For a uniform distribution from \(a\) to \(b\), the formula for standard deviation \(\sigma\) is:\[\sigma = \frac{b-a}{\sqrt{12}}\]In our problem concerning toothpaste, \(a = 0\) and \(b = 4.2\). Substituting these values into the formula, we find:\[\sigma = \frac{4.2 - 0}{\sqrt{12}} \approx 1.21 \text{ ounces}\]

This calculation shows that the amount of toothpaste remaining typically deviates about 1.21 ounces from the mean of 2.1 ounces. It illustrates how much variation we can expect when measuring random samples from this uniform distribution.

Probability

Probability is the measure of the likelihood that an event will occur. For a uniform distribution, probabilities for continuous random variables are calculated over intervals rather than specific points.

Let's consider the two probability problems in the context of our toothpaste example:- The probability of having less than 3.0 ounces remains is: \[\frac{3.0 - 0}{4.2 - 0} = \frac{3.0}{4.2} \approx 0.714\] Therefore, there is about a 71.4% chance that less than 3.0 ounces of toothpaste is left in the tube.- For more than 1.5 ounces remaining, we compute: \[P(X > 1.5) = 1 - P(X \leq 1.5) = 1 - \frac{1.5}{4.2} \approx 0.643\] So, there's about a 64.3% probability that more than 1.5 ounces is left.

These probabilities help us understand and predict behaviors of the random variable (toothpaste left).

Random Variables

A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. In simple terms, it's a function that assigns a numerical value to each outcome in a sample space.

For the toothpaste, the random variable represents the amount remaining in the tube. Since the distribution is uniform, each amount within the range from 0 to 4.2 ounces is equally likely.

Random variables can be classified into two types: discrete and continuous. In this case, the amount of toothpaste is a continuous random variable because there is an infinite number of possible outcomes. Such variables are often handled using probability density functions (PDFs) to evaluate probabilities over intervals.

Understanding random variables is critical to comprehending how real-world phenomena can be quantified probabilistically. They allow us to model uncertainty and make informed decisions based on statistical data.