Problem 27 - Calculus Concepts

Question

Frozen Yogurt Sales Let \(x\) represent the amount of frozen yogurt (in hundreds of gallons) sold by the G\&T restaurant on any day during the summer. Storage limitations dictate that the maximum amount of frozen yogurt that can be kept at G\&T on any given day is 250 gallons. Records of past sales indicate that the probability density function for \(x\) is approximated by \(y(x)=0.32 x\) for \(0 \leq x \leq 2.5\) a. What is the probability that on some summer day, G\&T will sell less than 100 gallons of frozen yogurt? b. What is the mean number of gallons of frozen yogurt G\&T expects to sell on a summer day? c. Sketch a graph of \(y\) and locate the mean on the graph and shade the region whose area is the answer to part \(a\).

Step-by-Step Solution

Answer

a. 0.16 b. 166.67 gallons c. Graph shows a line from (0,0) to (2.5,0.8) with shaded area 0 to 1.

Step 1: Understanding the Problem

We are given a probability density function (pdf) for the amount of frozen yogurt sold, defined as \( y(x) = 0.32x \) where \( 0 \leq x \leq 2.5 \). Here, \( x \) represents the amount sold in hundreds of gallons, so \( x = 1 \) corresponds to 100 gallons. We need to find probabilities and means using this pdf.

Step 2: Finding Probability of Selling Less Than 100 Gallons

To find the probability that less than 100 gallons (i.e., \( x<1 \) or \( x < 1 \)) is sold, we need to integrate the pdf from 0 to 1. \[ P(X < 1) = \int_{0}^{1} 0.32x \, dx \]Calculate the integral:\[ P(X < 1) = \left[ 0.16x^2 \right]_{0}^{1} = 0.16 \times 1^2 - 0.16 \times 0^2 = 0.16 \]Hence, the probability of selling less than 100 gallons is 0.16.

Step 3: Calculating the Mean Number of Gallons Sold

The mean (expected value) of a continuous random variable is calculated using:\[ \text{Mean} = \int_{0}^{2.5} x \cdot 0.32x \, dx \]First, simplify the expression:\[ = \int_{0}^{2.5} 0.32x^2 \, dx \]Calculate the integral:\[ \frac{0.32}{3} x^3 \bigg|_{0}^{2.5} = \frac{0.32}{3} \left(2.5^3 - 0^3\right) = \frac{0.32}{3} \times 15.625 = 1.6667 \]Thus, the mean number of gallons sold is approximately 1.6667 hundreds of gallons, or 166.67 gallons.

Step 4: Graphing the PDF and Locating the Mean

To sketch the graph of the pdf \( y(x) = 0.32x \), plot \( y \) as a linear function from \( x = 0 \) to \( x = 2.5 \). The line will start at the origin (0,0) and rise to (2.5, 0.32 \times 2.5). Indicate the mean at \( x = 1.6667 \) by marking this point on the x-axis.For part a, shade the area under the curve from \( x = 0 \) to \( x = 1 \), which represents the probability found previously. This shaded area is the region under the pdf from the origin to \(x = 1\).

Key Concepts

Continuous Random Variable

In probability theory, a continuous random variable is a variable that can take an infinite number of possible values. Unlike a discrete random variable, which can only assume specific values, a continuous random variable can take any value within a given range. For example, the amount of frozen yogurt sold by G\&T restaurant is a continuous random variable, measured in hundreds of gallons. This means that sales can be 1.5, 1.783, or any other number within the range of 0 to 2.5 hundred gallons.

• A continuous random variable is described by a probability density function (pdf).
• For a pdf, the area under the curve represents the probability of the variable within a specific interval.
• For our exercise, the pdf is given as \(y(x) = 0.32x\), defining the likelihood of different sales volumes.

Understanding a continuous random variable is crucial for many fields, such as statistics, engineering, and science, where more detailed and variable data measurement is necessary.

Expected Value

The expected value is a core concept in statistics and probability. It represents the average outcome that you would expect if an experiment or a set of actions were repeated many times. In mathematical terms, the expected value of a continuous random variable is calculated using an integral of the form:
\[ \text{E}(X) = \int_{a}^{b} x f(x) \, dx \]
where \( f(x) \) is the pdf of the random variable. For our problem involving frozen yogurt sales, we calculated this value to predict the average daily sales in hundreds of gallons.

• We computed the expected number of gallons sold, \( \approx 166.67 \), indicating a typical day's sales for the restaurant.
• The calculation helps businesses with resource planning, inventory control, and understanding customer demand.
• It is important because it provides a single summary measure of a set of possible outcomes.

For continuous distributions, integrating the product of the variable and its probability over its range gives the expected value, offering useful insight into the long-term average result.

Integral Calculus

Integral calculus plays a critical role in computing probabilities and expected values for continuous random variables. The integral of a function gives you an area under a curve in a specific interval, which in probability theory, translates to the likelihood of a random variable falling within that interval.

• In the frozen yogurt example, we used integrals to find both the probability of selling less than 100 gallons and the expected sales.
• The integral \( \int_{0}^{1} 0.32x \, dx \) was used to determine the probability of low sales.
• Similarly, \( \int_{0}^{2.5} 0.32x^2 \, dx \) helped us find the expected average sales.

Understanding integrals is fundamental for anyone involved with data analysis or statistics, as it allows us to quantitatively analyze the results of continuous data.

Probability Calculations

Probability calculations in a continuous context often involve functions known as probability density functions (pdfs). For continuous random variables, these functions don't provide probabilities directly, but instead, the area under the curve between two points represents the probability of the variable falling within that range.

• Calculating probabilities from a pdf utilizes integral calculus to find these areas.
• In our problem, the function \( y(x) = 0.32x \) defines the probability distribution of yogurt sales.
• The integral from \( x = 0 \) to \( x = 1 \) provided \( P(X < 1) = 0.16 \), demonstrating how to assess the likelihood of a specific sales volume.

Using pdfs and integrating over desired ranges makes it possible to gain insights and make data-driven decisions in various disciplines that require precision and accuracy.